~Six years ago, I wrote a blog entry about The Monty Hall problem called The Classic Monty Hall Problem Gets Goatsed and I continue to get comments on it. Recently, the entry was linked by a commenter on the Dilbert blog by fiftyfiftyman *AtlantaDude* who later recanted his statement and appears to have turncoated to side of the switchmen, and a new influx of switchmen have arrived to proclaim my profound insanity.

After ~six years of comments, I have held my ground and have gained even further insight as to *why* the authentic reality of the situation is 50/50, rather than 33/66.

To establish the ground rules, the Monty Hall problem in question (of which there are several) is strictly as follows:

You are a contestant on a gameshow hosted by Monty. Monty gives you the option of selecting one door from an available indistinguishable three. Behind one door is a new car, behind another is a goat, and behind the remaining door is also a goat. You select one from among the three indistinguishable doors. Monty now reveals one of the doors you have not selected to be a Goat Door. What is the probability of selecting the Car Door?

For ease of reference, I refer to those who believe that the elimination round is relevant as Switchmen, while those like myself who believe the elimination round as irrelevant as Fiftyfiftymen.

A switchman will advise to *always switch*, presuming that the elimination round impacts the stay/switch round, which is untrue. The final question is a diversionary tactic to create a false situation in which a certain branch of math *appears* to be applicable, but does not genuinely apply.

The explanation goes that, when conducted with a cup-and-ball experiment in which a ball is placed under one of three cups without your knowledge of which cup conceals the ball, and you play out the game an arbitrarily sufficient number of times (perhaps as with this simulation), that by empirical, statistical observation of wins and losses, wins by switching will gradually approach 66% of the time, whereas wins by staying will gradually approach only 33% of the time, so the safest bet would be to switch.

In order to arrive at that conclusion, however, the elimination round must be taken into account, even though the elimination round is irrelevant. The acceptance of the irrelevant elimination round as weight upon a 50/50 choice is the switchman’s error, when it genuinely bears no weight.

Allow me to offer another example of such diversionary questioning that creates a similarly false mathematical circumstance:

Three men require lodging, and at the chosen hotel, management asserts the cost for one room’s rent for the trio is $30, so each man pays his fair $10 share for the single room. Management is later crunching the numbers, and realizes that the price for a room rental for three men is actually $25. The next morning, management confronts the trio and offers a partial refund for the difference, giving back each man $1 each and proposing that $2 be offered to the bellhop as a tip, to which the trio agree. The trio therefore only paid $9 each for the room for $27 total, and $2 goes to the bellhop, which only totals $29. Where did the other dollar go?

The diversion within the question is that the $2 tip is added, rather than (properly) subtracted from the $27 total, to make $25. The question attempts to suggest that the original $30 total is still relevant, rather than the new total of $25 that is the only relevant total for application.

Similarly with the Monty Hall problem, the previous elimination round bears no effect on the most recent choice, to **stay** or **switch**. The math involved for Switchmen, like those fooled by the lodging problem, *requires* the now-irrelevant prior circumstances to remain within the body of evidence, when in reality both prior matters are newly and entirely irrelevant.

Further evidence to the contrary of the Switchman’s claim is when the figures are tallied for a situation involving four doors and three goats. Analysis of past wins indicates that switching wins 5/8ths of the time (according to this article) — but all of those require the belief that the prior choices *matter* in a new situation in which a 50/50 situation is *now present*.

There are several ways to explain how elimination rounds no longer matter to the final Stay/Switch round.

**One**

Imagine a large, blank, perfect-circle dartboard that can plane-rotate on its center point when spun. With a fine marker, precisely draw three radius lines from the center so that each radius intersection with the outer edge of the board create 3 pie-shaped regions of the board equal in surface area. Each region corresponds to a specific door, perhaps labeled 1, 2 and 3. The board is spun, the contestant throws a dart, and the mark made indicates which door is chosen in the elimination round, and that door is removed from play.

The elimination round dartboard’s layout is now wholly irrelevant and is no longer an acceptable instrument for selecting between whether one should stay with the door chosen by dart’s mark or switch to the other door, because of the presence of region that still represents the now-eliminated door.

A new dartboard is instead brought forth, which has a single bisecting diameter line that passes through center, dividing the board into two equal regions of Stay and Switch. The contestant still has no idea which door contains a goat door or a car door, and the new board is strictly a matter of Stay or Switch. Regardless of how many previous elimination boards that came into play before the 3-way board, such as a 4-way board, 5-way board, or 100-way board or more, the final decisive moment is a matter between Stay or Switch.

**Two**

The elimination round, itself, is an illusion, because no choice has genuinely been made. The host’s selection of elimination-round doors could have been entirely nonsensical — the question of, “What is the capital of Kentucky?” could be posed and depending on the answer right or wrong, one or the other of the two goat doors could be removed. An insider-audience vote of which person, either Monty or contestant *danced best* to the People’s Court theme song, could be the deciding factor whether which of the two goat door would be removed. The final option would still, in each above case, result in a single 50/50 choice remaining.

**Three**

The doors could be arbitrarily named Heads, Tails, and Wildcard, whereas the Wildcard will take the name of the door which is eliminated:

Regardless of what door the contestant selects, all three situations result in a Heads or Tails event.

**Four**

The technique to *choose* a door is decided by the roll of a 1D3 die — perhaps in the form of an imperfection-free, equilateral triangle extruded at a distance four times the length of one of the triangle’s lines, with ends that equally force a final resting position with one of the three triangle’s sides face down without the possibility of landing on the end (perhaps better visualized as a 3-sided pencil with both ends sharpened). The end facing down is the door that the contestant selects to keep, and one goat door is removed from among the two unselected doors. The 1D3 is now no longer a valid instrument for determining whether to Stay or Switch, so a 1D2, or simply a *coin flip* would be an accurate way to make the next decision since the prior options of the 1D3 are no longer relevant.

**Five**

There is a 100% chance that the car door is a car door, and a 0% chance that either of the goat doors is are a car door. One of the two goat doors (each with zero percent chance being a car door) is removed. The contestant must now decide whether to select one of two options — one door which is 100% the car door, or another door which is 0% a car door. Regardless of the choice the contestant makes in any elimination round, or how the choice was made, or how many doors that had 0%-car doors were previously been removed in how many elimination rounds existed previously, the final choice will be between a door which is 100% a car door, and another which is 0% a car door.

Closing Remarks

In an effort to resolve the dispute, I propose that the final question of the Monty Hall Problem be reworded:

Does the empirical

evidencethat the win/loss figureschangebased on the increase in number of elimination rounds preceding the final stay/switch round,substantiateorinvalidatethe belief that switching, when presented with the final stay/switch choice, is the most reasonable option?

Is a half-cup of water in a cup that could hold 1 cup, half empty, or half full? Math doesn’t lie, after all. Statisticians, however..

Pingback: The Classic Monty Hall Problem Gets Goatse’d | the ablestmage press

Why did you bother? You’ve just rehashed all the same fallacious arguments, misinterpretations and re-interpretations of the problem and illogical ramblings of your first blog entry. For the time you spent writing this (although most if is cut and paste from you other blog) you could have gainfully learned some basic probability theory to help you understand where you’re going wrong.

It seems pointless discussing this with you further when you won’t accept the evidence presented “…empirical, statistical observation of wins and losses, wins by switching will gradually approach 66% of the time, whereas wins by staying will gradually approach only 33% of the time”, and even provide a link to a simulation (which doubtless you haven’t tried). I suppose these statistics are all what… lies?

P.S You also mis-stated the problem initially (why am I not surprised?). You omitted the key assumptions that need to be made about the host’s behaviour (he must only reveal a goat door, always offer the option switch) and the question isn’t “What is the probability of selecting the Car Door?” (which depends on what you do and can be 33%,50%, or 66%) but “Is it advantageous to switch doors?” Still, don’t let a few important details get in the way of a poor argument, you never usually do.

The problem with you being incapable of grasping the issue, is that you can’t see the magnitude of the red herring/misdirection question of probability. The example of the lodging trick question is of like manner to the “probability” question in the MHP. It sets up an artificial calculation that, in itself, is erroneous, leading to a mystery about an erroneous technique to base calculations upon.

Could you offer your half-(full/empty) reasoning as to why the results differ when appending additional door elimination rounds prior? It is clear that only the final choice matters, not previous choices, because all choices will ultimately result in the identical option of staying or switching. All previous choices have no bearing on the 1-of-2 options in staying or switching. There is no process of elimination in heads or tails.

I could offer my reasoning as to why the results differ when appending additional door elimination rounds, but it would take too long to explain here. Instead I’ll direct you an article by two professors of mathematics from James Madison University. You might find it a bit ‘heavy going’ in places, but perseverance is the key to understanding! (http://educ.jmu.edu/~lucassk/Papers/ProgMonty2.pdf) .

Perhaps you can offer your reasoning that if (as you contend) any choice before the final 1-in-2 option is an illusion why the results would differ when appending additional door elimination rounds? Aren’t you contradicting yourself?

TLDR; I can randomly choose honda or porche all day. Thats a 50/50 guess. Doesn’t change who wins the race.

Your win% is based on ratio of prizes to doors, not your odds of selecting the doors randomly.

There are two events happening when you flip a coin and choose heads or tails to achieve a 50/50, We shall call the “coin flip” event A and “calling it” event B. In classic coin flip, note how both events are a 50/50 choice, and if either event remains constant while the other changes, the odds remain 50/50.

In the “Monty Hall Game” Event A1 (placing the goats/prize behind the doors) is a g33/g33/p33. The Event B1 is 33/33/33. Then a new event happens “the elimination of a goat”, which we call event A2.

If we walk into the game right now with no prior knowledge, are presented with two doors, g33 and p33, and are forced to choose at event B2 then the best odds you can get are 50/50. This is how a fiftyfiftyman approaches.

But a switchman has two additional pieces of information. He observed Event A1 and recognized that he only a 33% chance of selecting the prize. And he observed Event A2, where a goat is removed. NOW PAY ATTENTION.

What he is OBSERVING from Event A2 is literally watching the coin flip.

You’re argument is literally saying: If i choose heads or tails randomly, it doesn’t matter that I can see the weighted coin.

I concur.

You are correct. People make this too complicated. No matter how many doors there are, each door is the same and each door has an equal chance of having the car. The contestant’s guess can not change this. If there are 3 doors each door has a 1/3 chance of having the car. If there are 2 doors, each has an equal chance of having the car (50%). Just as the hotel example tends to confuse people, so does this problem. What bothers me is that, when the correct answer is explained in a simple logical way using valid statistical methods, so many people just don’t get it or just won’t allow themselves to believe they are wrong. They won’t allow themselves to see the truth; and worse than that is that they may be voting in November.

I can easily show you why it is NOT 50/50. First of all you must read what the previous responder said about you misstating the problem: You omitted the key assumptions that need to be made about the host’s behavior (The host must only reveal a goat door, and always offer the option to switch) and the question isn’t “What is the probability of selecting the Car Door?” (which depends on what you do and can be 33%,50%, or 66%) but “Is it advantageous to switch doors?”

If you NEVER switch doors, here are the only 3 ways the game can be played:

1. PICK CAR – DON’T SWITCH: WIN

2. PICK GOAT #1 – DON’T SWITCH: LOSE

3. PICK GOAT #2 – DON’T SWITCH: LOSE

YOU WIN 1/3 OF THE TIME.

If you ALWAYS switch doors, there are also only 3 ways the game can be played:

1. PICK CAR – SHOWN ONE GOAT, SWITCH TO THE OTHER GOAT: LOSE

2. PICK GOAT #1 – SHOWN GOAT #2 – SWITCH TO CAR: WIN

3. PICK GOAT #2 – SHOWN GOAT #1 – SWITCH TO CAR: WIN

YOU WIN 2/3 OF THE TIME.

Since you have a 33.3% chance of picking the car at the beginning of the game, if you never switch, you’re going to win only 33.3% of the time. If you do switch your choice you will win 66.7%

These are the only 6 ways the game can be played. Clearly it shows you that if you switch doors you will win 66.7% of the time.

I hope this clears it up for you.

The article itself is the very contention that prior choices are wholly irrelevant. You have yet to address any of those matters, and simply return to your tiresome and entirely-ineffective argument. The fact that the results differ is the very evidence that such lines of reasoning are fraught with error. It’s as if you’re still trying to calculate the lodging price for the missing dollar using the old $30 total from the misdirection question from the three-men problem. You just keep going back to the same false premise as if it has any merit whatsoever, and continue to calculate for an entirely unrelated event.

,”The article itself is the very contention that prior choices are wholly irrelevant.” No it isn’t, you’re just making stuff up now. You didn’t really read the article did you? Come on, ‘fess up.

The article proved (as in mathematically proved, not pick some lame-ass statement out of thin air and assert it to be true proof, which is your method) that the optimal strategy for winning the progressive MHP (multiple doors with multiple switch choices) is to switch doors once when only two doors remain, and that even switching every elimination round the probability of winning by switching never falls below 63.2% irrespective of the number of doors.

I’m sure Messrs Lucas and Stonehouse would welcome your rebuttal to their paper, and that the probability of winning by switching is in fact never greater (or less) than 50%. Be sure to include the five ways that show the elimination rounds don’t matter that you presented earlier. Maybe add some coloured diagrams of the dartboards you so meticulously described in example 1, what do you think? Oh, and don’t forget to mention the $30 hotel bill and the 3 guests in your “proof”- I’m sure that will be the clincher!.

LMAO

/EDIT

Messrs Lucas and Rosenhouse

Prior choices are only irrelevant if Monty randomly eliminates a door. But Monty’s behavior isn’t random. He has a strict set of rules he must adhere to.

“The fact that the results differ is the very evidence that such lines of reasoning are fraught with error”

No, the results differ because the advantage gained by switching increases with each additional door.

Marley 52 does not know or understand statistics. You are wrong and nasty to boot. I find it amazing that so many people are so convinced they are right when they are so, so wrong.

And your qualifications in probability and statistics are?

I’ve made no mistakes in what I’ve said, care to disagree about a particular point I’ve made, or are you just full of hot air?

You are still wrong and still nasty. I’ve studied and taught statistics in college, graduate school, Medical school and used it, probably for more years than you’ve been alive. For you to be right, the contestant would have to have some supernatural ability to pick the card which will be wrong, most of the time. The contestant’s chance of choosing correctly is the same no matter which door he/she chooses You are making a sophomoric mistake thinking that Monte Hall,s elimination somehow changes the fact that each door has an equal chance of hiding the car.

So Dr Girschick, answer this question since you’re so well-educated in math and stats:

If you play 300 games of the MHP, always pick Door1, and always stay with Door1 (when given the chance to switch doors) how many games would you expect to win?

Your first post has already been lampooned all over the internet for it’s laughable wrongness and you are back for more!

Buy a deck of cards, run the game and you will prove yourself wrong in less time than it took for you to write this. It is not rocket science.

I will bet you $100 that if we run through the game 50 times switching each time then the 33/66 will be the correct answer. Let me know when you want to do this, I need $100. We can even do it over Skype, and if you are willling to video it and post your loss on your site when it happens I’ll even give you 10 to 1 odds (so if you win I’ll pay $1000). C’mon Ablesmages, put your money where your mouth is!

Sean, I’ve come to the conclusion that Ablestmage is now just trolling. Why he’d do that on his own blog I’m not sure. On the other hand he could simply be incredibly dense.He is from Texas so it’s difficult to tell either way.

Yeah I’ve read this in AMAZEMENT. This has got to be a clickbait article, and let’s admit he has chosen a reliable one. It hurts to see others taken in by this but it’s clear that you can fall for the 50-50 fallacy only if a) you misstate the problem, or b) you refuse to understand the pure nature of the claim: that switching remains the optimal strategy. I don’t think ablestmage is stupid. I think he’s smart enough to generate clicks & comments. And I’m just the latest to fall for it 😉

I’m back for more because I am baffled that you Switchmen cannot discern the presence of the TRICK QUESTION before your very eyes. You are tasked with calculating probability of heads or tails, based on ILLUSIONS of choice before the final 1-in-2 option. The choice is only a matter of stay or switch, no matter if you danced a jig, quoted from Macbeth, or could name the capital of Kentucky in order to eliminate enough goat doors to make it down to a final goat-or-car round.

Let’s face it Ablestmage, you’re just baffled full stop.

‘Trick Questions’ and ‘Illusions’ – it’s a mathematical puzzle not an episode of David Copperfield’s TV show. Maybe it is all magic to you. “Be Amazed by Bayes Theory !!”, “Two Choices and it’s NOT 50/50. Can It Be True ??”. “See the Goat Disappear Before Your Very Eyes !!”

You are correct again.

The Monty hall game is not like flipping a coin. Flipping a coin has only 2 outcomes Head or tails, so it is 50/50. When you choose one door and DON’T switch when there are 2 doors, you’re not actually choosing anymore. You’re sticking with your first choice which was 1/3 at the beginning and is STILL 1/3 after a goat is revealed.

I’ve done it with a deck of cards. 100 x 3 cards, laid out in my living room. Shuffled repeatedly, over and over and over again. I went through them and found a red card (it was one black and two red cards per 3 cards) and flipped it over. I then flipped over the the first card on each deck and you knooow what?

It appears that 54 percent of the choices were the BLACK card. In that scenario I picked the car by NOT switching FIFTY FOUR PERCENT over the 46 percent “goat”.

So you took 3 cards ( 2 red, 1 black), looked at them and picked out a red card, and then turned over the 1st card from the 2 remaining cards? And 54% of the cards turned over were black? Sounds about right, you’ve got 1 red card and 1 black card and you’re picking 1 card at random. But what has this experiment got to do with the MHP, and why bother with 3 cards to begin with if you’re always going to remove a red card BEFORE you turn over a card?

Unfortunately, this will still not convince those who refuse to see the truth.

You’ve fallen for abletmage’s very sly (I applaud, seriously, well played) re-statement (mis-representation) of the problem. Argued like a true Republican 😉

unbelivable. “i have been wrong for 6 years..i have been proven wrong for 6 years..but now, after bringing nothing new to the table, im suddenly right.. “..

ablestmage, you must have no idea how stupid you sound. you are so hellbound to being right that you go out of your way to prove it. twisting facts and posting examples that has nothing to do with the scenario of MHP.

I too must come to the conclusion that youre either trolling or at least you need to read up on your math skills. either way i dont think you are not capable to understand facts or to understand when you just wrong.

im sorry for you, cause youre clearly suffer from some sort of illusion about your own skills…or just a troll.

It has been my experience that when someone is wrong and has no way to logically defend their position, they tend to resort to name calling and personal attacks. You are wrong, nasty and make no sense, But you sure can be offensive.

Please read the present post by van Savant at her website. Please examine closely the statement and crux of the problem. Please note that of all the people who abused her for her correctness, she has to date received ONE apology. Phenomenal.

I can’t believe that you are not able to see the facts, they are very simple to understand: Initially you have 33% probabilities to select the door with a car, and a 66% to select the door with a goat. Assuming you will always switch, in the first case you will finally get a door with a goat (as you are going to change it), and in the second case you will always get the car (the second goat will be discarded, and when you change your door you will get the car, as is the only remaining door). It is very simple: finally you have 33% to get the goat and 66% to get the car.

Don’t commit the error to compare to other experiments that are NOT the same (I only read the case one, the perfect circle dartboard with three sectors, then I stopped reading. In that case the process of discarding one sector is random, but in the MHP is NOT random, the one who discards one door knows witch door to discards. In other words, if you set the correct sector BEFORE discarding one (as it is in the MHP) you have 33% of probabilities to discard the right sector, but in MHP is 0% as the one who discards it knows witch one is the correct one).

Sorry for my grammar, I’m not english.

Yes it is very simple. However, no matter how simple anybody explains the solution. Ablestmage refuses to accept the proof staring him in the face.

You were right to stop reading Ablestmage’s cases after the 1st one, they only get more ridiculous as he continues with further fictitious examples.

No need to apologise for your grammar, it’s a lot more comprehensible than that of the writer of this blog.

Focusing exclusively on the elimination round, not the final round — would you say that Monty’s modification of the contestant’s random selection does or does not result, when combined, as a random result? The constant is “remove a goat door from (a/the) remaining 2 unselected doors,” with the (a/the) portion applying to both situations, not “or.” Monty’s move is constant, not random.

In the same way that a 1d6 die roll is random, and that a constant “adding +1 to whatever is rolled, whereas 6+1 loops back to 1,” the combined result is random… will the application of Monty’s constant “remove a goat door from (a/the) remaining 2 unselected doors” also result in a random, given that the contestant’s 1 in 3 choice is also random?

The contestant’s selection in the elimination round doesn’t get modified, it remains the same.

In multiple repetitions of the game you will get different doors selected by the contestant and different doors opened by the host in the elimination round. The contestant’s action is random, the host’s action is non-random as his is determined by the game-show rules. Does that answer your question?

The outcome of the contestant’s (random) door selection and the host’s (non-random) door elimination results in a random combination of 2 unopened doors remaining, if that better answers your question.

I’m not sure where you think you are going with this line of reasoning but I’m pretty certain it’s down another blind alley.

Actually, he’s relying on math instead. Since you’ve just agreed that the problem is NOT random.

Out of three options, you’re making the claim that “magically” the middle option is ALWAYS going to be GREATLY statistically higher of a chance to be the car.

Could you explain which magical pixies made the car the middle door far, far greater than 1/3 of the time? Hmm?

The last thing Ablestmage is relying on is maths. There’s no magic involved in the MHP just basic probability theory, which you also appear to misunderstand. I’ll explain it in simple terms:

Let’s say I pick Door 1 and Monty opens Door 3. Why is the car twice as likely to be behind Door 2 than Door1? After all the car was equally likely to be behind any of the 3 doors to start with, and we now know it’s not behind Door 3, so the car must be behind either Door 1 or Door 3 with equal likelihood, right? Wrong.

Here’s why. The car is either behind Door 1 or Door 2. If the car is behind Door 2 then Monty has no choice but to open Door 3. However if the car is behind Door 1 ( the door I picked) then Monty has a choice between opening either Door 2 or Door 3, and he picks one at random – meaning there’s a 50% chance he opens Door2 and a 50% chance he opens Door3. So, given that he has opened Door 3 it is more likely (twice as likely in fact) that he did so because had to than because he chose to. In other words it more likely (twice as likely) that the car is behind Door 2 than it is behind Door 1, so I should switch.

That is how probability theory works, it’s not magic, there’s no pixies involved. Ablestmage doesn’t (or refuses to) understand.

You are being confused by invalid logic. Keep it simple. Each door will have the same probability of hiding the car. This fact can’t be change by a contestant’s guess or how many doors there are.

When you switch doors, there are only 2 possible outcomes:

a) you pick the car,(switch) you win a goat, OR

b) you pick a goat, (switch) you win the car

Situation b) is twice as likely to occur than situation a), or to put it another way a) has a probability of 1/3, b) a probability of 2/3

That’s called logic (with some really basic maths thrown in)

No matter which door you choose, the chance of being correct is 1/the # of doors in play. If there are three doors in play the chance any one of the doors hides the car is 1/3(33.3…%). If there are two doors in play the chance that any one door hides the car is 1/2(50%). If one door is in play the chance that the car is behind that door is1/1(100%). We can explain this to you but, we can’t make you understand it. I hope you will eventually get it You sound like a bright person. Good night.

Dr. Joseph Girschick,

What’s that now,15-20 comments and not a single mathematically true or logically valid statement amongst any of them.

That must be some sort of record (apart from Ablestmage himself of course)

I notice you’ve studiously avoided answering any of the questions I’ve posed, and instead taken the coward’s way out instead.

No maths, no logic, no argument, no balls.

OH. hahahaha I thought you’d included ‘Dr.’ in your title because it bore some relevance to the fields of mathematics, statistics, game theory, probability, ANYTHING. No. That isn’t the case is it. You practice medicine in NY, if I’ve googled correctly? I was amazed someone who gave the impression of knowledge in this arena could actually be arguing in defence of such a glaringly incorrect result.

I’m so relieved.

Don’t apologise for your grammar – you have understood what those before you have not. 😉

Maybe I can provide a slightly different perspective as to why switching is advantageous.

Imagine instead of 3 doors there are 1000, and Monty eliminates 998 of them, leaving only your choice plus one other. Do you really think that out of the 1000 doors you were lucky enough to actually pick the car and that Monty had nothing but goats? Of course not, obviously its far more likely that Monty had to eliminate everything BUT the car. After all, you have 1 door. He has 999. Knowing that, do you really think you’re as likely to win if you stick?

Except that is precisely *my* argument as to why staying or switching is no more or less advantageous in the final round. The choice boils down to two options, regardless of whether doors or windows or portholes or trap doors or attic doors were opened, closed, added, removed, blocked by a huge rock, blasted apart by a canon, or not. The final round is its own game, and is completely unaffected by anything that goes on prior. There could have been a dancing round, a swimsuit competition, and then a final round of a choice between two doors. The final round is still a choice between two doors. You said it yourself that the game has been eliminated down to everything but the car and the door you’ve picked, so it is now between two options.

“The final round is its own game, and is completely unaffected by anything that goes on prior.”

Its affected by the previous round because Monty’s sort is based upon the contestant’s initial choice. Monty can never eliminate the contestant’s door, and Monty must always eliminate a goat if the contestant chose the other goat. Additionally, Monty can never eliminate the car. This changes the game entirely. You could only make your argument (and indeed you’d be right to say that its 50/50) if Monty eliminates a random door but this does NOT happen.

Of course it boils down to two options, but what you’re not understanding is that just because a proposition has two choices doesn’t mean its a coin flip. In the case of MHP, Monty’s behavior gives you some very important information with which to make an informed decision.

To reword my example:

You pick 1 of 1000 doors. At this point you have a 1/1000 chance of having picked the car. That means Monty will still have the car 999 times out of 1000, therefore 999 times out of 1000 Monty will be forced to leave the car behind his remaining door (he can’t eliminate the car). Hence, swapping will win 999 times out of 1000. Its quite simple once you wrap your mind around it.

And as has been said repeatedly in the past, just because there’s 2 options doesn’t make them equally likely. This is a concept that you singularly fail to grasp.

In the 1000 door example if you flip a coin you have a 50% chance of winning the car:

(1/2 * 1/1000) + (1/2 * 999/1000) = 1/2.

However if you switch you have a 99.9% chance of winning the car, because the final round IS affected by what has happened before, a statement that can be proven mathematically and empirically demonstrated to be true.

HOW do you ignore that this is not the entirety of the information that informs the probability of the outcome? This HAS to be deliberate deafness to the rest of the explanation! #godyoufrustratememan!! haha!

It is NOT boiled down to 2 options. it is only 2 options if you were offered the choice of 2 doors BEFORE you ever make a choice. If you choose one of the 3 doors, the host is essentially offering BOTH of the other 2 doors instead. The 2 other doors combine for a 66.7% chance of having the car. Your initial choice offers a 33.3% chance of having the car. Once thew host shows you one of the losing doors, those 2 doors STILL have a combined 66.7% chance of having the car, except that the open losing door has 0% chance of having the car. the other unopened door has 66.7% chance of having the car.

Erroneous statement: “Monty had to eliminate everything BUT the car.”

Correct statement: Monty had to eliminate everything but the car OR THE LAST GOAT. There’s your 50/50! If you switch and open Monty’s final door it has a 50% chance of a car. If Monty revealing every extraneous goat is a certainty -and they ARE extraneous- why do you include it in a question of probability?

Or think of it this way. Man chooses door 1. Monty reveals goat behind door 3 and Monty’s assistant leads the goat off stage. Music plays, TV audience looks at beautiful people advertising cars and floor wax. Red light clicks on camera and man faints when faced with the final choice. Man’s wife, having just arrived (traffic was terrible) rushes up onto the stage to finish the contest. She is faced with two doors. One hides a goat. One hides a car. How could that be anything other than a 50/50 proposition? Everything that happened before her husband fainted is no longer relevant. She has to choose between two doors, each of which hides one of two different items. How could changing from one door to the other change her odds? Can’t. Doesn’t.

#HBrown, you must be Ablestmage in disguise, since you show the same ignorance of probability and present the same fallacious and nonsensical arguments dressed up as witty and insightful observations as the author of this blog does.

Go and do some research, or even get 3 playing cards and perform the experiment yourself, (or try this website: http://www.nytimes.com/2008/04/08/science/08monty.html?_r=0)

Your situation is very different from the actual problem. The wife has only 2 doors to choose from and has no idea which door her husband chose. HER odds are 50/50. She doesn’t have the necessary information to improve her odds to 66.7. (The first choice, and which door was eliminated by being opened)

i tell you what this is the comment that made me understand.

Put another way: The weatherman says there’s a 90% chance of rain. But there are only two options: either it will rain or it won’t. Does that mean the weatherman is wrong and that there is a 50% chance of rain? No, it means that he has additional information that led him to the conclusion that one outcome is more likely than the other.

I heard that a 90% chance of rain means it will rain 100% somewhere, but only over 90% of the area.

50% chance of rain means it’s still raining, but only over 50% of the area. The rainman told me that. Yeah, def-def- definitely raining. yeah.

Sigh. That means you have a 9/10 chance of rain.

Which means that your are 90% SIGNIFICANTLY likely to have it rain.

It’s not 50/50.

What you’re claiming is that the MHP is NOT random. Which defeats any claim of 1/3, 2/3 or 1/2 or anything. If it’s not random, then the answer is irrelevant. The game show host could be trying to use reverse psychology, he could be trying to use reverse-reverse psychology, reverse-reverse-reverse psychology, hell he could just be being genuinely nice.

If the doors are truly random and you pick the first door only for the third door to be a goat, then STATISTICALLY it is now a 50/50 choice between doors 1 and 2. You have just as much chance to win or lose if you switch or don’t switch because now only two doors are in play.

You remind me of those ignorant engineering students who go on and on about the plane flying, yet base this on the claim that wheels spinning are what makes a plane fly instead of, you know, lift.

Monty’s actions are not random (at least not all the time). If your 1st pick is a goat (which is a 2/3 chance) then Monty’s actions as which door to open is determined, he has no choice. He only gets to make a choice if you’ve picked the car to begin with (a 1/3 chance).

There is no psychology involved, the rules of the game are clearly defined. You should be wary of talking about “ignorant engineering students” when it is obvious that you yourself are ignorant of basic mathematical probability.

OH MY GOD YES THIS!!!! THIS IS EXACTLY IT!!!!

To simplify even further:

Monty is always more likely than you to have the car initially. This forces Monty to leave the car behind the other door more often than not.

Its really that simple.

Let’s assume 100 doors. The question is whether the player has an initial odds of 1-100 to pick the car or 1-2.

Perhaps here is the way to make the point.

According to the rules, the player gets to pick 1 door. Let’s assign Monty Hall the rest. Thus, if the player had 1-100 chance of picking the car, that means Monty Hall must have a 999-1000 chance of having the car.

However, Monty Hall does not have a 999-1000 chance of picking the car because 998 doors CANNOT HAVE THE CAR.

Thus, the player must have a 1-2 chance of picking the car and Monty Hall has a 1-2 chance of having the car.

Wrong, your logic is seriously flawed. Monty does have a 999-1000 chance of picking the car because he’s got 999 doors, and he knows which ones have goats behind them (all 998 of them).

Or do you think you have a 50% chance of picking the car from 100 doors?

You and Ablestmage should form a club.

I’m here to campaign for the fiftyfiftymen, and debunk the switchmen.

We can simply re-name to game from “The Monty Hall Problem” to “Don’t match the host”. This is because the host always selects a goat, and if you match the host, you lose.

Host picks goat + you pick goat = you lose. Host picks goat + you pick car = you win.

********

Another way to look at the Monty Hall Problem is that the object of the game is to change a 33% chance of winning into a 66% chance of winning. But that is only possible in the following (3) ways:

1) Let the contestant pick 2 of the 3 doors instead of swapping one door for another.

2) Have the host swap a goat for a 2nd car, instead of removing it.

3) Let the contestant vote “False”, meaning he believes there is not a car behind his chosen door.

We know we aren’t allowed to pick multiple doors, so option 1 is out.

We know there aren’t two cars, so option 2 is out.

We can not point to the host’s door (goat) and declare “False” (which means we win 100% of the time), so option 3 is out.

But there is another way to look at option 3. When there are only 2 doors, one door is false, and the other is true.

This makes the chance of winning the car 50/50.

**********

Oh, and for the switchmen who want to use more doors in their argument, every time you remove a door, you reduce your denominator by 1.

Imagine a deck of cards. What are the chances of picking the ace of spades out of 52 cards? 1/52.

What are the odds that your card is NOT the ace of spades? 51/52.

But when the host removes a card that is not the ace, the odds that your card is NOT the ace of spades has changed to 50/51.

Remove 20 cards that are not the ace of spades, and the odds that your card is NOT the ace of spades has changed to 31/32.

Remove 50 cards that are not the ace of spades, and the odds that your card is NOT the ace of spades has changed to 1/2.

Hey Patty, you removed all your comments from Youtube, why was that? In your last comment brfore you deleted it you acknowledged you were wrong all along. But now here you are peddling the same BS maths and faulty “logic” . You’re in good company with Ablestmage though, he knows absolutely nothing about probability theory either.

Imagine a deck of cards, you pick 1 but don’t look at it. What are the chances of picking the ace of spades out of 52 cards? 1/52.What are the odds that your card is NOT the ace of spades? 51/52.

I take the remaining 51 cards (which have a 51/52 chance of containing the Ace of Spades) look through them all and DELIBERATELY remove 50 cards that are NOT the Ace of Spades. I have 1 card left in my hand. What are the odds that card is the ace of spades? That’s right it’s 51/52. It’s not 1/2, and this can be easily verified if you could be bothered to get a pack of cards and try it out for yourself. But you won’t do that because you’re way too arrogant (or stupid) to admit you could be wrong.

YouTube limits the length of replies, so a full explination can’t be offered without accusations of “trolling”.

Also, people were beginning to post insults instead of debating the math.

Finally, you can’t post web addresses, like the one for this simulator:

http://www.grand-illusions.com/simulator/montysim.htm

The debate is, “Who’s right about the probability?” Is it 33% if you stay? Is it 66% if you switch? Or does switching doors half the time make no difference (50%)?

The answer is YES to all 3.

Try it with the simulator. (let it run automatically, using “run __ times” at the bottom)

If you stick with your original choice, you will win 33% and lose 66%. (Right answer #1)

If you switch every time, you will win 66% and lose 33%. (Right answer#2)

But switching half the time, or saying it makes no difference, is calculated by adding both results together. That means you win 50% and lose 50%. (Right answer #3)

So the Monty Hall Puzzle is not about mathematical probability.

I can win the car 100% by voting “false” and picking the door the host opened.

The question is: “Does switching doors increase your chances of winning the car?” The answer is unequivocally YES.

The question is NOT: “What are your chances of winning if you switch 50% of the time and stay 50% of the time?” That’s something you’ve just made up in a lame attempt to lend some sort of credibility to your 50% argument (it doesn’t).

Let me jog your memory with some quotes from your 1st post:

“the object of the game is to change a 33% chance of winning into a 66% chance”. Yes it is and, although you say it can’t be done without picking 2 doors or having 2 cars, you can do this by SWITCHING.

“When there are only 2 doors, one door is false, and the other is true.This makes the chance of winning the car 50/50.” Brilliantly incorrect. Switching wins the car with a 2/3 chance.

There’s more:

“”every time you remove a door, you reduce your denominator by 1.” No you don’t, stop making maths up.

And finally:

“Remove 50 cards that are not the ace of spades, and the odds that your card is NOT the ace of spades has changed to 1/2.”. Not it hasn’t, the odds remain at 51/52, demonstrating once and for all that you know nothing about probability.

Wrong, wrong, wrong , and wrong again (just as you were on Youtube).

“So the Monty Hall Puzzle is not about mathematical probability.”. Yes it is, totally, but I’m not surprised you don’t think so.

LMAO at those people who pretend to understand probability, insist the answer to the MHP is 50/50, and then try to backtrack their position when they’re shown to be completely wrong.

I like to go to chat rooms and YouTube comment sections, and politely debate one side of an issue for hours. Sometimes for days.

Then I remove my comments and apoligize, just to see who might pop up and gloat about it.

It really forces assholes to reveal themselves.

Hey, does anyone know any good Athiest vs Christian sites?

When you’re debating an issue, it helps if you don’t completely ignore the arguments made by those on the other side of the debate which you repeatedly did when discussing the MHP on Youtube. Also, it’s generally an advantage to have some knowledge of the topic being discussed, and maybe do a bit of research beforehand. The old “Argument from Ignorance” position is unlikely to win you many admirers

*Douchebag Alert!* *Douchebag Alert!* You should really stop being so hostile to other people.

THIS. IS. AMAZING.

In your actual answer above, you have written the words that ARE the proof for the advantage to switching. AHHHHH this is killing me this HAS to be clickbait!!!! #whyyyyyyyyy

Obviously a lot of you switchmen don’t understand what odds means

odds is the desired outcome / undesired outcomes

chances are desired outcomes / total outcomes

if i want the ace of spades from a 52 card deck that is odds of 1/51

the chances are 1/52

then if i take away 50 goat cards, i have the odds of 1/1, or 50/50, with the chances of 1/2 to get the car, or 50%

And obviously you “fiftyfiftymen” have absolutely no understanding whatsoever of probability. Why don’t you get a pack of cards, pick 1 at random and place it face down in front of you (without looking at it). Now, pick up the remaining 51 cards, look through them and remove 50 cards that are not the A of Spades (alternatively just pick out the A of Spades if it’s there and discard the rest). You now have 1 card left in your hand. Which card is more likely to be the A of Spades, the 1 in your hand or the 1 face down in front of you? It’s not difficult to understand, even for a “fiftyfiftyman”, that it’s not a 50/50 chance.

so, if i have to pick a random door from a selection of 3, with 1 having the car, and no matter what i pick one goat will be removed, why does the first choice even matter?

if there are 3 cakes in a store all in separate nontransparent refrigerators, but only one has blue frosing and the others have red, and no matter which one i am thinking of picking my brother will take a red one first, the one i was thinking of picking wont help me pick the blue one from the remaining two

its like that third one is no longer a valid option because it has no cake, and another blue cake is not brought out to replace the taken red cake in it’s refrigerator

the same way switchmen think that the 33% from the revealed goat adds to the chance that the car will be selected

You should stick to eating coloured cakes, maths clearly isn’t your strong suit. Your 1st pick from 3 doors restricts the doors the host can choose to open, specifically if you pick a goat door (a 2/3 chance) the host is forced to open the other goat door, leaving the car behind the 3rd door. This will happen 2 times out 3, therefore 2 times out 3 switching to the 3rd door will get you the car – simple isn’t it?

Its been said a dozen times already: A proposition with two possible outcomes does NOT automatically equate to a coin flip. Probability isn’t simply a matter of outcomes, its a matter of information gleaned from the circumstances, specifically:

A. Monty still has the car two thirds of the time

B. Monty always eliminates a losing door

It is barely believable to me that people are still debating this issue when it was conclusively settled so many years ago.

The matter is conclusively settled by my arguments.

You don’t have any arguments. The only issue that you have conclusively settled is that you don’t understand maths in general and probability theory in particular.

I would also add to this that something happening within that (one being taken away) will never change, its going to always happen and you will still be left with two, so it cant be a dependant variable on the outcome, since it doesnt change anything, and the action itself never changes, therefore it should always be disregarded and the two left considered a seperate game entirely.

Very interested to find someone else with the same idea. It feels like the “switchmen” are truly trying to make something out of nothing.

Anyways, you stated exactly what I was thinking, but in a much clearer way than I ever could.

You do realise Ablestmage hasn’t got the slightest idea what he’s talking about? It is a very simple mathematical puzzle which Ablestmage either doesn’t, or refuses to, understand. It is demonstrably true that switching (given the conditions stated in the problem definition) increases your chances of winning to 66.7%. Look it up on the net, there are hundreds of sites that will explain it .

Don’t be fooled into thinking that just because there’s two choices they must be equally likely – it is a fact that the probability the prize is behind the door you picked to begin with is 1/3, and that the probability it is behind the other door (the one the host didn’t open) is 2/3. That is indisputable, it is not a matter of opinion. Get 3 playing cards (2 red and 1 black) and prove it to yourself – it shouldn’t take more than 10 minutes.

Pay no attention to this troll Marley — he consistently takes it upon himself to defy sound reason and instead cower behind his circular math logic, despite how I even demonstrated clearly and concisely that the “chances” change depending on how many previous rounds exist and can therefore be discarded.

Ablestmage let’s try you with a simpler version of the MHP to see if you can understand the logic involved in arriving at the correct answer.

Imagine there are 3 identical boxes. In one box there are 2 gold coins, in another 2 silver coins and in the third box there is a gold and silver coin. You pick a box at random and remove one coin. You see it’s a gold coin. What is the probability that the box you picked it from was the one containing the 2 gold coins?

I’ll await your reply and reasoning before posting the answer.

Your coinbox illustration is a divergent MHP that creates completely different rules, as if there were actually 6 doors, but gathered together in pairs, where behind one pair is a full goat, behind another pair of doors is a full car, and behind the remaining pair of doors is a half-goat-half-car hybrid and that by opening one door would reveal half of the pair’s identity. Abiding by the MHP rules we’re discussing, the selection of a pair of doors in your version would require opening both doors (or revealing the full contents) of a box you have NOT selected. Only in the final round is the contents of the box or pair of doors you have selected revealed. In all other situations, the contents of a box or pair of doors you have NOT selected is revealed. Your illustration changes the rules of the revelation process, and is thus inapplicable.

I notice with the 3 boxes problem you didn’t say what the answer is, why not? It’s not a trick question.

This particular problem is known as Bertrand’s Box Paradox, and is in fact the basis for the Monty hall Problem (as well as the 3 Prisoner Problem, the 2 Children Problem and various other similar probability puzzles) in that the logic to arrive at the solution is the same in both problems. I’ll repeat the reply I made to a previous poster to your blog regarding the MHP as it explains the logic for the 3 box problem too.

“Let’s say I pick Door 1 and Monty opens Door 3. Why is the car twice as likely to be behind Door 2 than Door1? After all the car was equally likely to be behind any of the 3 doors to start with, and we now know it’s not behind Door 3, so the car must be behind either Door 1 or Door 2 with equal likelihood, right? Wrong.

Here’s why. The car is either behind Door 1 or Door 2. If the car is behind Door 2 then Monty has no choice but to open Door 3 ( as per the rules of the game). However if the car is behind Door 1 ( the door I picked) then Monty has a choice between opening either Door 2 or Door 3, and he picks one at random – meaning there’s a 50% chance he opens Door2 and a 50% chance he opens Door3. So, given that he has opened Door 3 it is more likely (twice as likely in fact) that he did so because had to than because he chose to. In other words it more likely (twice as likely) that the car is behind Door 2 than it is behind Door 1, so I should switch”

You have only demonstrated that you don’t understand the problem and have absolutely no idea how to calculate conditional probability.

Not only that, you ignore the mountains of evidence that prove all your arguments and “logic” are categorically wrong (including the computer simulation that you yourself uploaded to your blog and then subsequently removed because the results didn’t support your position), and refuse to try the simplest of experiments (using playing cards) that would quickly verify your position if it were true (which it isn’t). Why not if you are so sure of yourself?

Again (but I don’t know why I bother), get a pack of cards, choose 1 at random and put it aside without looking at it, now pick up the remaining 51 cards, look through them and remove 50 cards that are NOT the Ace of Spades. You now have 2 cards remaining, the card you put aside to start with and 1 card left in your hand. Which one is the ace of Spades? Repeat until you realise that the 2 cards left are not equally likely to be the Ace of Spades. For most people that should take about 2 repetitions, for you ….. who knows.

You’d have more credibility arguing the world was flat than continuing with this charade.

I find it entertaining that despite your conclusion about my arguments, that you, yourself, insist on meddling in the affairs of those beneath your winsome charms. All you have been able to offer are simple restatements of the identical argument that I have offered mountains of evidence to the overt contrary. Instead of responding to each of the specific counterarguments, you have no choice but to simply and rotely mutter the identical mantra back at me.

There are two problems with your 51-card idea.

Firstly, is that in the MHP, if both cards were to be flipped over, one of them would always be an Ace of Spades. 1 out of 2, every time. Your illustration actually goes further to prove my point better than yours. There is no matter of “likely” involved at all. One of the two doors contains a goat, and the other of the two remaining is a car. It doesn’t matter if there are 52 doors like your card illustration, because Monty will always eliminate goat doors and never eliminate car doors, so that in the end, the game will always result in a choice between a goat and a car, or 1 out of 2.

Secondly, the problem with your 51-card illustration is that you are not the person who eliminates cards, but if there were a knowledgeable dealer of the deck of cards (as there is with Monty), and placed one card to the side unbeknownst to you whether it was the Ace of Spades or not, and it could even NOT be — when you pick a card, Monty eliminates one of the cards of the 51 that you didn’t select. He doesn’t eliminate the card that you picked. You’re actually not even making a choice in the first place, Monty/Dealer is making the choice for you, from an arbitrary selection on your part, with the illusion that what you’ve done is made a choice. Monty could just as easily require you to recite the first 5 letters of the alphabet and then eliminate a non-Ace/Spade card, and then have you spell “alphabet” and remove another non-Ace/Spade card, repeat until you’re back down to 2 remaining cards, one of which Monty/Dealer knows is definitely the Ace of Spades. The dealer now relinquishes control over to you about which door to eliminate, rather than Monty/Dealer controlling the elimination. Your choice decides the end of the game, rather than Monty’s choice of which door to eliminate for the entire rest of the game leading up to the final round, regardless of how many prior rounds.

Either you are a skillful troll, or you absolutely refuse to understand simple logic and probability. I will expand the original game so it should finally be abundantly clear that it is NOT 50/50. Pretend you and I are standing on a beach. I am thinking of one grain of sand. Let’s see if you can pick it. After you choose a single grain. I pick up another grain out of the billions of grains. And then I say, “The grain of sand I was initially thinking of is their the one you picked or the one I have in my hand that was whittled down from the BILLIONS of grains you didn’t pick. Do you switch? Or do you think think you picked the correct grain of sand at the very beginning of there game? You’re going to smugly assert that your grain of sand now has a 50% chance of being the grain of sand I was thinking of? If you do, then you’re a special kind of stupid.

.First to clear up one point, the question posed in the MHP is “do your chances of winning the car improve if you switch doors?”. And the answer is YES, they double. As I said in an earlier post the answer to the question “What is the probability of winning the car (after a door is opened)?” all depends on what you do: if you stay it’s 1/3 if you switch it’s 2/3 and if you flip a coin to decide it’s 1/2. But why flip a coin for a 1/2 chance when you can improve your winning chances to 2/3 by making an informed decision?

You have offered no evidence (for the simple reason that none exists) to refute the solution that switching improves your chances of winning to 2/3. As for your arguments:

1. With your dartboard example you just assert that’s it coin flip situation (stay or switch) and that because there’s only 2 possible outcomes they must be equally likely. Where’s the evidence that this is true?

2. “… the question of, “What is the capital of Kentucky?” could be posed and depending on the answer right or wrong, one or the other of the two goat doors could be removed.” Not true. The only time the host has a choice of which goat door to remove is when you pick the car in the 1st place (a 1 in 3 chance). When your 1st pick is a goat ( a 2 in 3 chance) the host has NO choice which door to remove, he is forced to remove the other goat door.

3. Heads, Tails and Wildcard? where Wildcard can magically change into either a heads or tails. This is absurd. No one is denying that the final choice is between 2 doors and that 1 contains a goat and the other a car. You call this evidence?

4. The same argument as No1. and the same assertion that any choice between 2 outcomes means they are equally probable. Again no evidence.

5. “the final choice will be between a door which is 100% a car door, and another which is 0% a car door.” Obviously, and if you knew which door contained the car you’d pick it, you wouldn’t flip a coin to make your choice. What evidence are you presenting here?

You also say “Math doesn’t lie, after all.” That is correct, this is a mathematical puzzle with a well documented mathematical proof (look it up). Statistics doesn’t come into it.

Regarding the 52 card problem, you say “one of them would always be an Ace of Spades. 1 out of 2, every time” Agreed, but which is more likely, (or don’t you believe in probability which is a branch of maths). You think it’s 50/50 ( a coin flip), so if you repeated this experiment say 10 times how many times do you think the Ace of Spades would be the card you picked to begin with, and how many times would it be the card left in your hand? You must reckon they’d occur with about the same frequency, wouldn’t you? If your life depended on choosing correctly between the 2 remaining cards which one would you choose? Be honest now. The rest of your comment is just a rehash of Argument No2 and is simply not true.

So this “mountains of evidence” you refer to basically boils down to: when faced with any choice between 2 possible outcomes the best result you can possibly achieve is to flip a coin.

Another question Ablestmage (actually 2 questions). Last year you uploaded a computer simulation to your other blog. The simulation worked like this:

1. You were presented with 3 doors

2. You picked a door

3. The computer opened 1 of the 2 doors you didn’t pick to reveal a goat

4. You were asked to choose between the 2 remaining doors

5. The computer then showed you which door the car was behind

I carried out 600 trials of this simulation, in 300 I stayed with my 1st choice and in 300 I switched doors. Of the 300 times I stayed I won approx 33% of the time. Of the 300 where I switched I won approx 67% of the time. I can’t give you the exact figures because you deleted all my posts.

Q1. How do you explain this apparent statistical anomaly? If your hypothesis is correct I should have won about the same number of times in each situation (i.e. around 150 times for stay and 150 for switch). Was i just on a “roll”? If I’d carried out more tests would they have evened themselves out? Or was the program “rigged” such that after I made my final decision the program repositioned the car to ensure that switching won twice as often as staying? Is there a worldwide conspiracy among computer programmers to “rig” all these simulations (and there are dozens of them including 1 on the NY Times website)?

Q2. Why did you remove the simulation from your blog shortly after I posted the results of my trials (which you also removed)?

In fairness to all the other “fiftyfiftymen” out there, and as the only blogger to support this 50/50 position, I think it’s incumbent on you to answer these very important question. What do YOU think Ablestmage? Come on, give it your best shot, the world is waiting.

You’re such an ignorant pompous jerk. LOL

http://www.grand-illusions.com/simulator/montysim.htm

Think of it this way: If you picked one door, and Monty asked if you wanted to swap that one door for both of the remaining doors, you would say yes, because it would double your odds.

That’s basically what he’s doing, except that he’s opening one of the remaining doors first. Since it’s obvious that both of the remaining doors can’t contain the car, showing that one of them holds a goat doesn’t change the fact that having two doors is better than having one.

The supposed odds actually are only based on 2. The original point of the game was never to have you pick from 3 at all. The matter of three doors is merely a smoke-and-mirrors trick to lead you to believe there was an edge to be had. Say for instance there is an envelope in Monty’s hand, and that you can pick between the certainty of a promised small cash prize in the envelope, or between a goat door and a car door on stage. If you chose the envelope, and Monty cruelly stated, “Nevermind, I was just joking, there is nothing in the envelope, now pick a door from the stage,” you now have the same 2 options on the stage as before as had you had 3 doors from before and 1 were removed.

And yet again you’ve fabricated a game that is nothing like the MHP.

You pick from 3 doors, so to say “…point of the game was never to have you pick from 3 at all” is patently untrue.

You steadfastly refuse to answer any questions actually pertinent to the problem, such as:

How do you explain the statistical fact that in multiple trials switching wins 2/3rds of the time?

Why did you remove the simulation from your blog shortly after I posted the results of my trials (which you also removed)?

Why won’t you do the experiment yourself with a pack of cards?

Come on Ablestmage, enough of the BS, it’s time to “put up or shut up”. LOL

What are the odds Marley52 will show up for more abuse the instant another comment is posted? 1:1 it appears..

I have already answered all of those questions.

#! How do you explain the statistical fact that in multiple trials switching wins 2/3rds of the time?

This is answered in the original post, this re-debunked article above. The 2/3rds tally is an illusion of advantage, because of the insistence of counting rounds prior to the final 1-in-2 option as if they were at all related. The sundry ways in which the initial elimination round is unrelated are explained plainly. The fact that the numbers change based on how many prior rounds are used, is further evidence that it is entirely the final-choice round that makes any difference whatsoever in the outcome.

#2 – Why did you remove the simulation from your blog shortly after I posted the result of my trials (which you also removed)?

That particular errand was a massive derail from the original premise of the post (which was from the earlier original debunking post, of which this is the re-debunking post, the prequel of this sequel in a sense). I briefly entertained the idea that the original elimination round did actually matter, before I fully realized just how much of an illusion that round is. I also explained how it is such a fool’s errand to go through those simulations because it is in like manner with this post’s examples of the hotel room. The MHP’s question itself is a ruse (like the hotel room bill’s question) designed to fool its victims into believing the initial round played any part whatsoever. Some fools, however, refuse to admit it is a ruse and continually stumble over it, pulling others down with them on their voyage to imminent faceplant.

#3 – I already explained why the pack of cards suggestion you made was foolhardy, in a comment above. All you’re doing is attempting to recreate the same illusion that the elimination round matters in any way whatsoever, when it doesn’t. You’re just rearranging the same non-argument as before, and simply saying that is is proof, without actually having any proof. You’re calling a Spade a Diamond and insisting that it is a Diamond, and base the entire authority of its Diamondness on the fact that you called it one ,rather than the fact that it is shaped nothing like a Diamond, nor even colored like a Diamond. It’s as if you are one of those “propeller plane on a conveyor problem” people who argue that the plane can never take off because of the conveyor belt speed, when it is clearly the propeller that causes lift, not the speed of the unpowered wheels on the conveyor.

This is not the same thing. In this example you eliminated the prize that the contestant DID choose. If given another chance to pick a prize THEN this you have a 50/50 chance of picking the car.

Again you fail to address the facts I presented and instead go off on an unrelated tangent and answer a completely different question.

The MHP is stated thus (whatever you may think it should be):

1. A prize is randomly placed behind 1 of 3 doors

2. You pick a door but don’t open it.

3 Monty opens 1 of the 2 doors you didn’t pick that he knows doesn’t contain the prize

4. Monty offers you the choice: stay with your 1st pick or switch to the unopened door

5. and the question is : “Is it to your advantage to switch doors?”

That is the MHP, and the answer to the question asked is: Yes, you double your chance of winning the prize if you switch doors.

According to your premise there is no advantage in switching doors, that you are in fact equally likely to win the prize by staying with your 1st pick as you are by switching to the other door. Would you say this is an accurate description of your argument?

We can test your premise experimentally as follows;

a) Play the game (as described above) 300 times and always stay with your 1st pick. If your premise is true you’d expect to win to 150 games. Do you agree?

b) Play the game (as described above) 300 times and always switch doors. If your premise is true you’d expect to win to 150 games. Do you agree?

How do you explain the FACT that when I played game a) I only won 100 games?

How do you explain the FACT that when I played game b) I won 200 games?

Either your premise is false, or the games were rigged. Those are the only two options. Which is it.?

You constantly refer to the initial pick and opening of the empty door to be an illusion, If that were so then those steps shouldn’t affect the outcome of the experiment, should they? But since they clearly do, it isn’t an illusion. QED

No surprise here… Ablestmage can’t quite explain why the simulations don’t support his hypothesis. There are multiple sites that allow people to play the Monty Hall Problem out, and they keep track of the results for the entire game. Here is one such site. http://www.stayorswitch.com/

Those who stayed won 33% of the time, and those who switched won 66% of the time. And that’s over >100,000 games.

Well Ablestmage even uploaded a computer simulation in order (I suspect) to support his hypothesis. He removed it, and all reference to it, a few months later after I’d run 600 trials (300 stay, 300 switch) and posted the results here. He also deleted all the posted results and associated comments.

It’s like it never happened. LOL

Um…. I think I’m the only one who agrees with the author, but that’s okay. I’m fine with being right, even if everyone else thinks I’m wrong.

🙂

Um….. why do you agree with the author? Have you looked into the problem elsewhere?

Actually… nevermind.

Tried the experiment at home, proved it wrong.

I’ve been struggling on this problem as well for years. As soon as one door is removed, there are two doors, one contains a goat, one contains a car, so it is a 50/50 situation. But the statistics (so, the reality) says otherwise. The only solution I had was to accept reality, and try to find what was wrong in my reasoning.

You state that the elimination state should not be considered, … without providing any evidence or explanation of that. Well, you try to, but it is a wrong one. You state that the door to be eliminated might be chosen randomly, either by a vote or a dart. This is where you are wrong. Because if the pre-elimination chosen door is a goat (which happens 2 times out of 3), then there is no choice for the removed door: you can’t remove either the car, or the door chosen -I should say “locked”- by the “player”. So if you choose a goat in the first step (67% chance), you have a 100% probability to get the car by switching.

Now… if you locked the car in your first choice, the presenter can choose randomly between the two remaining goat doors which one to kill, and you now have a 100% probability to get the car by not switching.

Since you don’t know which door you picked initially, you have a 67% chance of getting the car by switching, which corresponds to the probability of not locking the car before the elimination.

The flaw in your reasoning is that you think the elimination process is not constrained, but it is, as 2 times out of 3 the presenter has no choice on which door to remove.

The only way for you to be right is to slightly change the problem, letting the presenter remove one of the two remaining doors with no constraints. In this case, you will end up with your locked car and a switchable goat (33%), or your locked goat and a swichable car (1/2*67% = 33%), or your locked goat and a switchable goat (1/2*67% = 33%). If you choose to switch, you will get a goat (67%) or a car (33%), while if you choose to stay, you will get a goat (67%) or a car (33%). The two options are now equivalent, but it is still not a 50/50…

1 000 000 (one million) lottery tickets.

One of them is the winning ticket, 999 999 are losing tickets.

You pick one of them. All the other tickets are put into a pile.

You now have the choice to stay with your 1 ticket, or switch to the pile with 999 999 tickets in it. Should you stay or switch?

According to your logic, one of the pile has the winning ticket, and one of the pile doesnt, therefore it doesnt matter which pile you choose, you have a 50% chance of winning. However, according to math, probability, and common sense, the pile with a single ticket has a 1/1000000 chance of having the winning ticket, and the pile with the rest has a 999999/1000000 chance of having the winning ticket. Meaning that even though one pile has the winning ticket, and one doesnt, it isnt a 50% chance.

@ablestmage

you might want to read this….

http://en.wikipedia.org/wiki/Dunning%E2%80%93Kruger_effect

You have a 2/3 chance of picking a losing door from the start.

In this scenario, switching will lead you to the winning door.

You have a 1/3 chance of picking the winning door from the start.

In this scenario, switching will lead you to a losing door.

Therefore, switchers get the prize 2/3 of the time. 50/50 men get the prize 1/3 of the time.

What more do you need to be convinced then that?

Admittedly, I’ve only read a few of responses (about 4) but of those I have, only the first seems to be arguing correctly for the 2/3 scenarios, and even they haven’t stated very clearly.

The rule is that the host knows which door wins, an can’t pick that as the door that’s removed from the game.

The first choice of the contestant matters because it affects the behaviour of the host.

It restricts the host’s choice of a non-winning door in 2/3 of the possible cases.

In those 2/3 cases, information is exposed that the other door must be the winner.

Switching wins 2/3 times.

In the other 1/3 case, the host has 2 losing doors to choose from, so whatever he picks, the other door is a loser.

Switching loses 1/3 times.

Without the initial choice by the contestant, the host’s choice is unrestricted and therefore exposes no information.

Make no mistake, this is Game Theory. There are 2 players here.

The blogger here at on point casually mentions “the contestants first choice doesn’t matter, the initial choice is just playing the game” as if playing a game is somehow passive.

It’s not. Playing a game is active. The initial choice changes the game for the host (opponent) which leaks information about what the know.

This isn’t straight probability. One of the players knows where the winning choice is, and wants to keep it, but they are restricted by the ruleset. The ruleset causes a leak of what the knowledgeable player (host) knows.

That information leak is what takes this from classical probability into Game Theory, skewing the chances in favour of the non-knowlegdeable player (contestant) to 2:1 if they use the “switch” strategy.

p.s. I do periodically find myself revisiting this problem when something superficiously similar crops up (most recently in research where participants had to find the pair among 3 odors) but in the vast majority of scenarios the information leak is not present, so classical probability reigns supreme.

The MHP is a game contrived for entertainment and is of little application to similar looking probabilistic setups.

Its counter intuitive conclusion is, however applicable to a wide range of game scenarios, particularly where more knowledge is held by one player than the other(s), as it can present the non-knowlegeable players with valuable information leaked by opponent(s), in order to give the opponent a good kicking – the game is a foot!

Good illustration of the game theory aspect and the leaking of (a degree of) their secret knowledge through their actions:

http://en.m.wikibooks.org/wiki/Introduction_to_Game_Theory/Deal_Or_No_Deal#Comparison_with_the_Monty_Hall_problem

Note that we’re addressing the variant of MHP that means the house has to offer a switch every time. If the host is allowed to chose whether to our a switch, the odds swing back in their favour and it’s 1/3 to win vs. 2/3 to win if a switch is offered every time.

I did read someone exploring the real world MHP from the gameshow staying that the house is not only motivated by preventing the player winning, but increasing entertainment and ratings – as such, the strategies they employ may take the odds close to 50/50- anyone know the starts from the actual game show (switch not offered every time) as opposed to the simplified “switch always offered” version discussed here?

Starts=stats

Should we say the critical factor is…..

In this case, should the probability be static or dynamic (all time) for “ALL” items?

In the other word, should we re-calculate the probabilities for “ALL REMAINING” items after an event change in the original environment?

Please advise….

Thank you

Let me explain it to you:

When he reveals one door, he solves the “uncertainty” of what is located in the two doors you didn’t pick.

Everything else aside, you have a 2/3 chance that a car in located within one of the doors you haven’t picked. The only problem is that there is uncertainty as to which one you pick. The host solves this uncertainty by telling you which one contains a goat.

If you pick door A, and the host opens door B containing a goat, he is resolving uncertainty, because door C could have also contained a goat.

Since 3! equals 6, there are 6 ways the goats/car can be arranged.

ABC

CBA

BAC

ACB

CAB

BCA

Assuming A is the car and B and C are goats, and the leftmost letter (such as A in ABC) is the one you choose, you have a 4/6 probability of getting a car if you swap. Consider the six variations swapping:

A is chosen

Host opens B.

You swap

Lose

C is chosen

Host opens B

You swap

Win.

B is chosen

Host opens C

You swap.

Win.

A is chosen

Host opens C

You swap

Lose.

C is chosen

Host opens door B

You swap

Win

B is chosen

Host opens door C

You swap

Win

4/6=2/3≠ 50/50.

The key to understanding the MHP is not in the choice of doors, or the way those choices are made by either the contestant or Monty. It is in the way the doors are filled up in the first place. The door the contestant will eventually choose had been filled by a car 1 in 3 times, so it will always have a 1 in 3 chance of a car. The other two doors combined had a 2 in 3 chance of a car being behind one of them – there’s two chances the car went somewhere there. Seeing Monty’s goat tells us where it didn’t go (and it doesn’t even matter if he chose one of the two at random: he would just ruin the game one show out of three by winning his own car). Are you still there, ablestmage?

Sorry, ‘eventually’ was a silly word to use above.

The first choice is from the set of doors filled with the ratio 1 car to 2 goats. Chance of car 1 in 3.

The ‘switch’ option is to the remainder of a pair of doors filled with the ratio 2 cars to 4 goats after one door has been removed. 1 goat has been removed from the chance by being shown. That leaves one door filled with the ratio 2 cars to 3 goats. Chance of car 2 in 3.

The chance of chosing either door can be 50/50 if random (eg coinflip); the chance of a car being behind one particular door is, however, higher than the other because of the way they were filled.

I know my sums don’t stack up, but I’m trying to show that the ‘switch’ door has 2 chances of a car and 3 chances of a goat. It originally shared 4 chances of a goat with Monty’s door but he’s shown us one of those, leaving 3; it still retains 2 chances of a car (the combined chance of the two doors the contestant didn’t pick first). Monty has taken one goat out of the chances but not a car.

You’re wasting your time trying to explain the MHP to ablestmage since he can’t even understand the difference between choosing 1 from 2 and choosing1 from 3. He thinks the MHP is “There are 2 doors with a prize behind one of them. Pick a door, what are your chances of winning?”

How dumb can you get?

I read all of the comments, but you’re not describing a 1-in-3 chance or 2-in-3 chance. You’re describing wholly irrelevant lead-ups to a 1-in-2 choice. Regardless of whether you name all the US capitals in alphabetical order, or dance the I’m-A-Little-Teapot song before choosing stay or switch, your choice boils down only to stay or switch. Even if you are lead to believe that the previous selection in some way impacts the 1-in-2 option of stay or switch, for instance by suggesting that the first choice is actually part of the final round (when it is instead an illusion of impact), your final choice still remains stay or switch, and that is always 1-in-2. The fact that your calculations must change based on how many prior doors were involved is precisely how your angle is flawed. The choice will always amount to 1-in-2. All you’re doing is offering ways in which the illusion can trick someone into believing the prior rounds impact the final round, when they do not.

Incoherent nonsense as usual.

Incoherence is what you know best, Marley, so it’s no surprise that you would see it wherever you go.

If Monty Hall said “Stay or switch? And by the way the car is behind door three,” would you still contend that the odds are 50-50 by virtue of the fact that you’re given a binary choice?

If the doors are made of glass and you can clearly see where the car lies, is it still 50-50?

No? Why not? Because information is a pretty useful thing to have, right? Actually, it makes all the difference. The number of choices is only part of the equation and in some circumstances it might not matter at all. For example, if I’m holding a royal flush, I know I have a 100% chance of winning the hand even though the odds are apparently different from the other players’ perspectives. Simply having the option to fold doesn’t somehow reduce my odds to 50-50.

It’s not a 1 in 2 choice because there are still 3 doors – Monty’s open door is still in the game. You can see what’s behind it, you would probably not want to choose it, but it is still possible to say ‘What the hell, I know I will lose, but I pick the open door with a goat’. There are still 3 doors. Probability of car: Original choice = 1/3, Switch door = 2/3, Monty’s open door = 0/3. It is wrong to assume that Monty’s door is eliminated.

“… your choice boils down only to stay or switch” – I agree the action of selecting stay or switch is a 1-in-2 choice (50/50 if done randomly), but the MHP is not asking about that. It is concerned with the OUTCOME of the action of selecting staying or switching, which is 1-in-3 wins and 2-in-3 wins respectively.

You’re wasting your time – Ablestmage doesn’t ‘do’ math ………or logic for that matter.

“You’re wasting your time – ”

Oh well, at least it’s keeping my mind occupied. To be honest I still don’t fully grasp why the chance-of-car from the opened door is all bestowed on the non-chosen door rather than shared between the two unopened doors, although I know it isn’t. So maybe that penny will drop in my head if I keep at it.

Take this example: Play 300 games and pick Door1 each time

a) The car will be behind each door 100 times (in theory)

b) In 150 games Monty will open Door2 and in 150 games he’ll open Door3

c) In the 150 games Monty opens Door2 the car will be behind Door3 100 times, and consequently behind Door1 50 times.

d) In the 150 games Monty opens Door3 the car will be behind Door2 100 times, and consequently behind Door1 50 times.

e) Add up c) and d) and you get the car is behind Door1 50+50 = 100 times, Door2 100 times and Door3 100 times – which agrees with a) as it must do.

f) Pick ANY game at random, and the chance that the car is behind Door2 (or Door3) is double the chance the car is behind Door1.

g) Therefore switch from door1 to double your chances of winning the car.

Hope this helped.

Ablestmage says you only win 150 games by switching – he’d have more credibility arguing the Earth was flat.

No, what you’re examining is a false positive and you’re twisting my words.

You’re trying to draw a connection between the first round and the second round. The reward for the first round is merely the narrowing of the doors to two, but the reward for the second round is actually tangible. Trying to connect the two rounds into a single reward chance is the foolhardiness of that thought process. The game is actually only one round of a choice to stay or switch, with any number of possible arbitrary elimination methods to eventually reach a 1-in-2 choice.

The fact that the ‘probability’ changes with the increase in elimination rounds (starting with a larger number of doors and eliminating one at a time) is the reason the additional doors variable can be tossed out.

“… your final choice still remains stay or switch, and that is always 1-in-2”

This is stated in a post above, and is correct. However it is not correct to extrapolate from this:

‘Your final choice remains stay (and have a 1-in-2 chance of winning a car) or switch (and have a 1-in-2 chance of winning a car)’.

You can choose 50/50 between a brown bag containing a bottle of soda and a brown bag containing a bottle of finest sippin’ whiskey, and down your prize in one go. Your chance of getting drunk is not 50/50.

Correcting my own post above, you can choose 50/50 between a brown bag that has been filled by someone picking randomly from one bottle of whiskey and two of soda and a brown bag that has been filled by someone picking randomly from two bottles of whiskey and three of soda (the host having removed all other bags filled with the remaining bottles so as to leave one with whiskey and one with soda while not being allowed to remove the bag you picked first). Uneven chances of getting drunk. Not as elegant as before I added in all the manipulation, but that’s the key to the MHP – there is a lot of manipulation in Monty’s actions.

Ablestmage, in the 300 game example above, if you ALWAYS stick with Door1 when given the opportunity to switch, how many games will you win, 100 or 150?

It’s a simple question, what’s the answer?

It’s not even a valid question. It is identical in nature to the example in the article from the top from the room-renting problem: “Where did the other dollar go?”

It is a perfectly valid question, and you repeatedly refuse to answer it. Why? Is it because you know the answer can only be 100, thus making your “it makes no difference” argument nonsense? You’re a fool and an arrogant one at that.

(BTW, the answer to YOUR question is “there isn’t an ‘other dollar’ for it to go anywhere”. See, it’s not so hard.)

That’s avoiding the question, not answering it. At what time last night did you stop beating your wife? The only valid answers are a time. It’s a simple question, really.

You didn’t answer where the other dollar was, you explained a facet of the nature of the circumstance behind the question to negate having to ask it. That’s what I’ve been doing this entire time. you asking me to describe how many times I would win in whichever circumstance is an open-ended question designed so that the only way to answer is to confirm your position.

Here’s the answer to your question. $30 is paid initially,from that $3 is refunded to the 3 men and $2 is given to bellboy as a tip, leaving $25 which is the cost of the room. ( Or, the room costs $25, plus $2 is given as a tip – total paid out $27 or $9 each man) There is no missing dollar.

Now, stop avoiding my question, how many games will you win if you always stick with Door1, 100 or 150? Can you provide ANY evidence that your answer “it doesn’t matter” is correct? Of course you can’t for the simple reason you KNOW it is advantageous to switch doors as that gives a better chance of winning the car. You just haven’t got the balls to admit you were wrong 6 years ago.

(Your wife-beating question assumes I beat my wife to begin with, my question requires no assumptions to be made only that the rules of the game are followed)

That is not the answer to the question. Saying “there is no missing dollar” does not answer where a missing dollar is — that identifies an error within the question and goes against the rules of the game, as you say. I am identifying an error in the question, also, and will not answer it because yours also requires false assumptions.

The recent reference back to to the article at the top is helpful. This raises misdirection, similar to the case of the missing dollar in the hotel bill scenario, as a cause of people reaching a particular conclusion. In the MHP the third door is not eliminated as often claimed (even “destroyed” in one description), but remains as a choice. Its probability of revealing a car is known by contestant, Monty and observers to be 0/3, 0%, or zero; heck, even the goat probably knows that. However the contestant is not prevented by the rules from choosing it. Thus the total number of options is still 3, hence the 1/3 and 2/3 result for the other doors. Reframing the game as a 2-door one after Monty’s reveal is an example of misdirection.

The eliminated door is no longer a choice. It was a variable only in the elimination round, during which the object of the round was to get to the point that there were only two doors. The contestant *IS* prevented from choosing the eliminated door.

“The contestant *IS* prevented from choosing the eliminated door.”

By what?

The eliminated door is removed from play and shown to be a goat, and the remaining options are to either stay or switch.

And you are more likely to win the car by switching than be staying as you have already acknowledged

The door may have been removed in the way that it can’t be chosen, but it still factors in the probabilities at play.

You choose Door 1 (regardless of what’s behind it) It has a 33.3% chance of having the car.

The other two doors combine to have a 66.7% chance of having the car. Once one of the 2 non-chosen doors is opened,

those two doors STILL have a combined 66.7 % chance of having the car. Only now, the opened door has 0% chance of being a winner

and it’s 33.3% gets transferred to the other non-chosen door which is still closed. 33.3 + 33.3 = 66.7%

33.3% Your door (which you will not stray from)

0% Opened door you didn’t choose

66.7% Un-opened door you didn’t choose.

33.3 + 0 + 66.7 = 100%

Let me also put it this way. What are you chances of winning if you pick a door and having the host simply open your picked door to show you what you’ve won? It’s 33.3%, correct? How is that example different than making your initial pick, being offered to switch and declining?

THEY AREN’T DIFFERENT!!!!! Wouldn’t you agree that BOTH of these scenarios result in you having a 33.3% chance of winning the car?

How is declining the switch make your initial 33.3% chance of picking right increase to 50% by NOT switching? IT DOESN’T. At this point you’re NOT choosing between 2 doors, you’re choosing to keep your 1 door (which was and still has a 33.3% chance) instead of the other two doors (One opened, one un-opened, which have a 0% and 66.7% chance)

Yes there are 2 prizes: (the car or a goat) but there are 3 outcomes (pick car, pick goat #1, or pick goat #2)

How is it removed from play?

Is it dismantled and taken away? Or is a curtain drawn over it? Is a sign ‘No choosing’ stuck on it? Does Monty say ‘You must not choose the open door’? Is it closed again and padlocked?

Or is the phrase “The eliminated door” an example of misdirection?

Well, consider the prospect of not even using doors, and just having the goats and car plainly visible. If that were the case, and the object were to pick the car door, you could just pick the car and choose to Stay in the Stay/Switch round. That would jack up the stats that advise always switching. Making a prize visible eliminates it from the object of picking a car door, since it is revealed already to not be a car. There would be zero ‘probability’ that a goat were actually a car.

Again, how is the opened door removed from play?

While there is virtually no chance any contestant would ever pick it, its presence is contributing to the sum of 100% probability. Its contribution is vanishingly close to 0/3, but it is the reason that the other doors share the sum of 1/3 and 2/3.

Ablestmage has changed his position on the MHP. His new answer to the question posed by the MHP “Is it advantageous to switch doors?” is That’s not a valid question!”

I know, it’s a really, really lame excuse even for him but we’re not dealing with a normal human being here.

“Well, consider the prospect of not even using doors, and just having the goats and car plainly visible.”

Resulting probability of car = 100%, 0% and 0% (or 3/3, 0/3 and 0/3).

Having just one goat visible as in the actual MHP:

Resulting probability of car = 33%, 67% and 0% (or 1/3, 2/3 and 0/3.

Actually if the goat and car were plainly visible behind the doors, then according to Ablstemage since “… your final choice still remains stay or switch, and that is always 1-in-2″ you’d still only have a 50% chance of winning the car!

You’ve got to admire his impeccable logic. LOL

I contend that the MHP door probabilities are 1/3, 1/3 and 1/3 before Monty reveals a goat and 1/3, 2/3 and 0/3 afterwards.

This turns on the revealed door remaining in play (albeit a choice which virtually no contestant would ever make). To determine whether the MHP does or does not have a rule that the contestant must not choose the opened goat door, I return to the original Parade question: “You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, “Do you want to pick door #2?” Is it to your advantage to switch your choice of doors?”

This offers only Door 2, but does not explicitly prohibit Door 3. I contend that if a contestant nevertheless requested Door 3, Monty would allow it. It would make great TV as the audience would be incredulous – do they have a dope or a dupe before them?

The observer is therefore misdirected towards the pro-intuitive but inaccurate 50/50 solution by the implication that Door 3 is eliminated, when in fact it remains in play. Above, Ablestmage usefully draws attention to the misdirection in the hotel bill missing dollar question which I believe illustrates a similar state of affairs.

You’ve opened a can of worms with that comment Freddie. Ablestmage will probably write an entire book in response (none of it making any sense of course).

The opened door has (obviously) a probability of zero of containing the car, whether it’s still “in play” is immaterial, the unopened door is twice as likely to hide the car as the picked door, a fact which Ablestmage has tacitly acknowledged.

“whether it’s still “in play” is immaterial”

Ah, the third door. I can think the MHP through more easily when I assume Door 3 is in play throughout. However you may be right that it is immaterial from a certain stage in the game onwards. I prefer not to complicate things for myself – Door 3 must be in play until the goat appears. From then on it contributes nothing – literally – to the sum of probability by becoming 0/3. Though if only Doors 1 and 2 are left we cannot express their chances as 0.66/2 and 1.33/2 as we cannot play a fraction of a game so are left with 1/3 and 2/3. Even though /3 in probability (/2 in the ’50/50′ answer) refers to number of games not doors, my mind is susceptible to misdirection by these figures being the same.

Door 3 must be in play for Monty’s reveal, it must continue to have existed as it contributes to the /3 of its 0/3 and hence the other doors’ x/3. And the goat was, is and continues to be real – it must exist as a choice to be rejected. You can see the knots I’ve tied myself in above; I can do without the existential angst of a no-longer-material door cluttering up my mind as well.

While we’re waiting, imagine a show where everyone is incompetent. Incompetent Monty (let’s call him iMonty) invites his iContestant to choose one door from three. They pick Door 1. But iMonty forgot to tell the crew to fill up the doors – there’s nothing behind them, just a car and two goats left in the stockroom. iMonty hisses to a stagehand ‘Go get something from the stockroom and put it behind Door 1. And don’t turn on the light’. So the stagehand pulls out a random object from the darkened stockroom and puts it behind Door 1. Then iMonty announces ‘I’m going to open Door 2’ and mimes two horns to the stagehand who rushes off to the stockroom, switches on the light and grabs a goat which he installs behind Door 2. Monty opens Door 2 with a flourish and asks the iContestant ‘Wanna switch?”. To everyone’s amazement the iContestant says ‘Can I pick Door 2 please Monty?’. iMonty says ‘Sure. You must love goats’. 2 times out of 3 it’s a car left in the stockroom.

You’re only confusing Ablestmage even more than he already is. Stop it!

The author of this blog is a colossal retard.

Thanks for the great entertainment.

I agree. Right after I saw the “switch” explanation, i felt something is wrong!!! Finally found someone that hold the same opinion!

And that person who holds the same opinion as you has as much understanding of probability as a goldfish does. The answer is simply – switching has a 2/3 chance of winning.

IF you played 300 games and always picked Door1 and always stayed with Door1 how many games would you win – 100 or 150? I asked Ablestmage the same question and his response was “It’s not a valid question!”. What’s your response?

Tony – I might have interpreted your comment as agreeing with Ablestmage. If that’s NOT the case I apologise for my previous reply.

Wow this is still going.

Stumbled upon this site after introducing my girlfriend to the problem in conversation tonight. Know it’s counter-intuitive and a bit of a mind-fuck at first, but it’s astonishing that anyone denies the switchmen view after seeing the explanation and sheer weight of proof over the last 40 years. Absolutely unbelievable. This is not a debate: anyone who refutes the switchmen view has either misunderstood the parameters of the question, not spent enough time thinking about it, or is just plain slow.

The fact that it is so debated literally makes it debatable — in the most non-figurative way. The problem is a reasoning problem, but mathematicians for some reason (perhaps naturally) believe math overrides reason. It is in fact, the math explanation that is counterintuitive..

It’s not debateable, at least no more debateable than saying 1+1 =2.

It’s is a reasoning problem and a simple mathematics problem. – the two are not mutually exclusive. To artrive at the correct answer to the question (“Is it better to switch doors”) you can use reason (logic) or mathematics, they both give the same answer.

“or is just plain slow.”

You hit it in one.

Update: I am creating a video form of the article’s arguments, and it will be published on the following YouTube channel when it is ready. Please subscribe (free) to that channel if you wish to be notified when the video is ready to be viewed.

https://www.youtube.com/channel/UC0oP9ZAuQbA3zmYrVCjaJYw

I can hardly wait 🙂

I’ve decided to renew my intent in creating a video form of this series on YouTube, but with a different account than above. I’ve settled on a chalkboard for the medium of visual cues, although it may largely be spoken. The previously mentioned channel doesn’t like me for some reason, so the new one is below, and aptly named Switchmen Repent. I don’t have a reliable ETA, but I am setting aside definite time to focus on it.

https://www.youtube.com/channel/UCTEMJiBh1eJqGfQDydSvZpQ

Do you have a proposed date for when we can expect to see this video? It’s been over a year since you 1st put forward the idea.

Don’t forget to include lots of irrelevant stuff like; “What’s the capital of Kentucky?”, “3 men book a room in a hotel ……”, and of course “Imagine a spinning dartboard….”

I wait with bated breath!

Commenting as a Switchman. It’s a long one, but clarifies things.

As far as I have understood, Ablestmage’s argument is that there is one door out of the three which is irrelevant to the game. Therefore the game is effectively played between two doors, so chance of winning the car is 1/2.

This has a fatal flaw, because an irrelevant door is neither Monty’s nor the player’s decision, but in the actual game either Monty is choosing the “irrelevant” door to open, or the player choice compels Monty to show a door as being “irrelevant”. The conclusion is that both people consider none of the three doors as irrelevant to the game, but Ablestmage tries to make it appear irrelevant by focusing only on the second half of the game.

Surely a player, who is freshly presented with two closed doors and an open third one (with a goat), and is told that only one closed door contains the prize car, has typically a 50% chance. But this is just a description of half the game. The player makes choices in the game twice, the first one between three closed doors, and the second one between two closed doors; these determine the two halves of the Monty Game.

In the first half of the Monty Game there is one winning door out of three, so the door chosen by the player takes 1/3 chance for the prize, while the unchosen two take 2/3, and this poses a condition to the second half. It is not unreasonable: The same thing happens in other games, say basketball: Halftime score is changing the odds of the score at the end, so halftime score poses a condition to the winner.

Of course, the player has any freedom to make a choice. The player can ignore his score of the first half freely, and play as if a first half never occured. It is a question of strategy here, and comes from the combination of the desire to win plus the freedom of choice.

Till now, we can summarize three options for the player’s second-half strategy.

(1) Player sticks to his choice. No matter if he has considered the first half or not. Maybe he has, maybe no.

(2) Player discards all first half and decides to choose at random, say he flips a coin. Lucky guess, anyway.

(3) Player switches his choice. No matter if he has considered the first half or not. Maybe yes, maybe no.

There can be more than these three “strategies”: (4) He can use psychology and read Monty’s mind, (5) He can threaten Monty to confess etc. These include more than pure mathematical thinking though and are somehow unfair, invalid.

So what is the chance of winning the prize for each strategy option?

(1) 33% = 1/3

(2) 50% = 1/2

(3) 66% = 2/3

We cannot assign easily chances to (4), (5) etc. obviously.

The original problem is supposed to work with which option between (1) and (3) is best. But there is actually an additional strategy (2) which has 50% chance. It is valid for the mathematics of the second half-game if the first half is completely ignored. Such a case, there is an element of truth in Ablestmage’s claim. A claim which denies any probability that the second choice (second half-game) depends on the first. However, this claim has to be reasonably proven.

For me it is an amnesia claim, a memory distrust. How could I switch my first choice unless I can remember this? The intermission between the two halves of the Monty Game may last for long and forget my first choice and come indifferent to a freshman in front of a two-door dilemma.

Whatever the claim, however, there is undoubtedly no best strategy than switching the first choice, if you can. Lucky guess like flipping a coin is second-best, and sticking (especially if it comes out of stubborness) is worse.

I hope I helped.

Don’t expect that an appeal to reason will curry any favour with Ablestmage – he doesn’t know the meaning of the word. I once asked him: “if you played 100 games and always picked Door#1 and always stayed with Door#1, how many games would you win?”. His answer, believe it or not, was “That’s not even a valid question!”

Go figure.

Unless it is deliberately deleted, a reasonable argument is useful to reasonable people who read this blog.

Well Ablestmage deliberately deleted several of my comments around 2 years ago that provided empirical evidence to refute his ludicrous claims that staying/switching was 50/50. He even removed the app (which he’d uploaded himslef) that I used to gather the data to show his argument for the crap that it is.

The arguments Marley52 is talking about are not deleted, they are just moved to Moderation-Pending status so they aren’t publicly visible, but they’re still there. Marley52’s remarks took me down a flawed version of my own reasoning, and I now doubled back on that reasoning and take a different path in response to the same evidence presented. It will be discussed in the video upcoming. I have to make graphs and illustrations to time properly with the narration..

Additional edit: The “app” that Marley52 refers to was an offsite link, which is a java applet. My security settings no longer permit running anything java, so I can’t verify that it works, but I’ve never coded anything like that and don’t have (nor ever had) any access to that web space. This is the original link; use at your own risk. The original article had an addendum at the bottom that stated, direct quote, “

Edit (12/25/2011)– Here is a link to a (java) simulation that calculates the percentage of wins based on whether you stay or switch, and does not calculate previous choices as a variable. Staying or switching makes no calculated difference, because your choice is always between two doors — to stay, or to switch. Previous choices are completely irrelevant.” End direct quote.“Marley52’s remarks took me down a flawed version of my own reasoning,”

All your reasoning to date has been flawed. This video of yours is going to be great fun – don’t forget to include the rotating dartboards,

Marley52: 8/10/14 “Ablestmage will probably write an entire book in response (none of it making any sense of course)”.

It turns out to be going straight to the film of the book. I await the upcoming video with interest. I wonder if casting is complete – is it too late to approach Richard to play Ablestmage? Come to think of it, Richard is actually less wrong than Ablestmage; at least he doesn’t think it’s 50/50, so perhaps he could make script consultant.

And the film of the book will be going striaght into the remainder bin at your local video store. I’d pictured Forest Gump as Ablestmage, or perhaps Curly from The 3 Stooges who, although having died in 1952, probably has more brain activity going on in his corpse than Ablestamge has.

Richard of course can play one of the goats….or both.

Ablestmage, allow me to assist with the screenplay for your upcoming video. How about the scenarios from the top of this thread?

Scene 1 – the rotating dartboards (this answers the question ‘Is it to your advantage to randomly stick or switch?’ Answer: 50/50).

Scene 2 – the capital of Kentucky goat removal (this has a 1/3 chance of the contestant’s first choice being removed as a goat so is not the MHP).

Scene 3 – the Wildcard (if Monty opens the car door, he swaps it for the remaining goat. The contestant’s strategy is not stated so this is not the MHP).

Scene 4 – the Three-sided Dice (the contestant flips a coin. Answer: 50/50).

Scene 5 – the contestant wins either a car or a goat (yes, we know).

Hang on, we still need one for the actual MHP question, where the contestant calculates whether to stick or switch. How about this:

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice? (Answer: Yes; door No.2 now has 2/3 chance of the car).

Have your people check for any copyright issues (magazine rights etc) – you never know.

And the soundtrack could be Julie Andrews singing “The Lonely Goatherd” from The Sound of Music.

Riddle me this: What is the probability that you will select one of the two doors Monty will not reveal, and that both of those doors will follow you to the final round from which you must choose?

Can we have that riddle in English please? You’re the only one here fluent in Gibberish

That just describes the structure of the game, not the contents of the doors, so is a bit of a misdirection. These are probabilities the the game rules will stay the same. They don’t describe the game outcomes.

Split down to rules:

1> “probability that you will select one of the two doors Monty will not reveal” = 1/1 = Monty is never allowed to reveal a door you have chosen (or the car!)

2> “both of those doors will follow you to the final round ” = 1/1 = Contestant is allowed to switch in the final round.

And scenarios::

1.0> probability the car is still in the game after this point and Monty has revealed one of the 2 goats = 1/1, probability that the the contestant has already selected the car = 1/3

1.1> probability the car was behind one of the 2 doors = 2/3

1.1> probability Monty choosess goat#1 = differs depending on player choice: 1/3 times (when player chose car) = 1/2, 1/3 times (when player chose goat#2) = 1/1, 1/3 times (when player chose goat#1) = 0/1 (included these to illustrate that on player affects the choice of the other [host] player – the choice in round 1 is active, not passive)

2.1> player sticks with the 1/3 chance from 1.0 = 1/3 chance they have the car

2.1> player switches to the other (representing 2 doors, one of which would definately have resulted in a goat, but had been nullified) = 2/3

Given that there is a 1/1 probability that every possible outcome of the game’s rules results in a choice between two doors, there is no weighted probability that either door is the car door. You can dance as many dances as you wish prior to flipping a heads-or-tails coin, but no dance is going to influence the likelihood of heads or tails.

@Ablestmage

Here’s a “dance” you might want to include in your new video.

300 contestants play the game. You”d expect 100 of them to initially pick the car, and 200 of them to initially pick a goat, After Monty reveals a goat to each contestant (from amongst the unselected doors) what’s changed? Nothing. You still have 100 contestants who’ve picked the car and 200 contestants who’ve picked a goat.

Now ALL 300 contestants decide to switch (all being of moderate intelligence with a preference for winning a car instead of a goat).

What’s the result? Well the 100 contestants who picked a car unfortunately win a goat, but the 200 contestants who picked a goat win the car. 200 winners from 300 contestants, better than the 150 winners you’d expect to get if everyone flipped a coin wouldn’t you say?

If you want any more suggestions for your video I’m more than happy to help !

100%, 1/1, certainty.

One of the problems with the 50/50 argument is it implies the odds improve from 1/3 to 1/2 without anything changing.

Here’s a different description to show round 1 doesn’t magically improve the odds of a 1/3 selection to 1/2:

3 empty spaces into which 3 cards (in a 1 suit deck to remove draws) will be dealt face up. Highest card wins.

Bet on 1/3. Odds = 1/3.

Deal cards = win 1/3, visible all the time.

Subsequently remove one of the other 2 cards, leaving your card and one other card on the table and the odds of winning are….?

Still 1/3 as nothing has changed.

This is a logical equivalent of MHP without switching.

In terms of the 50:50 argument, at round 2 of MHP, removing a loosing door doesn’t affect the probability of a choice made in the past being right, so can’t increase the odds of a win from 1/3 to 1/2. It does improve the ability of a 3rd party observer to know where that car is, but cannot affect the chance of a win.

50:50 can not exist for a contestant.

My contention is that the odds are not 1/3 in the first round, because the removal of the “third” door was never a variable in the selection process to begin with. The fact that Monty will not remove the door the contestant picks, perfectly illustrates that only two doors are in play at any given time during the entire game. The idea of a third door itself is just misdirection.

The entire game consists of a single door that the player will choose, a door that Monty will remove, and a door that will remain influenced by either person. Monty will always remove the particular door based on a consistent algorithm throughout all possible choices, so the first round of removing the extra door need not even occur and is merely a formality conducted prior to the final and only actual round. The game can then proceed between a selection of the only two doors in play in the second round, which are still the only two doors that were playable in the first round.

When the final round begins, the contestant does not have a door to stay or switch from, he has a selection between two doors. The idea of “having” a door is an illusion, and only serves to represent an established variable according to which Monty must remove the extra unplayable door.

No door is ‘out of play’ for the entire game. Here are the stages of the game, using 1= first choice, 2 = Monty’s goat, 3 = switch door:

First choice – 1, 2 and 3 are in play,

Monty’s reveal – 2 and 3 are in play,

Stick or switch – 1 and 3 are in play.

Each door is in play for at least two stages, and each door influences all three stages of the game. However the most important stage is one you don’t mention, Ablestmage. It is this one:

Before the contestant is brought in, the game is set up by Monty’s stage hands – they fill three doors from two goats and one car. It is at this stage that the probabilities are cast.

“My contention is that the odds are not 1/3 in the first round”. and that statement is clearly incorrect since at the beginning of the game there are 3 doors and you can pick any one of them. Changing the rules of the game to fit the answer you want to arrive at is called ‘cheating’

Your contention is a “false premise’ – look it up if tou don’t know what the phrase means. Why don;t you just follow the rules of the game as they are laid out, it’s perfectly straightforward,

I hppe your fortcoming movie production is going to be a bit more rigorous in it’s approach to the the problem than simply making bold assertions with nothing to back them up – which is all you’ve done so far.

“My contention is that the odds are not 1/3 in the first round, because the removal of the “third” door was never a variable in the selection process to begin with”

Therein lies your problem. Why should it matter that he removes a door *later*? The contestant still has to make a choice from three doors, hence he has a 1-in-3 chance of picking the winner. Eliminating a door doesn’t somehow retroactively change a choice he already made.

’50:50 can not exist for a contestant’ … unless of course they pick randomly from two. Perhaps that’s where the misdirection leading to the 50:50 fallacy arises – since a random pick from two is 50:50 and there are two doors left, the non-random pick is also (incorrectly) assigned 50:50.

You’re totally correct Freddie. There is a 50:50 strategy in the game, but it needs a random selector to get there.

To summarise:

3 strategies exist (more may exist, but I can’t think of any that aren’t based on these/a combination of them)

A) pick 1/3, stick with that throughout = 1/3

B) once a switch is offered, flip a coin to decide between the 2 remaining doors to get 1/2 (not flipping a coin would mean you’re stick with the 1/3 choice from the start of the game)

C) always switch from your single

(Sorry – got cut off!)

C) always switch from your single 1/3 to the other 2 original doors to get 2/3

Side note:

I’m firmly in the “switcher” camp, but started off not fully understanding the issue, but now believe I do understand.

This make me far more curious as to what leads us to jump to 50:50.

I was looking around common cognitive biases to try to pin it down (and hopefully recognise similar issues elsewhere in life). Status quo bias will definitely be in play to some extent, but it seems to me more of a heuristic shortcut that’s at fault here. We shortcut stuff all the time to progress faster and more efficiently based on previous experience/observed/learned patterns.

It’s a standard facet of human performance that we use heuristics too arrive at an answer (most often correct/usable) more quickly, without needing to process the problem to it’s full extent.

People that flame the OP on this blog (and flame others that arrive at 50:50) haven’t understood this either, and for me, the reason people (apparently including PhD’s and a nobel lauriet or 2) arrive at 50:50 for sticking is more important than the probability calculation. It has a wider application to the way we do things.

The setup of the MHP is what triggers cognitive bias in the form of a heuristic shortcut. In behavioural economics, you learn about choice architecture – it seems that the MHP presents a facet of that – presenting choice in such a way as to trigger and exploit our heuristic bias – probably not designed in that way originally, but seems to have happily landed on a neat demonstration of the effect.

Interesting thoughts. I had a lot of trouble getting my head round ‘plane on a conveyor belt’ and found inaccuracies in some of the correct-answer explanations. I am wary of expressing certainty as I know I am far from infallible. On other threads, people argue that while 50:50 is wrong, they insist that 2/3 cannot be ascribed to one single door (though I believe it can) and must be shared equally between the two doors in the ‘switch zone’ – even when the goat door is open. Ablestmage illustrates misdirection in his missing dollar example but doesn’t accept it might be affecting his interpretation of the MHP.

Bit of a tangent – apologies for going here, but:

I’d never heard of “plane on a conveyor belt”, but having looked at it; I can’t see any issues with it taking off (*), other than it being more likely to get a landing gear failure as the wheels would be running at double speed (increases heat, may cause failure, ergo the plane might not take off due to mechanical failure!)

(*) thrust is against surrounding air, not the treadmill; more friction loss at wheel bearings as the wheels run at double speed, so some braking effect on the plane, but unlikely to prevent take off.

I suppose if you imagined the wheels replaced with rotating dartboards it might help …

You’re probably a big troll. I’ll reply anyway with a simple reasoning.

Suppose instead of 3 doors, you have 100, 99 are empty and one have a car.

If you pick door 1, what’s the probability of hitting the car? It’s 1/100, or 1%. What’s the probability of not hitting the car? 99%.

I believe we can agree up to this point.

Now, the host will open 98 doors and let one door remains. He will never ever reveal the car, so he will always open 98 empty doors.

So, let’s go through some examples.

Car behind: door 1.

You pick: door 1.

Host opens: 2-98.

Host asks: Want to switch to door 99?

You reply: yes.

Result: You lost the car, since door 99 is empty.

Car behind: door 2.

You pick: door 1.

Host opens: 3-99.

Host asks: Want to switch to door 2?

You reply: yes.

Result: You hit the car.

Car behind: door 3.

You pick: door 1.

Host opens: 2 and 4-99.

Host asks: Want to switch to door 3?

You reply: yes.

Result: You hit the car.

Car behind: door 4.

You pick: door 1.

Host opens: 2-3 and 5-99.

Host asks: Want to switch to door 4?

You reply: yes.

Result: You hit the car.

Car behind: door 5.

You pick: door 1.

Host opens: 2-4 and 6-99.

Host asks: Want to switch to door 5?

You reply: yes.

Result: You hit the car.

The actual question is: did you hit the car the first time?

Now, don’t let that fool you, if you come into the room after the host has opened all 98 doors and asks you to pick one of the two remaining doors, your chance will be in fact 50%. However, we have more information about the problem. We have previous information(the car is in one of the 100 doors) and we have new information(the car is not ain any of the 98 doors), so we can conclude that either the car is behind the picked door or behind any of the other 99 doors, but now they’re all open and only one is remaining.

Some observations. Not sure if they have been covered before but i didn’t see anything like it while scanning.

Here are some thought about the circumstance.

Its a game and games often gloss over problems that cant be glossed over in reality. Say you are driving down the highway and there is a 3 way fork in the road. One will lead you to your destination but you are not sure which one to take. you then drive down the right fork and then your friend calls you and tells you that he didn’t know which fork to take either but he knows that the middle fork is not the right road. This information doesn’t help in your previous decision as it has already been made and you are already driving down a road. yes you can now turn around and go the right way but that would be the equivalent of having exhausted all other options.. it would be like picking and opening door number 1 and then MH revealed door number 2. The certainty of the out come is know at this point. It is obvious at this point that this is a game and not based in reality. The game takes advantage of a temporal variable which includes past events. time travel is not an option.

Some may say that the situation can be applied prior to driving down any particular fork. In other words the driver has the intention of choosing the right fork and then gets the call from his friend. The problem is that if the driver chooses to take the right fork or choose to take the left fork based on the phone call does not change the fact that he is still on the same road before the fork. The new information changes the odds to 50/50 at that time. One must also consider that intentions are not actions or outcomes. If you include an intention in a calculation you are in essence including a non-action or a non-outcome or an non-existence (i.e. not reality).

This of course is not to detract from the merit of the MHP exercise. It is a great slight of hand (or time as the case may be) and also helps to sharpen cognitive tools as well as improve your chance when playing games ( which do exist in reality and do have real consequences ) even if you have to make-believe to get there. So i wouldn’t loose sleep over the counter intuitive nature of the problem as your intuition is correct and it is the game that is a false presentation of reality.

2cents

There is a concrete, quantifiable and observable outcome to the contestant’s first choice, demonstrating that it is no mere intention. It results in the action that Monty will avoid opening that door and will turn to the other two doors to open one of them which contains a goat. You can enact this in reality if you follow the script, and toy goats and cars are available to keep production costs down (though that might smack somewhat of make-believe). No sleight-of-hand, just math. And, as you point out, counter-intuition.

Math is a tool and it can misapplied. See the mars orbiter for an instance of the misapplication of math. Mathematically the calculations were sound but it was not tied mathematically to the proper restraints of reality. That result is what happens when ones “choice” to land a spacecraft and 125 million dollars are smashed into mars.

One can subtract 10 apples from 5 apples all day long playing games, but they will never encounter -5 apples in reality. There is nothing wrong with mathematically subtracting 10 from 5 but there is a big issue when one tries to apply this to reality. Math is not a magical application that will convert contradictions into truths. Setting up a math Problem in which you subtract 10 apples from 5 is not a mathematical representation of the real world. This is the case even if i can demonstrably dress up pretty women in bathing suits who hand contestants apples with a negative number painted on them as within the realm of a game show.

MHP is a game who’s analysis overlooks realities that are not possible. There is nothing wrong with the math but it is describing conditions which are a fantasy. This is why we are discussing a game show and not a physical phenomena.

Directly to your point: the choice the contestant initially made never manifests in the game. Consider what you would see if you watch the game show while it is muted, all you will see is a conversation and then Monty hall will open a door. after some more conversation the contestant opens a door. These are the only physical events of the game. All other constructs are interpretations. The only time the contestant directly affects the door is in an arena of a 50/50 option.

I’m not arguing that the interpretations are not interesting but i do assert that the interpretations are contained only in a game show who is not subject to the rules of reality. when you include the first decision in the accounting of events, you are doing the equivalent of assuming that you can alter a previous conditions by making subsequent decisions which is ludicrous. its a bit like claiming omnipotence. Literally what the interpretation of the calculation is saying is that if you include your first decision in the accounting of events, an un-chosen 1/3 chance is also a 2/3 chance simultaneously. Something cannot have two mutually exclusive attributes at the same time unless it exists as a game. i.e. being manipulated by the constraints of the game.

If i tell you i can make you arrive at the number five by playing a game it does not mean that i have created a situation which I can influence out comes.

Pick any positive number you like, any number, and do the following calculation:

Double it

Add ten

divide it by 2

and subtract you first number by your current number.

You arrive at 5.

If we were to draw an improper conclusion it would be that i can make people arrive at 5. Its demonstrable. if you don’t believe i have this power than obviously you don’t believe in demonstrable math. Its no sleight-of-hand, just math. In this situation the issue is not the math, its the conditions of the game in which i assert that i have the ability to make people arrive at 5. Just with the MHP, the problem is not with the math, the problem is that is math is being used to verify a false premise.

Take 100 doors. You pick one, the host will open 98 empty doors. If you picked the right door the first time, the car is in it(1%), if you didn’t pick the right door the first time(99%), the car will be on the only other door left by Monty.

This is pretty logical and completely rational. If you can’t understand that after several people pointed it out you might need a psychiatrist.

Your main point is that the 98 doors don’t matter since it all boils down to two choices, the thing is that it’s exactly this piece of information that leads to 1/99 instead of 50/50. If you ignore that knowledge obviously you will end with a 50/50 answer. That’s why probabilistic works, you CHAIN UP probability, you don’t treat them as separate events since they’re not.

You will find several good lessons about probabilistic on the web and an introduction to them may enlighten you giving you the foundations. While you don’t make an effort and seek them, let you be in the dark.

Fix: the piece of information leads to 99/100 by switching(not 1/99, not sure why I typed that in)

“Fix: the piece of information leads to 99/100 by switching(not 1/99, not sure why I typed that in)”

1/99 is still true for the first choice because of the 98 open doors plus the door remaining closed, though of course the 99/100 figure relates to the door remaining closed which is the point you were making. So it did still make sense.

Adam, another contributer who doesn’t understand the problem or basic probability. You shoud stick with fruit, you seem quite knowledgeable about apples.

I’m sorry you feel that way. Perhaps if you spent more time reading my response instead trying to figure out my deficiency you would have seen that actually i don’t have a problem with the math nor the game. I cant force you to read.

@Adam, you said ” The only time the contestant directly affects the door is in an arena of a 50/50 option. “. Since this is incorrect the rest of your comment has little relevance. I can’t force you to understand how probability works, but you could go and read up on some proofs as to why switching doors gives you a 2/3 winning chance.

And btw, what’s this “false premise” you refer to?

Once again the math in and of its self is not the problem. If you cant get past us agreeing on the math, i’m not sure where to take this.

The false premise is the only thing that makes the MHP relevant (aside form the interesting physiological studies around it). If one insists that there is a profound meaning of the counter intuitive out come of the MHP this is the false premise. For a corollary, if one is in disbelief of one of David Copperfield’s stunts it is not considered counter intuitive but an understanding of the difference between make believe and reality.

The events of the game are not real and can only manifest in a game just like Copperfield cant really make money fall out of your ear and that situation only exists with in the confines of the trick.

“Directly to your point: the choice the contestant initially made never manifests in the game.”

Yes it does. It determines that this door may not be opened for the goat reveal. Monty approaching one door while avoiding another is a physical event.

The choice its self is not a magical entity which maintains a physical being only in the moments between the time the contestant expresses it and the time Monty opens a door. The contestants “choice” is of no consequence. The only case that it would be is if the physical location of the door had any bearing on the game which it doesn’t.

Monty will always choose one of two doors, and will always reveal a goat. you could play the game in a fashion where the contestant secretly recorded the their “choice” and Monty randomly opened either of the doors. this would remove the choice from the equation. If you then decided to throw out any coincidence of Monty and the contestants doors being the same you will end up in the same statistical situation in which the contestants choice had no real bearing on the choice of the door. it is done in the fashion it is done in to facilitate the illusion.

T0 is when Monty opens the door. it doesn’t not matter the criteria that he uses to determine which goat to reveal.

If you change the rules of the game you’ll get a different answer. “it doesn’t not matter the criteria that he uses to determine which goat to reveal”, and that statement just goes to show tou don’t understand the problem or the solution.

You didn’t actaully look up any proofs did you – why not?

if you insist on dropping context i cant help you in your understanding. I gave example of how it is not changing the outcome of the game while illustrating my point.

If you are holding steadfast that the game must remain unchanged in order for this to “work” i would agree. This is my point is that this is only a game and does not have any bearing to the physical world.

There is no need to proof it. I have indicated that the math is OK for describing events which are not possible. Consider my point about negative apples. Negative apples can be proofed until the cows come home, but you will never encounter negative apples in reality.

Along with Freddie says, I want to address another point:

“when you include the first decision in the accounting of events, you are doing the equivalent of assuming that you can alter a previous conditions by making subsequent decisions which is ludicrous. its a bit like claiming omnipotence.”

No, it’s not omnipotence.

Well, I’m pretty sure you can’t(and neither do I) go back in time. However, going back in time in this case is replaced by monty allowing you to reconsider and take another door. He will let you change your choice after he made his, and there’s your “time-travel”.

In fact, if the host reveals an empty door and you could actually time travel to a point before you made you first choice, with the knowledge that door C is empty. Choosing between A and B would give you a 50/50 chance(since theoretically the host was allowed to pick the door you picked since you have picked it yet). This is all considering you can change what already happened. Yeah, sounds weird, thou this is some hypothetical thinking and has nothing to do with math.

Yes, re-reading Adam, who claims the MHP solution is ludicrous in allowing that “an un-chosen 1/3 chance is also a 2/3 chance simultaneously”, shows failure to separate temporal events. The chance is 1/3, 1/3, 1/3 before the reveal – when Monty physically and deliberately avoids one door, considers another and opens a third – and 1/3, 2/3, 0/3 after the reveal, never simultaneously.

Sorry for the sequence of my comments. The website seems to be placing comments in the wrong ‘reply’ The point about proving it to your self is for ash.

Monty’s actions are the only actions that are of consequence. He will always reveal a got. the contestants choice does not matter . Consider the statistical cases in which the contestant always chooses the door with the car in their first choice. All of these cases work against your proposition as Monty does not avoid any of those doors yet the game play is still the same.

Ill ceded that simultaneously is probably not the most apt description but im not sure how to describe two events where the event that happened first actually happens last.

“im not sure how to describe two events where the event that happened first actually happens last.”

I think that’s called ‘a discombobulation in the space-time continuum’.:). You are joking right?

“math is OK for describing events which are not possible.”. Since it is extremely simple to empirically varify the mathematics of this puzzle by using 3 perfectly possible playing cards it seems you still don’t know what you’re talking about.

” If one insists that there is a profound meaning of the counter intuitive out come of the MHP this is the false premise.”. Who ever said there’s profound meaning in the MHP.? And even if someone did how would constitute a ‘false premise’? Do you even know what a “false premise” is? Ablestmage doesn’t. You;’re not him by any chance are you (you sound an awful lot like him)?

“I think that’s called ‘a discombobulation in the space-time continuum’.:). You are joking right?”

– There is no word for it as there is no occurrence of in the real world. We don’t bother spending our time defining contradictions because they are of no use… so yea… sorry i don’t have a word that embodies a space time discombobulation.

“Since it is extremely simple to empirically verify the mathematics of this puzzle by using 3 perfectly possible playing cards it seems you still don’t know what you’re talking about.”

– you are verifying a trick. No matter how many times you see a magic trick, it does not make magic real. If you perform the trick you will verify that in the conditions of the game the 33/66 are the out comes. The problem is that the accounting of the game where you derive the 33/66 outcome hinges on including a mythical choice. yes it can make sense in contexts of the game but that all it is, a game, a trick. I challenge you to duplicate the exercise in a world where the choices are real and have impact on the physical world.

“Who ever said there’s profound meaning in the MHP.? And even if someone did how would constitute a ‘false premise’? Do you even know what a “false premise” is? Ablestmage doesn’t. You;’re not him by any chance are you (you sound an awful lot like him)?”

I didn’t fabricate this page or the Wikipedia page just to troll you into thinking the MHP has significance. I may be crafty but im not sure i could pull that one off. Im not sure where you are headed with this, If you plan on arguing semantics we can stop right now. Im not going to chase down what the definition of ‘is’ is. I don’t know Ablestmage, and i don’t care if he knows what a false premise is. But id be willing to bet he probably agrees with me on some things given his statement below:

“In order to arrive at that conclusion, however, the elimination round must be taken into account, even though the elimination round is irrelevant. The acceptance of the irrelevant elimination round as weight upon a 50/50 choice is the switchman’s error, when it genuinely bears no weight.”

@Adam. I don’t know what point you’re trying to make. The MHP is a mathematical game, it has a set of defined rules and can be replicated using 3 playing cards. At the end of the game the question is asked “Is it better to switch doors/cards in order to win the prize?”. It is mathematically and logically proven that switching doors gives you a 2/3 chance of winning, and this can be substantiated by repeated trials of the game.

You argue that in order to arrive at this answer some “false premise” is made at the outset – but you haven’t said what that “false premise” is. There is no magic trick involved – you pick 1 from 3, why do you suppose this is a “mythical choice” when your ‘pick’ restricts Monty’s subsequent actions? Explain yourself.

Ablestmage doesn’t have a clue what he’s taliking about, following his “logic” any choice between 2 possibilities is 50/50 – ‘Will the sun rise in the East or West tomorrow? 50/50″. “If I walk around in a thunderstorm will I get struck by lightening? There’s a 50% chance you will” etc. etc. The man is a complete ignoramus, pay him no heed.

im sorry you have missed the point im making. I can accept my faults in that breakdown of communication but this is with the caveat that you do seem to be more focused on the ad hominem more than would need be.

The point im making (which i believe is very similar to Adlestmage) is that the conclusion of the MHP which ends up at 33/66 is dependent on bad mathematical modeling of the real world. This is the false premise. If you don’t like the use of phrase ‘false premise’ we can use the words initial mistake, or preliminary bad idea.

The initial bad idea is that the contestants first choice (the one that is never represented in reality by the contestants physical action) should not be included in the accounting of events. If the world was modeled correctly, i.e. not modeled the way it is in the MHP, the derivation of the preference born by 33/66 would not be achieved.

If there is an assertion that the contestant’s first choice necessarily impacts the game as it restricts Monty choices of doors, I pose the question: What are we to make of the 1/3 of the time that the contestant picks the door with the car. This is an instance in the game where the contestants choice absolutely has no impact on Monty’s choice. So are we to believe that the game some how selectively works 1/3 of the time so as to not screw up our assertions that the choice is not a illusion the other 2/3 of game plays?

I think you have just accidentally proven my point to your self.

I think you have just inadvertently proven my point to your self

Yes, some comments do seem to appear out-of-sequence.

“Monty’s actions are the only actions that are of consequence. He will always reveal a got. the contestants choice does not matter” – this is false. Monty’s actions are determined by the contestant’s choice; he has to leave closed whichever door the contestant chose.

By the way, I owe you 5 apples.

i hope they are negative apples. they are delicious 🙂

If it is true that the contestants choice matters because it alters Monty’s choice then what are we to think of the cases where the contestant chooses the car first. by game definition this choice cannot impact on Monty’s choice as Monty has to reveal a goat and both goats are options.

“… what are we to think of the cases where the contestant chooses the car first”

The answer to this is in the MHP solution, which is that the contestant’s chance rises from 1/3 to 2/3 if they switch. It’s not 3/3 because of these cases. The contestant’s choice, every time, gives Monty a 2/3 chance of only having one goat to reveal. It’s an advantage, doubling the contestant’s chance of the car, but still not a dead cert.

The problem with that answer is the assertion that the contestants choice impacts Monty’s choice and thats why it has to be included in the post mortem accounting of events. But if the first choice does not affect Monty’s choice then there is no reason to include it in at least 1/3 of the game play.

In every game, the contestant’s choice gives Monty 2/3 chance of having to open a specific one of the two remaining doors and a 1/3 chance of having to choose one from those two doors to open. Without the contestant making a choice, Monty would have a 100% chance of having to choose one from those two doors. If that isn’t impacting Monty’s choice I don’t know what is.

The steady state, if you will, of the game is no action or choice and two goats available to Monty. only two thirds of the game’s play affect the options of the steady state. So by shear game play the contestants affect on Monty’s options are negligible 1/3 of the time. If the only reason we include the first decision in the postmortem is because the contestants choice affects Monty’s decision, we are explicitly including 1/3 of the game in which this assertion is not valid.

Mind you i assert that in reality there is no way to have a choice manifest in the game ( the time travel issue ) and as such there is no reason to include the first choice. To prove that it is indeed necessary to include this choice is because it affects Monty’s choice. the best that has been shown is that it should be included at least 2/3s of the time. The problem with this is that this 2/3 is the 2/3 which “prove” the conditions of the always switch mantra and the 1/3 that is negligible “prove” the contrary.

It would seem to me that the need to include the first choice is merely to prove an outcome. Which is exactly the problem. The only reason to include the first choice is to have a game in which 33/66 is the out come.

Correction: Without the contestant making a choice, Monty would have a 100% chance of having to choose one out of two particular doors from the three.

The circumstances for Monty are different when the contestant makes a choice compared to when they do not.

@Adam. You really haven’t got a clue have you?

“the conclusion of the MHP which ends up at 33/66 is dependent on bad mathematical modeling of the real world.”. No it isn’t. There is no false premise or initial bad idea – you’re frantically trying to make one up where none exists. Why don’t you try Ablestmage’s false “false premise”: that you only get to play the game once, he uses this ludricous notion to conveniently ignore the massive amount of statistical data which supports the MATHEMATICAL PROOF that switching doors is a 2/3 chance.

Can you really not see the difference between Monty ALWAYS having a choice of 1 from 2 doors, and 66.7% of the time having NO choice at all? Don’t you think this would affect the subsequent probabilities involved?

Why don’t you try the game yourself instead of making yourself look foolish. there are hundreds of online simulators.

from another reply:

The steady state, if you will, of the game is no action or choice and two goats available to Monty. only two thirds of the game’s play affect the options of the steady state. So by shear game play the contestants affect on Monty’s options are negligible 1/3 of the time. If the only reason we include the first decision in the postmortem is because the contestants choice affects Monty’s decision, we are explicitly including 1/3 of the game in which this assertion is not valid.

Mind you i assert that in reality there is no way to have a choice manifest in the game ( the time travel issue ) and as such there is no reason to include the first choice. To prove that it is indeed necessary to include this choice is because it affects Monty’s choice. the best that has been shown is that it should be included at least 2/3s of the time. The problem with this is that this 2/3 is the 2/3 which “prove” the conditions of the always switch mantra and the 1/3 that is negligible “prove” the contrary.

It would seem to me that the need to include the first choice is merely to prove an outcome. Which is exactly the problem. The only reason to include the first choice is to have a game in which 33/66 is the out come.

/

I have committed to stop using false premise as it seems to bother you. I’m not ablestmage so im not sure why you want me to discuss his propositions. I will say since you brought it up that on small distributions the trend, if you will, has not set in. you probably have to hit some where on the order of n=5 games before you get away from the noise floor when empirically proving this. Im not familiar with the game show but if they operate like regular game shows you probably wouldn’t make more than 1 or 2 appearances and subsequently not be out of the realm of statistical noise. so yea on the distribution level of 1 game i can see some merit but i agree switching might not be a bad idea due to the fact that i recognize that it is a game.

@Adam

Perhaps if I lay it out step-by-step you’ll be to identify precisely what the “false premise” is, or exactly where the problem is ‘badly modeled mathematically speaking’

1) Take 2 red marbles and 1 white marble

2) Put all 3 marbles in a bag

3) You pick out 1 marble, but you’re not allowed to look at it

4) The host looks in the bag and is obliged (by the rules of the game) to remove a red marble. If there are 2 red marbles in the bag he can remove either one.

5) The host shows you the red marble he removed, and then discards it.

6) There is now 1 marble left in the bag, and you have 1 marble in your hand

7) Which marble is more likely to be the white marble?

If, as you assert, it;s 50/50 between the 2 then you need to explain how you have a 50% chance of picking the white marble when there are 3 marbles in the bag.

Edit

“how you HAD a 50% chance of picking the white marble when there WERE 3 marbles in the bag”.

Ill admit the act of holding the marble does make it feel like you are really doing something, but you still have not committed to a choice. Monty always pulls a goat marble. its the insistence on including a choice which at least only impacts 2/3 of the game play that is the problem.

So, essentially you want to change the game so you can win. I think we’re done here gentleman. You either play by the rules or go make your own games somewhere else.

@Adam

said “but you still have not committed to a choice”, and then in the very next sentence says ” its the insistence on including a choice”

Contradict yourself much Adam?

How is it a problem? Where is the “false premise”? Where is the “bad mathematical modeling”. Come on explain yourself.

Does it help if instead of the contestant nominating a door, it happens like this:

The doors each have a tiny spyhole. The contestant goes up to one of the doors, takes out a key and locks it (without peeping). Monty immediately follows on to peep through the spyholes and open one of the doors, but he is not allowed a key. Does that make first-round choosing feel more like an action with an effect?

@ john,

I’m not changing any rules. I’m explaining why the rules only illustrate the game. and i agree we are running close to a epistemological/metaphysical issue and may quickly reach an impasse.

@ Marley

The contradiction I’m explaining is the contradiction inherent in the MHP. But I’m happy to see you have at least acknowledged its existence.

I have explained my self several time. This response is yet another example of the issue.

@ freddie – see my response the to the marble bag scenario.

@Adam

I don’t acknowledge any contradiction in the MHP, because there isn’t one. Where did you get that idea from, and what is this supposed contradiction in any case?

Why won’t you answer my questiions?

1) Which marble is more likely to be the white one, the one left in the bag or the one in your hand?

2) Will you play the card game?. According to your assertion you have a +ve expectancy of $10 per game. You should ‘bite my hand off’ to take those odds – but you won’t of course because you realise that in reality you have a -ve expectancy of $20 per game

Stop fudging the issue. Switching cards, marbles, or doors doubles your chance of winning “the game”. You’re obviously one of those arrogant “know-alls” who isn’t prepared under any circumstances to admit it to be wrong . You’re absolutely full of it.

(Oh, and btw, you still haven’t said what the “bad mathematical modeling” is. in the MHP. Please elucidate, we’re all dying to know)

I’m afraid you have not understood probability as a concept, Adam.

@ Marley – the time travel thing might be where the error in reality modeling comes in… just a hunch.

@ Freddie – I understand them better than you think. probabilities do not make truths, they are just an analysis of optional variables as it relates to a given situation or out come. You could include a question as to whether or not the contestant wore boxers or briefs in these probabilities and you would not violate any stipulations of probabilities. Math is a tool and you can use it in several different ways. You can use a hammer as a paperweight, but that does not mean that the paper will become nails.

They are not optional variables, they are possible outcomes. I’m afraid you have not understood probability as a concept, Adam.

What “time-travel thing”? What are you talking about? I don’t think you know what you’re talking about. The variables in the MHP aren’t “optional”. Perhaps you should learn some basic probability theory instead of posting nonsense.

And you still haven’t answered the question (why not?) : “Which marble is more likely to be the white one, the one left in the bag or the one in your hand?” Perhaps you don’t know the answer or can’t work it – do you want me to explain it to you?

Freddy – you have that confused with probability distribution.

“What “time-travel thing”? ” Seriously have you read anything iv posted? My first foot in the door here illustrated the temporal illusion of the uncommitted choice of the contestant. Every one fired back indicating that the non-manifest choice impacted Monty choice, but as i countered that is at best only a relevant event in 2/3 of the game play. so to include it in all game play is to falsely assert its relevance and it is explicitly un-manifest in 1/3 of the game play. No fully integrated defense for the inclusion of the first choice in the post mortem calculation has been proposed as of yet.

I still have not answered your loose marble question because i reject the premise of the question. Its the same reason you don’t have an answer for the following question: Why have you stopped slaughtering gerbils while wearing a tutu. … Perhaps you don’t know the answer or can’t work it – do you want me to explain it to you?

“Freddy – you have that confused with probability distribution.”

OK Adam – the possible outcomes within the probability distribution after the contestant’s first choice are as follows:

Monty opens one goat door and leaves a car unopened = 2/3, and

Monty opens one goat door and leaves the other goat unopened = 1/3.

The distribution if the contestant makes no choice are:

Monty opens one goat door and leaves a goat and a car unopened = 1/1.

I’m afraid you have not understood probability as a concept, Adam. One starting point might be the idea that ‘no action’ or ‘no change’ can be a possible outcome just as ‘action x’ can be.

@Adam

I’ll make it even simpler for you.

This is a card game for 2 people. 3 cards, the King Of Clubs,the 2 Diamonds,and the 2 Hearts. The winner of the game is the player who has the highest card

1) I deal 1 card face down to you, and 2 cards face down to me.

2) I look at both my cards and discard the lowest one (or pick a random card to discard if they’re both the same rank)

3) You look at your card

If you win I give you $100. If I win you give me $80. We’ll play the game until

a) you go bankrupt, OR

b) you realise that perhaps you’re not as smart as you think you are.

How about it, want to play?

Ill play that game as soon as you can tell me where the missing dollar went from the game in the opening post.

What missing dollar? There isn’t one, if you believe that you’re even less smart than I think you are….. and I don’t think you’re particularly smart at all.

Exactly. there is no missing dollar. This is the point that ablestmage made and you apparently agree with. So it can be established that even though there could be the appearance of math backing up the missing dollar, the dollar is not indeed missing. It’s observed absence is due to the situation created in the story or game. So i pose to you why is it so easy to see this in the math mathematical trickery of the dollar story yet so difficult to see it in the MHP?

@Adam

“Exactly. there is no missing dollar. This is the point that ablestmage made”. Ablestmage didn’t make any point, and never has (I’m beginning to think you’re actually him though). Change the game, the 3 men hand over a $100 note, the hotel guy returns $75 and they give him a $2 tip. Where’s the missing $71? “even though there could be the appearance of math backing up the missing dollar,” Only if you’re mathematically illiterate and can’t differentiate between addition and subtraction Totally absurd.

There is no “mathematocal trickery” in the MHP., and you’ve yet to explain how you arrive at that conclusion.

“I still have not answered your loose marble question because i reject the premise of the question.”. The question posed in the MHP is “is it advantageous to switch doors?”, and you reject the premise of the question, why? What’s the premise of the question? It’s a simple question, which both you and Ablestmage repeatedly refuse to answer, because you KNOW the answer is YES, it is advantageous to switch doors/marbles. You’re a joke. Expalin how, if you played 300 games, always picked Door1 and and always stayed with Door1, you’d win 150 games. Go on, give it your best shot.

So now I’ve answered your question how about playing the card game? Got plenty of money handy? Or do reject the “premise” of that game as well?

@Adam, and btw I haven’t stopped slaughtering gerbils while wearing a tutu. What gave you that idea?

@Adam, when did you stop posting on forums about the MHP where people demonstrated your contentions on probability were flawed?

@ Freddie, you can lead a horse to water but you can’t make them drink, especially if in principle they epistemologicaly deny the existence of water in favor of the theory of water that all the other horses are delusionally invoking.

@marley, you have not answered my question. I have also concluded that you have not the ability to recognize it in abstract. I won’t be able to continue the conversation as this is an crucial obstacle to cross.

Math models the real world. If the conditions of the model are not possible in the real world the consequences of the model have tangible meaning in the real world. Math does not dictate reality.

@Adam, there are a number of misconceptions in the arguments you advanced earlier. For example, that there is some kind of steady state when Monty has no car to avoid revealing (ie there are two goats left after the first choice). But this is not a steady state, it is the possible outcome, 33% likely, that the contestant loses by switching. Also that the first choice is not an action or event, even though Monty’s actions (and their outcomes) are dependent upon it. The sum in the probability space is 1, and possible outcomes may include those where there is no change or no effect. I’m afraid you have not understood probability as a concept Adam.

@ Freddie, yea, Im pretty much done with this. you are far too worried about your perception of my understanding. I understand thoroughly enough to pose what you think my misunderstanding is. Via the Bayesian approach you think I’m failing to understand that the incorporation of new knowledge (as it would be seen) is not a separable event. By separating these events Im not showing that it fails the second axiom (your sum of probability space), Im showing that it is out side of a reasonable model of reality. so in summation I know what Im talking about and you have missed my point. in leaving [in stead of repeating my self again and again] ill re-post the last part of my first foray into this time wasting hive of ad hominem.

“This of course is not to detract from the merit of the MHP exercise. It is a great slight of hand (or time as the case may be) and also helps to sharpen cognitive tools as well as improve your chance when playing games ( which do exist in reality and do have real consequences ) even if you have to make-believe to get there. So i wouldn’t loose sleep over the counter intuitive nature of the problem as your intuition is correct and it is the game that is a false presentation of reality.”

Not seperable Adam, conditional. The solution is in dependent probability.

@Adam

said “If the conditions of the model are not possible in the real world the consequences of the model have tangible meaning in the real world”. The MHP is easily modelled in the real world, just get 3 playing cards and follow the rules (and it doesn’t have any meaning – it’s just a math problem – other than to demonstrate a simple probability concept)

” I know what Im talking about and you have missed my point”. Regarding the MHP you certainly don’t know what you’re talking about, and you’ve never explained your point (though you have done a lot of hand-waving and obfuscating)

I’m pretty much done with you too, the whiff of smugness that permeates your every post is becoming pretty nauseous.

Hey Ablestmage – how’s the video coming along? I just thought up another screenplay to offer you – picked up a few ideas above.

Set up three doors and randomly hide a car and two goats, one behind each door. Bring on a contestant and get them to choose a door. Open their chosen door Ablestmage, open it and give them whatever’s inside. Now run it again with another contestant, then another and another (etc etc). On average you are giving away one car to every two goats aren’t you Ablestmage? 1/3 car, 2/3 goat. Next, stop one contestant before you open their door. Ask them to throw a three-sided dice at a rotating dartboard, name the capital of Kentucky and write ‘heads’ or ‘tails’ on the doors while dancing to the ‘People’s Court’ theme. Then open their door and give them their prize. Nothing changed behind that door – still 1/3 car 2/3 goat I’d say. Keep going, and stop another contestant and have a sly peep behind the other two doors – open one to reveal a goat. Then open the contestant’s door and give them their prize. Nothing changed behind that door – still 1/3 car 2/3 goat. Now stop yet another contestant, reveal a goat as before … and invite them to switch their door. Perhaps they turn the invitation down and you open their first door and give them their prize. Nothing changed behind that door – still 1/3 car 2/3 goat. Pause the music Ablestmage, this is the dramatic climax: Nothing … changed … behind … that … door. For the final scene, have some contestants who switch. But how to calculate their chances of winning? Maybe try subtracting 1/3 from 1. Run credits.

Here’s a game to play at home:

Try to pick the ace of spades from a deck of 52 cards. You make a starting guess, then have a friend remove 50 of the other 51 cards (they can only remove ones that AREN’T the ace of spades). Then you get the option to either keep your original choice or switch with the remaining one.

Have fun and be sure to post your results!

The author of this blog (and some of his acolytes like Adam) would argue that you never actually pick a card in the 1st place, it’s just an illusion. The single card you selected and placed face down on the table in front of you is a figment of your imagination, and the card is in fact still in the deck. 50/50 makes perfect sense if you think about lthat way. LOL

Hi ablestmage,

I don’t understand your solution.

But since you are sure that 50/50 is still king could you please point out my mistakes in understanding this problem?

1) There are 3 doors and 2 goats, so the chance that a contestant will pick a goat is 2/3.

2) When a contestant picks a goat, Monty will have to choose between a goat and a car.

3) Since Monty can only reveal a goat, the car will always be behind the door Monty doesn;t open.

Where am I wrong?

Asking Ablestmage to explain the MHP is like asking a young Earth creationist to explain evolutionary theory

“The doors could be arbitrarily named Heads, Tails, and Wildcard, whereas the Wildcard will take the name of the door which is eliminated: ”

The doors are not arbritarily named, since the names Heads and Tails reflect your bias that MHP is a 2 door rather than a 3 door problem..

Rather than polluting the problem with your bias, you should ask yourself: the following questions.

What happens when a contestant picks goat1?.

What happens when a contestant picks goat2?

What happens when a contestant picks the car?.

This what happens.

If the contestant picks goat1, Monty will remove goat2. Switching wins.

If the contestant picks goat2, Monty will remove goat1. Switching wins.

If the contestant picks the car, Monty will remove either goat1 or goat2. Switching loses.

Frankly, the problem is so trivial that it’s almost embarrassing.

You are wrong. This is what happens: If the contestant picks goat one, then Monte takes away goat 2. Switching wins. If the contestant picks goat 2, then Monte removes goat 1. Switching wins. If the contestant chooses the car and if Monte removes goat 1, switching looses. BUT if the contestant chooses the car and Monte removes goat 2 switching looses again. There are 4 possible and equally likely possibilities. Each has a 1/4 chance of occurring. Switching will win 1/2 of the time (1/4 +1/4). Not switching will win 1/2 of the time (1/4+1/4).

The probability that you pick Goat1 and Monty opens Goat2 = 1/3 * 1 = 1/3 . Switching wins

The probability that you pick Goat2 and Monty opens Goat1 = 1/3 * 1 = 1/3 . Switching wins

The probability that you pick the car and Monty opens Goat2 = 1/3 * 1/2 = 1/6 . Switching loses

The probability that you pick the car and Monty opens Goat1 = 1/3 * 1/2 = 1/6 . Switching loses

You should just give up before you embarrass yourself even more than you have done already.

Monty Hall could potentially run in one round with two doors and hit 2/3 vs 1/3.

I was stuck on round one being Bullshit (as are many other people).

the option to switch in round two could create a virtual third door, which happens to be an existing door.

Follow this: two doors, A or B? B. Okay, do you want to switch?

This essentially creates the opportunity to choose A twice. It’s that choice that matters… And nothing about round one. It’s just by coincidence that round one had 1/3 odds. This would still work with 100 doors down to one with the last round still having 2/3 vs 1/3 and resolve insane odds of the 1 billion doors example having two doors at the end and one with 1 billion to one odds as well as the problem with retaining odds based on three from round one, when one choice is removed (which is the logical block for most hold outs)

It only make odds based on what currently exists.

I need to code this.

“This essentially creates the opportunity to choose A twice.”

Offering a switch after a choice out of two doors essentially creates the opportunity to choose A twice and B twice. Each iteration offers the same chances and each round is independent. However in the MHP the first round offers the opportunity to choose A, B or C and the second round the opportunity to choose A or B having been shown C was a losing choice that you did not choose (door probabilities therefore being dependent on round one). Neither this nor 100 or a billion doors work in the same way as your two door scenario.

Freddie, did you see that Ablestmage eventually conceded on YouTube that over multiple trials your expected win rate by always switching “tends” towards 66.7%. Of course, being the mathematical ignoramus that he is, Ablestmage doesn’t realise the contradiction between that admission and his nonsensical assertion that any individual game however is a 50/50 chace. You’ve got to laugh.

You are wrong. No matter how may doors you exclude, when you have two doors left and one hides the car, the chance of choosing the door hiding the car is 50%.

“It’s just by coincidence that round one had 1/3 odds.”. No it isn’t, it’s because you start with 3 doors. If you start with 100 doors then after Monty opens 98 losing doors you get 1/100 and 99/100 for the respective probabilities of the 2 remaining doors. Similarly with 1 billion doors, you have a 1/1,000,000,000 chance if you stick with your original pick.

Some people have already coded it here: :http://rosettacode.org/wiki/Monty_Hall_problem

“… conceded on YouTube …”

Wow, I think I missed that. Some of those MHP uploads have 10,000s of comments. No sign of the new Ablestmage MHP video yet; I was wondering whether silence may be taken for assent.

So by this logic if there were 100 doors and you got down to 2 doors switching would increase your odds of success to 99/100. So dumb. Of course the last pick is always 50/50.

So dumb (but you’re in good company here with Ablestmage who doesn’t understand the problem either)

Take a deck of cards, pick 1 but don’t look at it. The chances you’ve picked the Ace of Clubs are 1/52 right? Give the remaining 51 cards to a friend and ask him to look through them and remove 50 cards that are NOT the Ace of Clubs. You both now have 1 card each – who’s more likely to hold the Ace of Clubs, you or your friend? (Ablestmage refuses to answer this question for some reason)

Holy crap… Ok I see how this trick works. You pick at random but Monty picks based on inside knowledge. That’s why it’s not 50/50. The odds that you will pick the goat are 2/3. If you get the goat there is a 100% chance that Monty will pick the car because he knows where it is. So because he has inside knowledge you always go with his pick. The more doors there are the more likely that Monty is right and you are wrong. If there were only 2 doors from the beginning it would be 50/50.

I make that a 38-minute turnaround (quicker than mine I have to admit). Ablestmage is on about nine years and counting …

Ok, ive been under a rock, and only just heard of this problem. My two cents. I’m still in the 50 percent camp at the moment. Here’s why (and I haven’t seen this particular argument exactly, so sorry if I’m beating a dead horse).

I think the switchmen are inadvertently leaving out 3 possibilities from all of the possible outcomes. Namely, the TWO possible doors that the host can reveal when you DID pick the right door originally. (If I pick door 1 and I’m right, he could reveal door 2 or 3, and the same applies for the other 2 initial picks, if right).

This may be hard to read in this format, but here goes. The numbers below represent:

actual location of car / initial pick / door revealed by host / win if switch / win if stay

1 1 2 0 1

1 1 3 0 1

1 2 3 1 0

1 3 2 1 0

2 1 3 1 0

2 2 1 0 1

2 2 3 0 1

2 3 1 1 0

3 1 2 1 0

3 2 1 1 0

3 3 1 1 0

3 3 2 1 0

As you can see, there are 12 possible scenarios, 6 stay, and 6 switch.

Interested to hear thoughts on this.

Interesting way of thinking. I don’t know how to refute it properly (using structured/logical argumentation), but I’ve a feeling the 2 options in the “right first time” aren’t relevant (there’s only ever one event there?) – the simulations seem to disprove what you’re suggesting too.

but it would be worth exploring restating it in different ways to see if any flaws become clear.

P.s.

If I’m understanding your notation, you’ve typod 2 of the “car is in #3”

3 1 2 1 0

3 2 1 1 0

3 3 1 0 1

3 3 2 0 1

Hi EricG. You are inadvertently assuming the frequency of all possible outcomes is equal. There are indeed “TWO possible doors that the host can reveal when you DID pick the right door originally”, however the frequency is as follows:

When you pick the car door first (1/3 of the time), the host will open one of the other doors to reveal a goat.

When you pick one of the goat doors first (1/3), the host will open one of the other doors to reveal a goat.

When you pick the other goat door first (1/3), the host will open one of the other doors to reveal a goat.

I guess that’s the crux of the whole argument, which information is relevant and which isn’t.

So I ran my own computer simulation of this problem (I’m stubborn), and I am 100% wrong. I can’t quite wrap my brain around it, but running samples of 10,000 at a time, I came out with almost perfect 66 / 33 splits every time (66% chance if you switch, 33% if you don’t)..

I did the same thing, thinking that it was surely 50/50. I began writing a simulation but I eventually realized my own error in reasoning as I was thinking my way through the algorithm. I didn’t even compile the program before I realized why switching is advantageous.

With all the videos out there explaining why switching is advantageous, I don’t think there’s much else to say other than if you still hold on to the 50/50 notion, you are stupid.

I gotta comment on this, cause I think the exchange between Adam and the others is honestly the most bizarre argument I’ve ever read on the internet.

First though, I’ll say what I think of AblestMage v. Marley52: it’s obvious that they’re thinking of two games with different rules. In AblestMage’s version of the game, Monty doesn’t know where the car is. He reveals a door at random, and if by chance he accidentally reveals the car he goes, “oops, that’s not how this is supposed to go” and they start over again, repeating until Monty gets a goat. In that version of the game, the remaining doors have a fifty-fifty chance of concealing the car.

In Marley52’s version of the game, Monty knows where the car is. If by chance your first choice of door is a goat (a two-thirds chance), then Monty actively avoids opening the door with the car, thereby revealing where it is (by opening the other one). If you got the car the first time (one chance in three), then Monty opens whichever of the other two doors. Two times out of three, you’ll win by switching, so in this game you should always switch (because you don’t know whether you picked the car in the first place or not).

Adam’s chain of comments is an entirely different beast. I’ve never seen any discussion quite this weird:

ADAM: I understand the game, but I’d just like to point out that it doesn’t apply to reality.

OTHER COMMENTERS: You obviously don’t understand the game, let me explain it to you.

ADAM: I already said I DO understand the game. I’m just pointing out that it doesn’t apply to reality.

I don’t know why Adam didn’t just explain what he was talking about. Maybe he gets a kick out of showing off his ability to use flowery language to obscure the meaning of obvious and rather unimportant points? Still, I think it’s fairly obvious that all he was really trying to say was that Monty’s behavior would make no sense if he wasn’t following arbitrary game rules. Since nobody seemed to get what Adam was talking about, I’ll go ahead and belabor the point just a little.

You’re choosing between three alternatives and Monty knows which one is correct. If he wants to help you, he’ll just tell you which one it is. If he wants to hinder you, he’ll lie to you about which one it is. Either way, he’ll try to persuade you to believe him. If he doesn’t care, he won’t tell you anything.

If you play the MHP according to Marley52’s rules, then Monty is knowingly helping you if you originally guessed the wrong one, and knowingly hindering you if you originally guessed the right one. Outside of a game, there’s no reason for Monty to do that.

If Monty doesn’t know which choice is correct, but he wants to help you anyway, he could investigate one of the choices. Assuming that neither you nor Monty initially has any idea or preference, he’ll pick one at random. If he happens to get it right, he’ll tell you and you’ll pick the same one. Otherwise, you’ve got a fifty-fifty chance like in AblestMage’s version of the game. Overall, his assistance improves your chance of getting the right answer from one in three to two in three. On the other hand, he could hinder you by investigating and then lying about his results. If you believe the lying Monty, you have zero chance of getting it right.

Bottomline, Monty’s behavior in the game would make zero sense in real life, but that point is so obvious and unimportant that I don’t know why Adam even bothered to point it out in the first place, and then his convoluted defense was just ludicrous. Like I said, I think this was the most freakishly weird discussion I’ve ever seen.

Interesting comments. I note that in Ablestmage’s examples ‘One’ through ‘Five’ above, he describes the removal of a goat door as if deliberate. There is no indication that a random door is ever opened with the attendant possibility of revealing the car, it’s always a goat door. I therefore believe Ablestmage does NOT mistake Monty’s reveal for random. I think Ablestmage actually confuses the independent variable with the dependent variable, hence confusing the chance of switching (the 50/50 error) with the chance of the car’s position (the MHP). His recent silence on the matter despite having promised a video suggests to me that he may have finally realised this error.

Marley52’s ‘version’ IS the MHP.

And Adam has not understood probability as a concept.

I’m putting the video project on indefinite hiatus because of the planning process involved. I’d rather do it in a way that isn’t rinkydink-looking, with good audio/etc and can’t find the right mix (I originally had in mind a style similar to the popular “Ten Dimensions Explained” style). I’ve experimented with several both similar and different styles and can’t figure out something I’m satisfied with in terms of presentation/delivery, and I think simple text accomplishes this far better.

Monty will always remove a goat door, and never a car door, and isn’t wholly random. Which of the two goat doors is revealed is irrelevant. While I still hold my original position, conversation here and on the numberphile video about MHP has prompted me to revise my ‘delivery’ of the original position to be described as “no advantage” rather than “50/50,” since 50-50 only seems to promote the red-herring thought process rather than rightly disregarding it as a red herring (and there is already discussion on numberphile’s video about whether 50/50 and no-advantage are genuinely different, which I maintain it is). The failure for switchmen to understand why they are different may be a kind of systemic obstacle of why they cannot properly see the no-advantage answer.

My further responses will be scattered, if any. One day in the future I may make another anniversary re-re-statement with the revised phrasing of the original statement, but the video version is on indefinite hiatus for stylistic reasons, not change of reasoning or position.

Yeah right, I believe you Ablestmage., “stylistic reasons”. The rest of us prefer substance over style when it comes to maths and logic.

“No adabantage” or “50/50”?. Well there’s definitely no advantage in a 50/50 proposition, and a 50/50 proposition presents no advantage for either choice.

It’s all beside the point though, The MHP isn’t a 50/50 proposition and there is an advantage to be gained by switching doors. I’m tempted to ask Ablestmage why he thinks the two concepts are different but he’d probably reply with “That’s not even a valid question!” (again).

One day in the future maybe you’ll see reason.

From Ablestmage: “Which of the two goat doors is revealed is irrelevant”.

What, even in the 2/3 of games when one of the goat doors is the contestant’s first choice?

There are three possible combinations:

CGG

GCG

GGC

If you choose door 1 and Monty picks a goat door (it doesn’t matter which door), then one of the possible combinations goes away. (If Monty opens door 3, then there is no way combination 3 can happen). It ‘s now a completely different game and becomes 50/50. All the switchers try to apply the beginning probabilities after Monty has opened a door, but the probabilities have changed.

You appear to have made the mistake of counting the number of possible combinations, not their likelihood. The probability of GGC is indeed now zero, the probability of CGG remains 1/3 (as at the start) and the probability of GCG is now 1 – 0 – 1/3 = 2/3.

Ok, I’ll explain in more detail. If the 3rd combination is now not available, and door 3 is now not available, then row three and column three go away. This changes the 3×3 matrix below:

CGG

GCG

GGC

into this 2×2 matrix:

CG

GC

therefore, 50/50. When Monty opens a door, it’s now a different game – it changes from 1 out of 3 situation into a 1 out of 2 situation.

I suggest you learn how conditional probability works before suggesting the “switchers” have got it wrong.

If you play 300 games , always pick door1 and always stay with door1 when offered the chance to swap doors, how many games will you win?

Conversely, in those same 300 games if you always switch, how many games will you win? According to you the answer to both is 150.

Good!

Thanks, but it’s important to learn how to apply conditional probability correctly. In conditional probability, the numerator and denominator reduce by the amount that has been eliminated each time a choice is revealed. If P(G) = 2/3, then P(C I G) = 1/2

The probability of a goat in the situation “GGC” is 2/3. The probability of a car, given a goat chosen first is “GC” or 1/2. You can’t apply a 3-part probability to a 2-part situation.

Check the website https://www.mathsisfun.com/data/probability-events-conditional.html. The example of red and blue marbles illustrates this clearly.

Let the probability of getting a car on the first choice be C1, that of a goat G1. Let the probability of getting a car by switching be C2, that of a goat G2. Thus:

P(C1) = 1/3, P(G1) = 2/3.

P(C2 | C1) = 0, P(C2 | G1) = 1.

P(G2 | C1) = 1, P(G2 | G1) = 0.

P(C2) = 1/3*0 + 2/3*1 = 2/3.

Thanks, I know perfectly well how to apply conditional probability – your comment clearly indicates you don’t. Saying ” In conditional probability, the numerator and denominator reduce by the amount that has been eliminated each time a choice is revealed” is complete nonsense.

THIS is how you calculate the conditional probability that the car is behind the door Monty didn’t open

Assume we pick Door1 and then Monty shows us a goat behind Door2.

Let A be the event that the car is behind Door3 and B be the event that Monty shows us a goat behind Door2. We want to find Pr(A∣B).

Bayes Theory tells us that: Pr(A∣B)=[Pr(B∣A)×Pr(A)]/Pr(B)

Pr(B∣A) = 1 (Probability Monty opens Door2 given the car is behind Door3)

Pr(A) = 1/3 (Probability car is behind Door3)

Pr(B) = (1/3×1/2)+(1/3×0)+(1/3×1)=1/2 (from the Law Of Total Probability)

Probability the car is behind Door3 given Monty opens Door2 = Pr(A∣B) = (1*1/3)/1/2 = 2/3

Your July 9th response really looks impressive but it is faulty as you have not accounted for the fact that no matter how many doors you have, they all have the same chance of hiding the car. The answer to your last question is: If you switch you will be right 50% of the time. If you don’t switch you will be right 50% of the time. It is superstitious to think anything that Monte or the contestant does will change the location of the car. The chance that any one door hides the car is, and always will be, 1/the # of doors. Nobody and no theorem, formula or equation when used correctly can change this fact. You are making this too complicated for yourself.

I take a deck of cards, deal 1 face down to you and keep the remaining 51.

Who is more likely to have the Ace of Clubs, you or me?

I sort through my 51 cards and deliberately discard 50 cards that are NOT the Ace of Clubs; We both now have 1 card

Who is more likely to have the Ace of Clubs, you or me?

Come on brains, impart you vast knowledge and tell us your answer.

Happy 94th birthday 8/25/15 Monty Hall

So I’ve done a lot of thinking. This game has a problem with definitions. Odds, chances, likelihood, and all that. Think of it like this:

If I have 2 doors (Door A and Door B) and I tell you that a car is behind Door B and nothing is behind Door A, then how likely are you to win a car?

Well, if you’re smart enough to listen, you’ll win 100% of the time. However, the fact remains that there are 2 doors and 1 car. So, in reality, you have a 50/50 chance of picking the right door. You COULD still pick the wrong door. It hasn’t been removed from your options. You could also leave it to fate and flip a coin. Just because you were told which one had the prize, doesn’t change the fact that there are 2 doors and 1 car.

The same goes for your deck of cards ideology. Yeah, the odds are against you on your first pick, but what if you made a list of every possible move the game could play out?. There’s 2601 possibilities – 51×50 (you pick a losing card, Monty can’t pick winning card) + 1×51 (You pick winning card, Monty can pick any remaining card). This number of choices is doubled by the final choice of stay or switch (5202), and wouldn’t you know, half of those choices end in victory, and half in defeat.

You pick a losing card first:

51×50 = 2550 possible initial moves

If you switch, there are 2550 winning moves.

If you stay, there are 2550 losing moves.

Total possible moves = 5100

You pick the winning card first:

1×51 = 51 possible initial moves

If you switch, there are 51 losing moves

If you stay, there are 51 winning moves.

Total possible moves = 102

Grand Total = 5202

2601 = win and 2601 = lose.

So even though it makes sense to switch, there are 50% winning moves, and 50% losing moves. It’s not about what’s smart, it’s about what’s possible.

I guess there’s no edit button, but I’d like to point out in my previous post that by switching you’ll win in 2550 scenarios, and lose in 51 scenarios. So clearly, you should always switch if you want to win. However, statistically, you’re still looking at a 50% win 50% loss ratio in total possible scenarios.

However, the question asked is “Is it advantageous to switch doors?”, which as you demonstrate it clearly is. The author of this blog thinks it makes no difference.

” However, statistically, you’re still looking at a 50% win 50% loss ratio …” Yes, but that’s their proportion of the number of possible outcomes, not their likelihood.

@Freddie

It is just as likely that you can switch as you can stay. 50/50. Just because you have prior knowledge doesn’t change your possible options. In a truly random test, a computer would pick to switch at the end 50% of the time and pick to stay 50%. This would create 50% wins and 50% losses. You, as a human, are purposely choosing to ONLY switch, which skews the results in your favor.

@Freddie

I reread what you said, and you should reword it to be more clear.

The likelihood of winning or losing the game is 50%.

The likelihood of winning the game by only choosing to switch is some percentage higher than 50% (I didnt calculate that, and it varies based on the number of choices).

Nov 13 “It is just as likely that you can switch as you can stay. 50/50.”

The likelihood that you ‘can switch’ is 100%, and the likelihood that you ‘can stay’ is 100% as both are possible. The likelihood that you will switch OR stay is 100%, there being no other proper options. The likelihood of winning by switching is 66.6…%. The likelihood that you WILL switch depends on your understanding of probability; for example in one report 8% switched and in another 56%.

“I reread what you said, and you should reword it to be more clear. The likelihood of winning or losing the game is 50%.”

It’s not my wording that isn’t clear. The likelihood of winning the game by switching is 66.7%. This gives the solution to “Is it to your advantage to switch your choice?” which is the Monty Hall Problem question. It may cause confusion that it is paraphrased incorrectly at the top of this page: “What is the probability of selecting the Car Door?” cannot be resolved without defining the parameters within which such a selection is made.

ablestmage, a very long way to convince yourself of something that is false. The rest of us weren’t so easily fooled.

Here is your answer dumbass. In the beginning it is a 1/3 chance. So now you picked a door and it has a 1/3 chance of being correct. This means that the other two doors COMbINED have a 2/3 chance. When he eliminates one of those goats you are combining the odds of those two unpicked doors. So that door just went from 33% to 66%. you have to pick it. The initial pick stays at 33%. Besides the bullshit you are know this. There is now way in hell you really believe the shit u are saying. You are just getting kicks out of acting stupid. A troll if you will.

@Diggs,

If the Monty Hall Problem teaches us one thing, it’s never to underestimate the stupidity of people.

?: your pick

1/3: chance of car

[]: combined chance of car

xxx: revealed

1/3(?) [1/3 1/3]=2/3

1/3(?) [xxx 1/3->2/3]=2/3

Don’t switch: 1/3(?) 2/3

Switch: 1/3 2/3(?)

Problem solved.

##

To be honest, this blogpost looks more like a lesson in sophism than an actual debate.

I can’t believe that the 50%ers still don’t get it. This probability model is very simple to understand. There are 3 doors, 1 has a car behind it, 2 each have a goat. The contestant guesses a door, that probability to win the car is 1/3. There is a 100% chance that at least one of the remaining doors has a goat behind it. Monty knows where the goats are and opens 1 of the 2 doors not chosen to reveal one. There is no new information given to the contestant about the door chosen, therefore that half of the equation remains unchanged, a 1/3 chance. The only change due to revealing the goat is that the 2/3 probability of a car being there has been narrowed down from 2 doors, to 1.

I would also like to know from a 50%er as to why with millions and millions of test results, that the switchmen are correct in doing so 67% of the time. Shouldn’t it be, according to you, 50% of the time? Or perhaps its just a coincidence and over time the numbers will somehow balance themselves out.

Monty..”Congratulations Mr. 50/50, you are the show’s finalist and I will give you the opportunity to win our grand prize of the day. See here, there are 3 doors, 2 of which are hiding a goat, and one door hiding a brand new car. Now all you have to do is guess which door you think it is. After your guess I will show you one of the doors you didn’t choose that hides a goat. And as a bonus you will have the chance to switch your guess to the other closed door at that time.”

Mr.50/50… “OK Monty, I choose door 2 and I don’t need you to open another door showing a goat, because I’m going to stay with my guess since I will have a 50/50 chance irregardless. After all if you opened door 3 and it shows a goat then the odds are 50/50 with my guess and door 1. If you show door 1 has a goat, then I will have a 50/50 chance with my guess and door 3. So it’s 50/50 either way, and that’s why I’m going to stay and you need not open any doors other than my guess.”

Monty….”Wow, who you have thought of that?!!!!. So door 2 then?”

Mr.50/50….”It’s okay Monty I’m staying with my original guess because I have a 50/50 chance no matter which door you would have opened to show a goat!!!!, Just open the door 2 please”.

Monty….”Oh Geez, sorry Mr. 50/50 I was going to show you door 3 hiding a goat, but you thought you had a 50/50 chance to win the car so you stayed with your choice of Door 2. Sorry you just weren’t lucky (stupid) because the car was behind Door 1.”

What helped me realize you should switch was playing the game backwards. Maybe not backwards, but I’ll explain. You get on stage and they ask you to pick 2 doors instead of 1. So from the very beginning you have a 66% chance of winning. Now they open one of your doors to reveal a goat behind it. Now they ask you if you want your one remaining door or the one you didn’t originally pick. You now DON’T switch because the door you didn’t pick only had a 33% chance of winning when you chose your doors. No one would lower their odds of winning. That would be foolish.

Itisnt5050, it is kind of satisfying to get it no matter what it takes to ‘get there’. Those that are in the 50/50 mindset view the second phase of the problem as a complete reset, a NEW contest consisting of 2 doors this time, and of equal probability since a goat has been shown. They need to realize that if the original guess is a 1/3 chance to win, then the chance of the OTHER 2 doors hiding BOTH goats has to be 1/3 as well. So if they were looking at the second phase as a new problem it should perceived like this by the contestant….There are 2 closed doors. Monty tells the contestant there is a 1/3 chance that the object behind each are the same. Then he opens a door to reveal an object reminding the contestant that there is a 1/3 chance the other door hides a similar object. Would the 50/50 player see the error of his ways at this point? Probably not because he wouldn’t have read more than two lines of this post!!

I repeat the offer I made on the other post.

I will pay for a first class ticket from anywhere in the country, and a room at a five-star hotel of your choice, if you will agree to the following:

Meet with me. We will each bring $100k. I will play Monty. You make your choice and stick with it every time. If you win, I’ll pay you $110. If you lose, you only have to pay me $100. We’ll play until one of us has all the money.

If you are correct, and the odds are 50/50, then this will be an easy $100k for you, plus a free vacation.

I still await your response.

I will also extend the same offer to anyone who agrees with the author’s position. Put your money where your mouth is.

Alternatively, I will happily fly to you instead. And if you’d rather play Monty, that’s fine, too – I promise to switch every time. I’ll pay the same odds.

I am 100% serious, and again, I extend this offer to anyone.

Ablestmage; having been a member of the switchmen group for several years I have always thought of you as a moron, retard, idiot, etc. etc. Well let me tell you, after the experience I had last night I have come closer to taking some of those thoughts back.

I was with a few other friends and playing a game similar to Trivial Pursuit. One of the questions asked of me was “What is the capitol of Kentucky”? As a Canadian my reply was “How the hell should I know?” Anyway when the answer was revealed to me everything you have said about the logic behind the Monty Hall problem hit me like a ton of dart boards, excuse me, bricks! I saw the light, and it was good. I know you must be delighted to read this.

I do have one teenie weenie little question though. I’m may be nit-picking and the question probably isn’t important anyways, but here goes. If the odds of the original guess went up from 33% to 50% is it because there is now a greater likelihood that it hides a car since less doors remain? I just need to know because when I explain the Monty Hall problem to others, they might ask some stupid question like “Why is it easier for the original guess to win when the amount of doors has been reduced, at the same time HARDER for the other door to win with the same reduction in doors?” Anyway I will just tell them that the odds for that door went up from 67% to 84%, throw in a safety tip like don’t drive in Louisiana and we should be good to go.

In closing, my sincerest thanks.

I love your sense of humor (really!) but, the chance of any one door hiding the car is always 1/the # of doors left after the elimination by Monte. Nothing that the contestant or Monte can do or say will change the location of the car.

Working through this problem and reading all of the “solutions” on both sides of the issue has been interesting and enlightening. When I first encountered the problem I was absolutely convinced the there was no advantage to switching and that the odds of me winning was 50-50 if i held to my first choice. As I read through the blogs I was impressed by several things. There always seemed to be a tendency to either complicate the problem rather that simplify it or to throw nasty barbs and cloud things. I found myself dismissing those who made nasty comments even though I now see that some of them had the right solution. I am now convinced that switching improves your chances to 2/3 rather than the 1/3 chance that you get if you hold. I will try and make my reasoning clear and simple but I am open to considering a new explanation to the contrary. If I know anything after this experience, it is that I am not infallible. Here is my analysis:

The Monty problem is asking us to identify the best strategy to maximize the chances of picking the door with the car behind it rather than the goat. There are three doors (A, B, C) – one with a car and two with goats. There is an equal chance of the car being behind any one of the doors – hence 1/3 chance of a win behind door A or B or C. Two strategies are being evaluated. The first says that I will pick a door and stay with it – for example A. The second says that I will pick a door for example B and after seeing one of the doors with a goat behind it, I will switch. The essence of the problem is that if I pick door B and agree to switch I am actually picking doors A and C ( much better odds than A alone ) and I will win if the car is behind either A or C. This gives me 2/3 chance to win. Simple probability 2/3 by picking door B and agreeing to switch – i am covering doors A and C with my bet. I am looking forward to your comments.

Charles Green.

Of course the other “strategy” is to flip a coin when there are 2 doors remaining which is where the 50/50 answer comes from.

Only Ablestmage in his “profound insanity” (his own words not mine) believes this is the optimum strategy.

I am surprised that no one supporting the 50-50 solution has replied. Has my analysis resolved the issue?

I’m pretty much the main defender of the no-advantage position, and I’m working on a YouTube version of the whole thing, so my energy is not really be spent here. Plus, the article itself is from 2013 and there isn’t a lot of buzz at the moment about the MHP in general so as to draw in commenters. Plus, it’s a bit daunting to insert oneself into a well-established, lengthy thread, even if there is agreement regarding no-advantage.

I quit responding due to the lovely and wonderful personalities that rather worry about insults as opposed to communicate meaningfully.

Your two sentences contradict each other. In the first sentence you imply that the chances are 1/2. The only way that can happen is if the car does move!

The probabilities of the two remaining doors are 1/3 and 2/3. Over the course of 100 games, the contestants would choose about 67 wrong doors. Those doors will STILL be wrong after a goat is revealed. You cannot win 50 games out of 100 with only 33 correct doors.

It is NOT 50/50. You cannot win 50 times out of 100 from each of the two remaining doors. About 33 out of 100 first guesses will be correct, 67 wrong. THOSE NUMBERS STAY THAT WAY WHETHER A GOAT IS SHOWN OR NOT!! So how can you win 50 times out of 100 with 33 correct doors??? DID 17 manage to spawn from the 67 WRONG doors? Give you head a shake, please!!!!

Dr. Joseph Girschick…..

There are two methods towards winning the car. 1. pick the right door from three and staying 2. pick EITHER of the two wrong doors and switching. For someone whom calls himself a doctor you are not too smart.

Well, I swore back and blue that the switching theory was very obviously wrong. But reading all the arguments to b the cotrary made me look into it.

30 minute of walking and thinking, and I have had to overturn my thinking.

They are quite correct. The trick is, look at it from the perspective of the the second door opener. He is constrained. He must pick a door other than yours, and which does NOT have the prize.

One time out of three, he has a random choice of two doors, as you have picked the prize.

Two times out of three, he only has one door he can open, because you have picked wrongly, and the prize is behind one of his apparent options.

Therefore, if you always pick the door he has left closed, two times out of three, you get the prize. One time out of three, you have wrongly switched from your prize pick.

So, no doubt whatsoever, you have a 66.6% chance of winning if you switch.

I’m still amazed, but convinced.

markx

Congratulations for making the effort to logically think it through and come up with the correct answer. Now if you can convince Ablestmage of the same …… well that would be truly remarkable

It is impossible to change a person’s mind if that person hasn’t had a single educated mathematics learning experience.

You role a dice, you get a 6, or you don’t, so you have a 50% chance to get a 6! If the Monty Hall problem is 50/50, then that previous sentence is true.

In fact, Ablestmage himself said something so utterly ridiculous, that I might suicide because of how stupid someone can get. He said that if you had 100 doors and 99 goats, and 1 car behind 100 doors, open 1, and reveal 98 goats once at a time, that switching still gives you a 50% chance! My desk broke from the air pressure of my facepalm.

Ablestmage doesn’t realize that the word “switching” means you have something to base your choice on, which is an original choice. And to get that you must go through the elimination round where you choose and a goat is shown. Hold on, that must mean that the elimination round effects the final choice! By giving you the original choice that you yourself chose, and that choice is most likely a goat, 66.666…% chance to be a goat, then it is probably a goat!

Now the host always reveals a goat, never eliminating a car, so if you got a goat as the original choice, there is 2 goats, the one you selected (most likely, 66.666…% chance) and the one that has been showing, that means the one which you can switch to is probably (66.666…% chance) a car.

I coded something which automatically makes 1000 random scenarios by selecting a car door, and an originally selected door, now we all know that if your originally selected door ISN’T a car door, then switching makes you win. Just scroll to the bottom for results: http://ideone.com/ygPH5n

I agree that you have reached the correct solution. However, your explanation and my earlier explanation is lengthier than it needs to be. Since my earlier explanation, I have tried to simplify the issue. A careful analysis of the problem reveals that there is nothing the host does that changes the probabilities. The die is cast the moment the contestant makes a choice.

The 50-50 Man picks a door and stays with it. His chance to win a car is 1 out of 3.

The Switch Man picks a door knowing that he will switch. He wins if the car is under under either of the 2 doors that he did not initially pick. His chances of a win are 2 out of 3.

I hope that this simple explanation convinces some of the 50-50 advocates to change their minds as I once had to. Admitting you have been wrong, as I did, is not always easy. Hoping to hear from some of you.

Hey Ablestmage!!!! Answer this…….

You are a student and given a test with 100 questions. Each question is a multiple choice with three possible answers to which you circle either A, B, or C. The passing mark is 50 correct out of 100. After the teacher examines your test and says you have only 33 of the 100 right. But the teacher is nice to you, and although he never marked any of the 100 answers as to whether or not it was right or wrong, he eliminates one wrong answer, not including any of your original answers, from each of the 100 questions and offers you to retake the test. So now you have the same test with 100 questions but instead of three options to choose from, you have only two.

So the question is, why don’t you answer each question of the rewrite test with the same ones you did the first time? Do you think your mark can go up from 33 to 50? Or do you simply choose the OTHER answer to each question knowing you will get 67 right??

Hey Ablestmage!!!! Was my previous post irrelevant? Did I forget to include a missing dollar? Well try this one out…….

Let’s play against each other 100 times. We both choose the same doors, and we both stay with our guesses. The difference between our games is that my rounds end as soon as we choose a door from three available, without you seeing what’s behind it of course. Your rounds end after a goat is revealed. So explain to me how you would win 50 times while I win 33 when we have the same doors.

Below I’ll post a simulation in ruby that proofs switching gives you a significant edge. I’ll post the same simulation again, changing 1 line that randomly selects whether to switch or to stay (your fifty fifty man hypothesis).

The first simulation gives you roughly a 66% edge for always switching.

The second simulation (where you randomly choose to stay with your choice or switch) is a neat 50%.

## Simulation 1 – Stolen from somewhere else

def createDoors

doors = [false, false, false]

doors[rand(3)] = true

doors

end

def openADoorThatHasADonkey(doors, userChoice)

([0, 1, 2] – [userChoice, doors.index(true)]).first

end

def changeUserChoice(choices)

([0, 1, 2] – choices).first

end

def play

doors = createDoors()

userChoice1 = rand(3)

hostChoice = openADoorThatHasADonkey(doors, userChoice1)

userChoice2 = changeUserChoice([userChoice1, hostChoice])

doors.index(true) == userChoice2

end

games, wins = 100000, 0

games.times { wins += 1 if play() }

print 100*wins/games, ” %”

## Simulation 2 – The fifty fifty man, slightly modified

def createDoors

doors = [false, false, false]

doors[rand(3)] = true

doors

end

def openADoorThatHasADonkey(doors, userChoice)

([0, 1, 2] – [userChoice, doors.index(true)]).first

end

def changeUserChoice(choices)

([0, 1, 2] – choices).first

end

def play

doors = createDoors()

userChoice1 = rand(3)

hostChoice = openADoorThatHasADonkey(doors, userChoice1)

userChoice2 = rand(2) == 1 ? changeUserChoice([userChoice1, hostChoice]) : 0

doors.index(true) == userChoice2

end

games, wins = 100000, 0

games.times { wins += 1 if play() }

print 100*wins/games, ” %”

Conclusion: Both hypotheses are true.

(It is still smarter to switch. If you always stay you’ll win 1/3rd of the time and lose 2/3rd of the time. Randomly switching merely balances out 1/3rd vs 2/3rd)

So, I wrote a program on this for school and initially assumed it would be 50/50. When i wrote out the steps logically, it kept returning 33/66. I asked my professor what i was doing wrong. He told me that my problem was that I was counting the elimination round and once the other door was removed It was only 50/50. This did not sit well with me. So I started reading about it. Mostly I found people claiming the 33/66 view. I went to a part and posed the question to my brothers and explained why I thought 33/66 was the truth. They listened and then told me repeatedly that I was wrong. They failed to produce convincing arguments for their point of view that I had not already heard. I was torn. Today I was on a mission to prove that 33/66 was wrong as I do respect the opinions of my older brothers. I found your article and thought, “finally someone who believes differently”. I read through some comments and decided screw it, time to test it. I grabbed a deck of cards, took two red and one black, shuffled them, and chose the card 1 every time. I flipped the other two over and looked at them, removed a red card and picked the other every time. After the first time it seemed so overwhelmingly obvious that I cannot believe anyone would disagree after giving it a try, but I continued on. at ten flips switch had won exactly 6 times. I carried on and by the time I reached 50 flips switching had won 36 times. I was going to do 100 but it did not seem necessary at a 72% win rate. I am firmly in the switchmen camp now, and I cannot see any reason why anyone would not be.

Here is a way to visualize why you have a 2/3 probability of winning by switching. The possible ways to order the doors are as follows:

Scenario 1: Door 1- Goat, Door 2- Prize, Door 3- Goat

Scenarios 2: Door 1- Prize, Door 2- Goat, Door 3- Goat

Scenario 3: Door 1- Goat, Door 2- Goat, Door 3- Prize

Now suppose you pick Door 1 in each scenario.

In the first scenario, Monty will always open Door 3. Even though you don’t know it, switching to Door 2 in this scenario will result in a 100% win rate.

In the second scenario, Monty will either open Door 2 or Door 3 (it doesn’t matter which one he opens). Even though you don’t know it, switching to the other Door (the one of Door 2 or 3 that he did not open), will result in a 0% win rate.

In the third scenario, Monty will always open Door 2. Even though don’t know it, switching to Door 3 will result in a 100% win rate.

3 separate scenarios with 100% win rates for 2 of them by switching, and 0% win rate in the other by switching, results in a 2/3 probability of winning by switching when you don’t know how the Doors are ordered (i.e., which scenario you are playing).

Marley52, I admire your perseverance in the face of Ablestmages’s continued circular reasoning (i.e. anything before the final choice doesn’t affect the outcome because anything before the final choice is irrelevant).

I’m not sure anything will change his/her mind as this point, but we can of course keep trying. Assuming this isn’t just one big troll session of course. But even if it is, it at least helps others see the right answer.

Fiftyfiftymen seem to be treating this the same as if someone came into this choice at the last step (i.e. without knowing what Monty did), in which case they do indeed have a 50% chance of guessing the right door. Swapping doors will make no difference. So then, how can the chance be 50% for one person and 66% for another person, when they are looking at exactly the same problem? One of my friends raised this exact objection.

Well of course the difference is that the two people have varying degrees of knowledge. Consider a sports final between two teams – the Bears and the Badgers. If you were asked to pick a winner, but had no knowledge of the sport or the teams, you’d have to choose randomly, and you’d have a 50% chance of getting the right answer. However someone who has been watching the whole season will have a lot more information. They’ll could estimate from their knowledge of previous games, player lineups, injuries etc. that the Bears have a 67% chance of winning. So assuming their analysis is pretty good, they will choose the Bears and have a 67% chance of being right. This is obviously not the same thing as the Monty Hall Problem, but it illustrates the the point that two people can have different probabilities of guessing the right answer to the same problem, based on having differing amounts of information about what happened in the past.

In the Monty Hall Problem, the player has information about how we arrived at the 2-door problem. He knows that if he originally picked a goat, the other door WILL have the car. And since there’s a 2/3 chance that he originally picked the goat, there’s a 2/3 chance that the other door has a car. So if he wants the car, he should always switch.

Too many comments to read here, so I apologize if I repeat an idea already expressed.

Anyway, here is my take on the problem.

The odds of winning were NEVER 1/3.

Since the game is set up from the beginning to leave the player with the choice between two doors, the one originally picked and one other, the probability of winning was always 50/50.

The choice of door X, where X is 1,2, or 3, was not locked in. The player’s first choice has no bearing whatever on the final result. Likewise, the revealing of door z, where z is any door but door X, has no bearing on the eventual outcome.

The choice of X by the player, and the elimination of Z, are only the SETUP of the actual game, which doesn’t really begin until this point. Since the player will certainly be given the option to select door X or door Y, where Y is the door neither chosen nor eliminated in the setup, the choice is really meaningless.

Now, when only door X and door Y remain, is the actual start of the game. There is one chance in two that the prize is behind either door X or door Y. The player is given at this point the freedom to choose either door. It’s 50/50 either way.

Any notion that the probability was EVER 1/3 for any door is held only by those who ignore the fact that the game is predetermined to be a choice between two options and always ends with one of two options being the winning choice.

Troy

Clearly you don’t understand basic probability. If you pick from 3 doors you have a 1/3 chance of picking the car. If you ALWAYS stay with your 1st pick you are NEVER going to win more than 1 game in 3. It is impossible to win 1 game in 2 by staying.

Troy,

Although your solution to the problem is wrong, it is helpful to the discussion because it clearly and concisely identifies an error in thinking behind the 50/50 answer.

I’ll restate the problem as simply as possible. What is the probability of winning a car for each of two different strategies?

Strategy 1 is to select a door (“original choice”) from three doors, each of which has an equal probability of having a car behind it, and to stay with that selection “lock it in” regardless of anything that happens in the game from that point on. In the long run a person using this strategy will win one time out of three – therefore the odds of winning are 1/3, contrary to your statement that the “odds of winning are NEVER 1/3”. Since you either win or lose, the odds of losing using this strategy are 2/3.

Strategy 2 is to select a door (“original choice”) with the absolute requirement of switching away from that door “locking it out.” The odds that a car is not behind that door are 2/3. A person using Strategy 2 loses only if the car is behind the “original choice” (odds of 1/3) and therefore wins if the car is not behind the “original choice” (odds of 2/3).

Troy’s explanation clearly identifies an error in thinking when he states “The choice of door X, where X is 1, 2, or 3, was not locked in”. This statement illustrates a fundamental misunderstanding of the problem. The essence of the problem is “What is the probability of winning if a player chooses a door and agrees to stay with the “original choice” (in essence “locks it in”) versus the probability of winning if a player chooses a door and agrees to switch (in essence “locks it out”).

A person using Strategy 1 wins if the car is behind the door of his “original choice” (probability 1/3). A person using Strategy 2 wins if the car is behind one of the other two doors – not his “original choice.” That person has covered two doors to win (probability 2/3).

An error in thinking by the proponents of the 50/50 solution arises from the failure to recognize that the problem involves “locking in” or “locking out” the “original choice.” “Locking in” (Strategy 1) covers one door and “locking out” (Strategy 2) covers two doors – hence a winning probability of 1/3 versus a winning probability of 2/3. Troy has contributed to the discussion by clearly identifying this error in thinking by the 50/50 proponents.

i dident read this entire conversation because its giving me a headache from laughing , are there not exactly 24 possible outcomes where exactly half result in a victory through switching? ive drawn out the chart numerous times in my life.

just draw a chart, im using a nintendo to write this message so i have no means of uploading mine, just make 3 collums , door 1 , 2 & 3, then bellow them draw out every possible outcome entailing the positions of car (A), goat (B), goat (C), the door you initially selected, & finaly the goat initially revealed. there are exactly 24 possible configurations, and exactly 50/50 win lose ratio , its real life, alternativelly you can decide not to diferentiate between the 2 goats by labelling them (B) & (C), the gameshow hoast must still however pick 1 of 2 distinct doors in the circumstances where you initially select the car, this leaves you with exactly 12 configurations and a 50/50 chance regardless, you might think your right but id like to see you visually prove this wrong

I don’t know where you’re getting 24 possible configurations from. I suspect you’re picking the door with car twice as often as you should be (i.e. in 50% of cases)

I’m done. There is a more intelligent conversation on the following web site. http://marilynvossavant.com/game-show-problem/

I don’t know where 24 combinations comes from when there are exactly 3. The following table allows you to simulate the game by running through both strategies. It clearly shows that switching wins 2/3 of the time, while staying wins 1/3 of the time

Door 1 Door 2 Door 3 Result

Game 1 Auto Goat Goat Switch and you lose

Game 2 Goat Auto Goat Switch and you win

Game 3 Goat Goat Auto Switch and you win

Game 4 Auto Goat Goat Stay and you win

Game 5 Goat Auto Goat Stay and you lose

Game 6 Goat Goat Auto Stay and you lose

Run through the games first choosing door 1, then choosing door 2, and finally choosing door 3. You will quickly realize that switching wins twice as often as staying..

there are 3 doors and you dont always pick door number 1, so picking door number 1 (& not yet to your knowledge ,it contains the car) then he reveals goat (A) behind door 2, & gives you the option to switch to door 3. so now imagine you started by picking door 3, and that it contains the car instead, he shows you goat (A) behind door 2 & asks if you want to switch to door one, these two scenarios, while putting you in the same position are none the less two equally probable distinctly diferent possible configurations, if you keep this in mind while making a chart you will find that there are 24 diferent configurations with 50/50 odds if you diferentiate the identities of the individual goats with goat (A) & goat (B), but if you deny their individual identities and only account for the remaining factors being (initial location of car & goats, initial selection, initial reveal) then there are 12 possible equally probable configurations still with 50/50 odds on the switch. one last example would be that the car is behind door 2, you blindly initially select door 2, wether or not you diferentiate the goats identities, there is a distinct diference between him revealing door 1 & door 3, regardless of the fact that it puts you in a seemingly to you identical position, its none the less in reality 2 diferent configurations, equally probable, peoples mistake i believe is in eliminating these configurations on the false assumption that they are identical, but in this real life situation they are not and u cant ignore this truth

There are 9 possible scenarios: the car is behind Door A, B, or C, and for each one the contestant can pick Door A, B or C.

3 x 3 =9, and in 6 of them switching wins.

your oversimplifying it, your ignoring factors that you have to take into account, what i think is happening is that this is infact what makes it paradoxical, were using 2 diferent logically correct methods of obtaining the answer, making no mistake in our logic, yet ariving at 2 diferent answers, (otherwise why would they call it a paradox)

for the people who bring up the arguement about a scenario with 100 doors, while it is true that if you all at once remove 98 doors it would be a safer bet to switch, for your arguement to actually remain equivalent to the original you would need to remove 1 door at a time and give the contestant the option to switch each and every time, removing 1 more door and offering an aditional switch regardless of wether or not he switched before until you arive in the end at the same puzzle of 2 remaining doors and wether or not to switch in the end will infact still be 50/50

further thought on the 100 door scenario is that you would be better off to switch to an alternate door in the end if throughout the game no matter how many times you switched the hoast chose never to reveal your initial door, this would leave you to rightly still believe it only has the initial 1% chance of being correct

il upload my chart to facebook l8er today once i get access to a computer, please come & check it out in a few hours, youl find me under Caleb Drroog Hudson

k its up, top post on my wall, Caleb Drroog Hudson

let me know what you think and if you see a flaw, im open to the possibility that im wrong, it would honestly blow my mind if you could show me a mistake in my reasoning, i use no equations its all visual, im confident that it shows every single distinctly diferent configuration

I know you’re wrong.There’s 9 possible scenarios, in 6 of them switching wins.

why am i lookig at 24 distinctly diferent scenarios then lmao, there are no duplicates and they are all valid look on my facebook page its there

i sent a pm to numberphile’s facebook page and asked them to have a look, if he shows me that im wrong ill admit it but i really think my proof is irrefutable

Door

State 1 2 3

X A G G

Y G A G

Z G G A

There are 3 possible States for the Auto (A) and Goat (G)

Contestant who stays with his original choice:

If he chooses Door 1, he wins State X and loses State Y and Z

If he chooses Door 2, he wins State Y and loses State X and Z

If he chooses Door 3, he wins State Z and loses State X and Y

Clearly, he wins 1 time out of 3 and loses 2 times out of 3.

Contestant who switches to the door not revealed by the host:

If he chooses Door 1, he loses State X and wins State Y and Z

If he chooses Door 2, he loses State Y and wins State X and Z

If he chooses Door 3, he loses State Z and wins State X and Y

Clearly, he wins 2 time out of 3 and loses 1 time out of 3.

your skipping configurations that affect the real life odds for the convenience of your formula , look at my chart on facebook youl find me under Caleb Drroog Hudson, its my most recent post, i have visually drawn the 24 possible configurations that each leave u with the choice to switch or not, half in whitch you win, none of them are wrong, none are excluded or can be denied, look at it and leave me a comment on my post describing what you see wrong with it please thats all i ask

He’s not skipping any configurations. There’s only 9 possible (3 for the car x 3 for the pick). You’re counting too many

i believe that it is 2/3 now, but i still count 24 distinctly diferent outcomes, only that the entire half of configurations that end in a loss are only each half as likelly as those that end in a win, because in initally selecting the car is the only casse in which the host has 2 options, both of witch cause a lose if you switch, i was giving each of these outcomes equal probability in, but u infact still only have a 33% chance of initially slecting the car, so then from there you can simply deduce that while it being the only scenario inwhich the host has 2 options , they are each only half as probable so they count as a half lose but then adding all up in the end makes it 2/3, i was wrong about that but im glad i figured it out my own way, i still stand by there being 24 possible games, they just arent equaly probable

Caleb, the contestant chooses a door that can hide either a car, Goat A, or Goat B. From there you only need three ‘staying’ or three ‘switching’ results to confirm it’s benefit.

Staying….

1. Choose car, host shows either goat, stay, win.

2. Choose Goat A, host shows Goat B, stay, lose.

3. Choose Goat B, host shows Goat A, stay, lose.

Staying wins one out of three times. Simple.

Troy:

The odds of winning were indeed 1/3 at the beginning. Imagine if you randomly pick a door. The odds of winning are 1 in 3. Now pretend that you leave the room at that point and say “I don’t care, just tell me what happens”. You then won’t hear the switch or stay option and won’t get a chance to switch. So, you will win the car only if you selected it initially, for which you have a 1 in 3 chance. This is actually the same as if you’ve heard the switch or stay option and decided to stay with your original choice, since in both cases you decided to stay with your original door.

Now imagine there is another contestant on the show (Bob), who will get the car if it’s in either of the doors that you don’t choose. Bob has a 2 in 3 chance of getting the car. Wouldn’t you rather be Bob? Yes, of course.

Well, if you switch, you are basically becoming like Bob! You are only allowed to pick one door, but the host has already shown you that one of Bob’s doors has the goat. So if you choose to switch, you are essentially becoming Bob and getting his 2/3 chance. Why? Because there was a 2/3 chance that the car is behind one of Bob’s doors. Revealing one of the doors doesn’t change the probability. There is still a 2/3 chance that the car is behind one of Bob’s doors, though in this case there is only one of Bob’s doors left. Bob’s remaining door has a 2/3 chance of winning the car.

Any arguments that “the chance is 50/50 because the first round doesn’t count” are faulty because they are assuming that the first round doesn’t count without actually saying why. Just because you have two choices, it doesn’t mean they have a 50/50 chance of being correct. The elimination of one door is affected by the choice that you made in the first round, so the first round does indeed count.

I always thought it was a 50/50 chance until figuring it out for myself today.

Do not think of the initial choice as 33% chance to win, think of it as 66% chance to lose.

When one door is eliminated, the probability that you chose right the first time is still 33%

As you were 66% likely to choose wrong, switching is the way to go. Probably.

In the end, it is a 50/50 choice, but you had a 66% probability to get it wrong on your first guess.

Well said – clearly and concisely!!

Don’t worry, Ablestmage will still assert that the first round doesn’t count, so it will still be 50/50.

“In the end, it is a 50/50 choice, but you had a 66% probability to get it wrong on your first guess.”

Bram, are you sure you don’t want to do an edit? 50/50?

Since the contestant chooses from three doors that have two cars, then 2/3 of the time the host reveals a goat from a goat/car combination of doors. Therefore 2/3 of the games the host will leave a car door for the contestant to switch to.

Now I have it wrong!!! the three doors have two goats, NOT two cars, lol!!

I vesaid t before and I say it again, the”goat Problem”is a pompös Hoax, that shouldn’t actually being considered being discussed. It amazing though, how many grown up people, not just Kids, talk about this filth. Monty Hall is a fraud, the goat Problem is just a funny and interesting Name, but it’s empty, easy to Trash (when using common sense) and irrelevant.

” the goat Problem is just a funny and interesting Name, but it’s empty, easy to Trash (when using common sense) and irrelevant.”

The problem is easy to trash by those who’s common sense is trash. If someone cannot accept the fact that choosing a door hiding a goat initially occurs 2/3 of the time, then to claim they have an understanding of the problem is in itself fraudulent.

You cannot win half of the time with the original guess if that guess is wrong two-thirds of the time to begin with. One’s measure of common sense would have to be extremely low in order to disagree with that.

It seems to me that the problem between 50/50 and 33/66 is one of determining whether or not round 2 is contingent and conditional on the choice in round 1. It seems pretty clear from OP that it is, in fact, NOT conditional. If round 2 is not conditional on round 1, then round 1 is irrelevant. I totally understand the logic and math behind the switchmen, but I am inclined to disagree. I don’t believe round 1 to be relevant at all.

You have 3 doors. Behind 2 doors are a goat and behind 1 is a car. NO MATTER WHAT DOOR YOU PICK, the host will open one door and reveal a goat. This changes nothing of the information that you have. It is in effect the same as walking into round 2 blind. Why? Because regardless of what door we choose, the outcome is the same. A door is opened and a goat is revealed. Since the host cannot reveal our door, how is this a choice at all? We glean no information from round 1. All we know is there are two doors in round 2, one with a goat and one with a car. Not choosing a door would effect precisely the same inter round outcome as choosing an arbitrary door. We enter round 2 with 2 doors, 1 with a goat and 1 with a car. We have no idea which one. The fact that the host revealed a goat does nothing to change our minds.

Since round 2 is in no way conditional on your choice in round 1 (this has been logically proven multiple times) then why are we bringing in the false 1/3 probability in round 2 at all? There aren’t two separate games. There is only one game—-round 2—-with an irrelevant starting round that changes nothing. It gives an illusion of an edge.

It’s quite easy to prove yourself wrong John, just get 3 playing cards and try it out. After a few turns you’ll realise your mistake.

Im not questioning the math or that the scenario as described gives a certain mathematical result. That much is certain. Obviously if you ask a question such as this one with round 1 implied as part of a conditional probability equation, the math will come out how the math comes out. I’m questioning the question itself. The question entails a choice in round 1 that is not actually made.

” imagine that Monty has no knowledge of what’s behind the doors”

Why? – that’s a different problem.

The qestion askked is “Is it advantgeous to swtch doors?”. The answer is YES

What I mean is this: imagine that Monty has no knowledge of what’s behind the doors, picks your 1st door for you, and is then shown the results, and then continues on with the problem as before. This leads directly to the precise situation we have in round 2 in the original question, except that you haven’t chosen the first door, you only choose the second. Are you claiming that now the ‘switch or stay’ part gives you a 33.3/66.6 % chance of winning?

It is precisely the same situation. Monty has no knowledge at the beginning of round 1, same as you.

Then Monty looks behind all 3 doors and opens one (not the one he picked for you) with a goat behind it. You enter at the beginning of round 2 and have to ‘switch or stay.’ Precisely the same situation, except you as a subject did not pick the initial choice.

The point is, and the point that OP was making, is that it’s a trick question. The original choice does not matter who makes it, does not matter if its you, or you are in a soundproof booth when it happens and have no knowledge. The empirical conditions are precisely the same in any example you can come up with, and if that is the case, then the original choice is illusory and has no actual bearing on the question, and so attributing the conditional probability of 1/3 is empirically false, despite the math coming out the way it does.

As I said, get 3 playing cards (2 red and 1 black) and try it out. Pick 1, look at the other 2 and remove a red card. keep count of how many games you’d win if you were to switch, and how many you’d win if you were to stay. Get back to me when you’ve completed a reasonable number of trials (say 100)

I doubt you understand anything about conditional probability.

I don’t think you understand what I am getting at, but that’s fine. Thanks for your replies.

You seem to be saying it makes no difference whether you stay or switch – which of course is incorrect.

I am amazed. Just when you think its all over, someone changes the problem and then tries to explain why the correct solution to the original problem doesn’t fit the new problem that they have invented.

I’m sorry you can’t see that the choice in round 1 is arbitrary, but it is. It doesn’t matter if you make it, Monty makes it, some other person makes it before you even get there, or nobody makes it. Monty ALWAYS removes one door with exactly one goat behind it, no matter what you pick. This means the decision is arbitrary, and the ONLY solution to the equation is 33/66, but it’s priorly arranged for you. If you give Monty a choice, or you allow him to open your door sometimes, or look at some of the possibilities then it’s interesting. But always opening a door with a goat behind it isn’t a decision.

I don’t know what you mean when you say that the choice in round 1 is arbitrary. The only decision being evaluated is the decision of whether to stay with your original pick after Monty reveals which of the other two doors has the goat behind it or whether to switch to the third door. The correct decision is to switch to the third door which maximizes your chances of winning a car. Switching gives you a 2/3 chance of winning while staying with your original pick gives you a 1/3 chance.

When you make your original pick there is a 1/3 chance that the car is behind that door and there is a 2/3 chance that the car is behind one of the other two doors. If you stay with your original pick you will win 1 time out of three – nothing that happens after that makes any difference. If you decide to switch you will win 2 times out of three because you are covering two doors and Monty is telling you which one of those two doors wins by revealing the goat behind one of the two.

I don’t know why so many people have trouble with this. We can simplify the problem by re-stating it this way. There are three doors. There is a car behind one of the doors and a goat behind each of the other two doors. Do you want to pick one door to win the car or do you want to pick two doors and win if the car is behind either of the two doors. Clearly 33/66 not 50/50.

No one has trouble understanding the scenario. My issue is with the ‘choice’ that takes place in round 1. There isn’t one. Round one is simply a card removal scenario, there is no choice at all being made until round 2. The odds in round 2 are INFLATED by the card removal scenario in round 1. The actual odds of picking between 2 doors in a vacuum is 50/50. The odds here are inflated to 33/66 because there is a card removal scenario in round 1.

My other issue is that the question is not clearly enough defined. Monty has many options in an open state problem. But here the problem is closed state, meaning Monty acts like a robot every time. The question would be more interesting if Monty was free to act. It is a gambling question after all. If Monty does the same thing every time it’s a dumb question.

I’m not sure based on your comments which side you are taking on this problem. Let’s make this a multiple choice test.

Given the problem are you:

1. More likely to win if you stay with your original pick.

2. More likely to win if you switch from your original pick to the third door after Monty reveals the door with the goat.

3 It doesn’t matter because they are equally likely 50/50.

Is your answer 1,2,or 3?

John, there is a clear difference between the MHP which requires the host to know where the car is and must reveal a goat as opposed to a similar problem with the exception he doesn’t know where the car is a reveals a goat randomly. These are two different problems altogether with two different outcomes in terms of probability of winning by switching.

First the MHP. To ignore the relevancy of the initial guess is to ignore the frequency of an event. Two out of three times the contestant will pick a goat. So two out of three times the host will have a goat/car combination of doors from which he leaves the car door available to switch to. Contestant wins by switching, two out of three times. You can determine the probability of winning by switching easily by just knowing the remaining probability divided by the doors available to switch to. Take total probability which is 1 (one car) minus original probability of 1/3 (contestant’s guess) and the remaining 2/3 is divided by the one door available to switch to. So 1-1/3=(2/3)/1=2/3.

If the host does not know where the car is and reveals a goat randomly, then the switching/staying chances are the same. The probability of winning the car by switching is determined by the chances of the contestant and the host both choosing goats. So 2/3×1/2=1/3. The probability of winning by staying is determined by the chances of the contestant choosing the car and the host a goat. So 1/3×1=1/3, same as in switching. The other 1/3 of the games are lost without a chance to continue when the host reveals a car.

In summary the contestant wins in the MHP when he guesses right from three doors available and stays, which happens one-third of the time, or if he chose either goat door and switches, which happens two-thirds of the time.

“It seems to me that the problem between 50/50 and 33/66 is one of determining whether or not round 2 is contingent and conditional on the choice in round 1. It seems pretty clear from OP that it is, in fact, NOT conditional. If round 2 is not conditional on round 1, then round 1 is irrelevant. I totally understand the logic and math behind the switchmen, but I am inclined to disagree. I don’t believe round 1 to be relevant at all.”

I said it before and I will say it again…”To ignore the relevancy of the initial guess is to ignore the frequency of an event.” TWO out of THREE times the host MUST remove the goat from a goat/car combination of doors which allows the contestant to win by switching TWO out of THREE times because the car is left behind the door he can switch to.

Memorize this and move on…..

The contestant picks door 1.

Scenario A: Door 1 contains a car. The host randomly reveals goat 1. Switching yields goat 2.

Scenario B: Door 1 contains a car. The host randomly reveals goat 2. Switching yields goat 1.

Scenario C: Door 1 contains goat 1. The host deliberately reveals goat 2. Switching yields a car.

Scenario D: Door 1 contains goat 2. The host deliberately reveals goat 1. Switching yields a car.

Boom. 50% chance of getting a car if you switch.

Thrilled Oaf, when the contestant selects the car the host can show either goat. Choosing the car still occurs 1/3 of the time.

1. Pick car, host shows Goat A, switch lose. (1/6 of the time)

2. Pick car, host shows Goat B, switch lose. (1/6 of the time)

3. Pick Goat A, host shows Goat B, switch win. (1/3 of the time)

4. Pick Goat B, host shows Goat A, switch win. (1/3 of the time)

Boom! I fixed it for you. Switching wins 2/3 of the time.

You showed four scenarios, half of which lead to a win. Ergo 50% win by switching. I don’t know why this is so hard for you to grasp, it’s basic math. Slapping random fractions on the end of the scenarios doesn’t change their probability of occurring, Klaus. Perhaps you should return to grade school to study arithmetic again?

WRONG!!

Thrilled Oaf, the contestant’s chances of selecting the car is 1/3; same as for Goat A, and Goat B. When the car is selected the host can show either goat, it doesn’t matter which one, nor does it matter if he divides the choice between Goats A and B equally. It doesn’t change the fact that the car is selected 1/3 of the time. So switching wins 2/3 of the time.

If the host does NOT know where the car is and happens to reveal a goat by way of randomness, THEN it’s a 50/50 chance between the two doors. The chances to win by switching in that case is calculated by the probabilities of both the contestant and the host picking goats…so 2/3×1/2=1/3. The chances to win by staying is calculated by the probabilities of the contestant picking the car, and the host a goat…so 1/3×1=1/3, same as in switching. The other 1/3 of the games are lost early due to the chances of the host revealing the car 1/3 of the time as well.

In the MHP the probability of the host revealing a goat is 1, not 1/2 as in a random game, therefore the chances to win by switching is calculated, again, by the contestant and the host both picking goats…so 2/3×1=2/3.

I know my math.

See, Charles agrees with me.

Wrong again! Klaus is absolutely correct. There is no doubt about it.

Thank you Charles for your comment. A lot of 50/50’ers think they are clairvoyant enough to be able to select the car half of the time when there are twice as many goats. I like reading their (rare) responses when I ask of them “So why didn’t you just pick a goat instead and then switch?”

For some reason it is easier to understand the correct solution to the problem than it is to explain it to someone who doesn’t.

Since there is one car and two goats, a person is going to pick a car 1/3 of the time and a goat 2/3 of the time. Clearly, the person who stays with his pick will win 1/3 of the time (if he picked a car) and will lose 2/3 of the time (if he picked a goat). The person who switches will lose 1/3 of the time (if he picked a car) and will win 2/3 of the time (if he picked a goat).

I have noticed that the conversation generally stops when the 50/50’er realizes he is wrong.

There wouldn’t be a difference if you picked a goat and switched, since there’s only one goat to pick, as the other will always be disqualified.

What???

“There wouldn’t be a difference if you picked a goat and switched, since there’s only one goat to pick, as the other will always be disqualified.”

Ablestmage would be proud of your logic. You are clueless as to the difference the knowledge of the host makes with the probabilities. Let me spell it out for you, not that you will accept it anyways…..

It’s 50/50 if the host does NOT know where the car is and reveals randomly a goat. In a sample size of 99 games….

1. Contestant will select car 33 times.

2. Host will select car 33 times.Games end immediately.

3. Neither do, the remaining door has the car 33 times.

It’s 50/50 in this case because only 66 games CAN continue, 33 cars behind the contestants doors and 33 behind the ones he can switch to.

In the MHP ALL 99 games can continue as the #2 variable above does not apply. Contestants will have 33 cars behind their doors and 66 will be behind the doors he can switch to.

I know you will try to discredit such proof yet again with your trusted “but with two doors left it’s 50/50” argument.

Oh and btw, here’s a classic common theme of your 50/50ers. After the contestant chooses a door your next step in your argument is “after the host reveals a goat, blah, blah, blah” and you’re smack in the middle of the two door debate. Have you even bothered considering what happens BETWEEN the time of the contestant choosing a door and the host’s reveal? Probably not because it requires more thought. Well I will explain….Since there are two goats and one car, 2/3 of the time the contestant will pick select a goat. That leaves a car/goat combination of doors from which the host reveals a goat. That leaves the car available to switch to 2/3 of the time.

By now you must be just itching to write back and tell me it’s 50/50 because there are two doors left. But think about this for a minute….if we played 100 games with the SAME doors and if you always stayed with our guess after the reveal and my scores were counted BEFORE the reveal do you really think you will win 50 times while I win 33 times WITH THE SAME DOORS?????????

Sorry if I wasted so much time of yours. Looking forward to hearing from you yet again about the the ‘two doors’ thing…..

I wish people would read. I never questioned the math being correct.

The fact of the matter is that the puzzle is nothing but a trick question. It’s the same thing as those 2(3/2) + 6 x 2 questions

If everyone here would read then we would all agree that switching wins 2/3 of the time; instead we have statements such as this one…..

“No one has trouble understanding the scenario. My issue is with the ‘choice’ that takes place in round 1. There isn’t one. Round one is simply a card removal scenario, there is no choice at all being made until round 2.”

It’s true. As written the odds are 2/3 for switching. However, picking the car correctly or incorrectly has no bearing on the fact that Monty removes exactly one goat prior to your switch, and so all the people saying you have a choice or this is a game are deluded.

The Monty Hall Problem gets its name from the TV game show, Let’s Make A Deal, hosted by Monty Hall in which contestants play the GAME. Some win and some lose. By every definition, it has all of the elements of a game; and you absolutely have a CHOICE. The choice is whether to stay with your original pick or to switch to the door that Monty does not reveal. At that time you don’t know if you picked a goat or a car. If you CHOOSE to stay with your original pick, you will win the game 1/3 of the time. If you CHOOSE to switch to the door that Monty does not reveal, you will win the game 2/3 of the time. No CHOICE?? Real contestants have played the game and either won a car or lost. Ask them if they had a choice. For more information watch The Monty Hall Problem – Numberphile on YouTube.

Your right, the whole argument is based on a logical fallacy people make when collecting the data here’s the mathematical explanation if you want it. (This is based on the host always revealing a door that’s incorrect)

The chances of you guessing correctly are both (2/3) and (1/2)

Explanation: You start out with (1 chance of being right/3 total doors), regardless of which door you picked when one door gets revealed you can either boost your chances of the remaining doors being right by 1/3 for both of the remaining doors or decrease your chances of being wrong by reducing the total amount of chances to 2, aka one for each of the remaining doors.

If your asking for the likely hood of “your” choice being right it’s (1/2)

(3)x – (1)x = (2)x this is the same as (2)X =(1) this means X = (1/2)

If your asking for the likely hood of “a” choice being it’s X = (2/3)

X = (1/3) +(1/3) = (2/3) this is the same as X = (2/3) this means X = (2/3)

The chances of you guessing wrong are both (1/3) and (1/2)

Explanation: You start out with (2 chances of being wrong/3 total doors), regardless of which door you picked when one door get’s revealed you can either reduce your chances of being wrong by 1/3 for both of the remaining doors or increase your chance of being right by reducing the total amount of chances to 2, aka one for each of the remaining doors.

If your asking for the likely hood of “your” choice being wrong it’s (1/2)

(3)x – (1)x = (2)x this is the same as (2)x = (1) this means X = (1/2)

If your asking for the likely hood “a” choice being wrong it’s (1/3)

X = (2/3) – (1/3) = (1/3) this is the same as X = (1/3) this means X = (1/3)

The fallacy is that people assert irrelevant data (the likely hood of a door being right prior to the reveal) as the reason why the probability changes to a specific extent when a door gets revealed instead of the fact that “a door just got revealed”. In the end the probability of all the remaining doors being right is 2/3 and the probability of them being wrong is 1/3, but it’s up to the participant to choose one of the two equally likely doors which means the participant’s odd’s 1/2.

@Sole

said “In the end the probability of all the remaining doors being right is 2/3”

Obviously totally incorrect as the total probability must = 1

Please look at my later explanations it’s not incorrect you just misunderstood what I wrote, each door has a 1/3 chance of being right plus a fallacious/debunked 1/3 chance of being right from the revealed door regardless of whether you switch or not is what causes the 2/3 chance. Regardless though, if you persist in the fallacy you end up with a choice of one out of two doors that both have the same 1/3 chance of being correct making it 50:50, and if you don’t commit to the fallacy you get 1/2 vs 1/2 which again is 50:50.

Hey Sole, what a coincidence. Your original post is word for word of that of a troll on YouTube!!!

Another way of explaining it is to show people what their own work means. Example:

(1/3 chance from the one that’s given to you + 1/3 chance from your choice = 2/3 chance of you getting a prize)

(1/3 chance from the one that’s not chosen = 1/3 chance of you not getting a prize)

Edit: This means you have a 2/3 chance of getting your prize if you choose “a” door it doesn’t matter which(or whether you stay or not) and a 1/3 chance of not getting a prize. Both doors add a 1/3 chance so it doesn’t matter which one you pick you’ll always have a 2/3 chance of getting your prize, but only a 1/2 chance of choosing the door that has the prize.

@Sole

said “each door has a 1/3 chance of being right plus a fallacious/debunked 1/3 chance of being right from the revealed door”

If there are 2 doors remaining they can’t BOTH have a 2/3 chance of hiding the car – it’s mathematically impossible.

Your argument is, if anything, even more nonsensical than that of Ablestmage himself

The whole point is that you start with x/3 + x/3 + x/3 = 1 and when a door gets revealed this is what your left with (x-x)/3 + x/3 + x/3 = 1 the fallacy that people like to commit is pretending that the X = 2 and 1, 2 for whatever is not picked because it’s one of the 2 that you didn’t pick and 1 one that you picked since it’s the 1 that you picked. You can’t apply math selectively with such an arbitrary(irrelevant) reason (1.5 – 1.5)/3 + (1.5)/3 + (1.5)/3 =1 aka: (1/2) + (1/2) = 1 Your chances are 1/2 for each door, anyway if you don’t get it at this point your probably not going to, bye.

In case they try to pull mental gymnastics here’s what you can show people

This is what your doing: (1/3 chance from the door that’s revealed”this part has been debunked and therefore is no longer part of the equation” + 1/3 chance from door doesn’t matter whether you switched or not = 2/3 chance of getting the prize”fallacious because 1/3 of this chance is a false the true value is 1/3″)

(1/3 chance from the door that’s not picked regardless of whether you switched or not = 1/3 chance of not getting the prize)

1/3 vs 1/3 in the end you have only two choices both of which have an equal chance therefore it’s 50:50

This is reality: (0 chance from the door that’s been revealed”this chance was debunked” + 1/2 from the choice that’s made regardless of whether you switch or not = 1/2 chance of choosing the prize)

(1/2 chance from the one that your not picking regardless of whether you change or not = 1/2 of not getting the prize)

1/2 vs 1/2 in the end you have only two choices both of which have an equal chance therefore it’s 50:50

“This is reality: (0 chance from the door that’s been revealed”this chance was debunked” + 1/2 from the choice that’s made regardless of whether you switch or not = 1/2 chance of choosing the prize)”

No, this is the reality. See your doctor, get referred to a neurosurgeon, and have an EEG of your brain. It will confirm what we already know of you; there is no electrical activity whatsoever ‘upstairs’, none, nada, zilch, flat-lined!!!

Your posts are a copy and paste from those you mounted on YouTube. Get a life with a new hobby until you get your lobotomy which btw swapping for you would be a HUGE advantage!!!

I don’t know why this has to be made so complicated. Let’s take it in small steps.

Is the probability that the car is behind the first door that is chosen 1/3? Yes or No?

Is the probability that the car is behind one of the other two doors 2/3. Yes or No?

Assuming that your answer to each of the above questions is Yes and that you really want to win the car, I ask you to pick any one door and tell me which door it is that you picked (A, B, or C).

Next, I give you a new choice. Do you want to stay with that door and win the car if it is behind that door or would you like to switch and win the car if it is behind either of the other two doors?

Only a fool would choose to stay with the first pick thinking the chance of winning is one out of 2 (1/2) when switching obviously increases the chance of winning to 2 out of 3 (2/3).

Mr. Green, Sole will not answer that. He was asked nine times this question on YouTube…..Which one door wins by staying, which two win by switching. His excuse for never answering it was that he already debunked the MHP numerous times!!! He’s just trolling.

Could it be that Sole has realized that he was wrong and joined the others in SILENCE?

No, lol.

Charles have you ever heard of a gentleman named Thomas Finkle? He’s been a 50/50 advocate for years and always on a mission with various mathematical proofs. Anyways, here is a response he gave to someone who wrote to him after doing an experiment confirming a 2/3 swapping advantage.

—————————————————————————————

Thomas Finkle

Quality Engineer at Marvin Engineering

HI Matus,

You have performed a simulation as it is most commonly done.

However, there are two decisions (i.e. factors) that the contestant makes. To get the most accurate picture of the simulation, we need to incorporate the percentages associated with the first decision into the simulation.

Then we can calculate the conditional probabilities: What are the percentages somebody wins if they selected the correct door then choose to stay/switch? What are the percentages somebody wins if they selected the wrong door then choose to stay/switch?

I ran a simulation of 100 trials with the following results: Choose the correct door on the first guess: 31 out of 100 Choose the wrong door on the first guess: 69 out of 100 Out of the 31 times the contestant choose the correct door, the contestant stayed 16 times and won.

This gives a percentage of 51.6% of winning given the contestant selected the correct door and stayed with the original choice.

Out of the 69 times the contestant choose the wrong door, the contestant switched 35 times and won.

This gives a percentage of 49.3% of winning given the contestant selected the wrong door and switching to the other door.

I use simple Excel formulas for this simulation. I would be happy to share this Excel based simulation with you if you are interested.

Of course, this simulation uses random numbers and the results will change when you open the file.

Regards,

Tom

—————————————————————————————

……..I should have warned you to go to the bathroom first before reading this, sorry.

You are right. You should have warned me.

The Monty Hall Problem is an exercise in deception, and ultimately, BULLYING.

Whenever someone attempts to offer their solution, they eventually become the victim of name-calling.

First of all, I am in the 50/50 group.

The 50/50 group falls under attack 66% of the time!

(I’m joking.)

But let’s take another look at the problem:

Imagine 3 contestants are invited on stage.

Each player stands in front of a door.

Monty reveals a goat, and that contestant is sent home with nothing.

The two players remaining have the option of staying with their first door, or switching places with each other.

Which player gains the advantage? Why?

The 33/66 group’s reasoning can’t apply anymore.

Each remaining contestant has a 50/50 chance of winning. Switching is irrelevant.

Explaining this, using MHP math, is impossible.

That’s because it is not a math problem. It is a word problem.

This is why the 33/66 group and the 50/50 group have such a hard time communicating, and the insults begin to fly:

“I have a degree in math, so I am right, and you are stupid.”

(Logical fallacy #1: Argument from authority.)

“You can’t explain why the math is wrong, so you are wrong.”

(Logical fallacy #2: Argument from complexity.)

“The answer is given by Monty Hall, and Monty Hall is always right.”

(Logical fallacy #3: Circular argument.)

“You sound like a Republican.”

(Logical fallacy #4: Playing the Trump Card. -er- I mean – Sidetracking.)

What’s your point Roy? The problem you describe iSN’T the Monty hall Problem. There’s no such thing as ‘MHP math’,there’s just math, and math proves it is better to switch doors. It’s also easy to verify experimentally, but I guess you haven’t tried 8that

“Explaining this, using MHP math, is impossible.That’s because it is not a math problem. It is a word problem.”

Roy, it IS a math problem. The chances to win by switching is calculated by the probabilities of both the contestant and the host selecting goats. Therefore 2/3×1=2/3.

If the host does not know where the car is, and the door he opens happens to reveal a goat when it could have been the car, then it’s 50/50. The chances to win by switching is calculated in a similar fashion as the MHP but now it is 2/3×1/2=1/3, same as in staying.

To understand the difference between the two problems better, consider this sample size of 99 games, first at RANDOM….

1. Contestant picks car 33 times

2. Host picks car 33 times

3. Neither do, 33 cars will be behind doors that can be switched to.

Because of line 2, only 66 games can continue and there are 33 cars behind the contestants doors, 33 behind the doors they can switch to. So it’s 50/50.

In the MHP line 2 does not apply and all 99 games can continue and there are 33 cars behind the contestants doors, 66 behind the doors they can switch to.

ablestmage, I have read this post and replies, and your Nov 2007 post. Just one question for you: Does Monty Hall act randomly or not in revealing a goat?

Yes. In the event the contestant should pick the car initially, Monty is left with two goats. He needs to eliminate one. Can you give any proof (in this situation) that Monty’s choice isn’t random?

Agreed, in one of three possibilities. The other two possibilities where the contestant unknowingly chooses a goat, does Monty act randomly?

Brad, all I pointed out was that the whole “Monty Hall acts randomly” argument does not apply in all cases. Therefore, that argument is debunked.

…or should I say, I debunked the “Monty Hall does NOT act randomly” argument?

Not debunked. If he does not act randomly in any case in the initial pick by the contestant, then more information is available. If more information is available, the best way to determine the probability is to enumerate the cases. Case #1: Choose car, switch, lose. Case #2: Choose one goat, switch, win. Case #3: Choose other goat, switch, win. Win in 2/3 cases. OR Case #1: Choose car, STAY, win. Case #2: Choose one goat, STAY, lose. Case #3: Choose other goat, STAY, lose. Win in 1/3 cases. Therefore your probability of winning is better if you switch than if you stay.

Brad, the question is “Once the contestant picks a door, does Monty Hall then pick his door at random?” If the contestant picked a goat, and the remaining doors are a goat and a car, then Monty Hall does NOT pick randomly, and I would agree with the argument. But as I have demonstrated, if the contestant picked the car initially, then Monty has two goats remaining. Whichever goat he chooses is completely random. So that specific argument (Monty does not pick at random) is debunked.

It does not matter if you “debunk” an argument (that I did not make), I was just asking questions…. Now I have enumerated the cases, which clearly demonstrates that your chances of winning are better if you switch. Is that not the original debate?

Brad. I did not say you were debunked, I said the “Not Random” argument was debunked, so don’t take my comments personally. As far as the original argument, “There is a 2/3 chance of winning if you switch”, is wrong. The reason is that one of the three choices is removed. If the puzzle had not removed any of the available prizes, and the contestant was allowed to change his answer from “I think I am picking the car” to “I think I am NOT picking the car”, then the odds of winning would increase from 1/3 to 2/3. But that’s not what happens in the Monty Hall Game. The parameters were changed from three doors to just two. So your odds changed from 1 in 3 to just 1 in 2. 50/50. It’s amazing to see how this basic fact slips by so many people.

In your scenario, where the contestant is allowed to change his answer, how do the odds increase? There are still three doors, he has reconsidered, the odds remain 1/3. Just as if we were flipping a coin, and I change my mind, it does not increase the odds. Please show me what is wrong in how I enumerated the cases? If I enumerate the cases for flipping a coin: Case#1: Choose tails, flip is tails, win. Case #2 Choose tails, flip is heads, lose. Odds are 50/50 if you choose tails. OR Case#1: Choose heads, flip is tails, lose. Case #2 Choose heads, flip is heads, win. Odds are 50/50 if you choose heads. I enumerated the Monty Hall example same way, how is it flawed?

Brad, if Monty does not remove a door, and allows you to pick again, then no matter what choice you make, your odds of winning remains at 1 in 3. But when Monty removes a door, then whatever choice you make, your odds of winning becomes 1 in 2. (The third option would be to ask for the goat that Monty just revealed. But we know that that choice has been taken out of the game.) Therefore you have two choices, and two choices only. 1=Keep your door. 2=Take the other door. There is no 3rd option.

Brad, when you wrote “In your scenario, where the contestant is allowed to change his answer, how do the odds increase? There are still three doors, he has reconsidered, the odds remain 1/3.” I misread your question. I was offering an example of the contestant changing his decision from “Car” to “Not a car”. The odds are 2 to 1 that any of the three doors is not a car at the onset of the game. Your statement involves simply giving up the door he picked (which has a 1/3 chance of being the car) to another door (which also has a 1/3 chance of being a car). Once again, when the number of doors to choose from goes from 3 to 2, then the odds of winning for each door increases from 1 in 3 to 1 in 2. 50/50.

Roy, so you cannot point out the flaw in enumerating the possibilities then? Did we start with three doors or just two? If we started with three doors (which is my understanding), then your first choice includes three options. You skip straight to two doors which is not what happened in the game. The contestant was not given a choice between two doors from the start. They were given three doors and then one was removed.

Brad. We did start with three doors. Behind one of the three doors is a car. The chances of picking a door with the car behind it is 1 in 3. But if you pick a door and claim you think it is NOT the car, then your odds of being correct is 2 in 3, because two of the doors do not have a car behind them. **But you are not given that option in the Monty Hall Game.** You are only allowed to guess which door has the car. When you start with three choices, your chances of being correct is 1 in 3. When you are given only two doors to choose from, your chances of being correct is 1 in 2.

Roy, are you familiar with enumerating possibilities in the study of statistics? You must start with the original given scenario. If we cannot start there, then I am done debating.

Brad, what are you talking about? I enumerated the statistics several times for you. Maybe you’re confusing the number of prizes with the number of approaches the contestant can make. As I explained, if the contestant is only allowed to select from the stage of doors, then his odds are defined by the number of doors. (Initially 3, then reduced to 2). However, in my attempt to sympathize with those who believe in a scenario in which his odds would convert from 1/3 to 2/3, then the argument must focus on his approach (Yes, this door is a car = 1/3 chance of being correct) / (No, this door is NOT a car = 2/3 chance of being correct).

Here is how a statistician would enumerate the possibilities for flipping a coin. Case #1: choose tails, flip is tails, win. Case #2: choose tails, flip is heads, lose. Would you agree?

Brad, what about case #3? What if the result is neither Heads nor Tails? You may believe that the opposite of Heads is Tails. But that is a false dichotomy. The opposite of Heads is actually “Not Heads”. For example, I flip a coin, and you call “Heads”, but the coin doesn’t lie flat. It bounces off of the table, or it comes to rest leaning against something. Clearly it is not Tails, but more importantly,it is not Heads, so you made the wrong call. [Are you beginning to see how a different approach is not the same as a different physical option?] The Monty Hall Problem focuses on the physical number of doors. Not the dichotomy of “car/NOT a car”. The physical number of doors is reduced from 3 to 2. So, statistically, your odds are 1 in 2.

Roy, let’s talk about pure stats, and have a computer flip the coin. Do you agree or not?

Brad, a coin and a computer are two different things. Maybe you don’t like knowing what a true dichotomy is, so you need to forego the coin to a computer-generate choice? Okay then. I will choose “Not Heads” and unplug the computer or catch the coin and put it in my pocket before it lands.

Roy Shafer said “So, statistically, your odds are 1 in 2.”. No they’re not. Statistically when you switch you’ll win 2 games in 3, and statistically when you stay you’ll win 1 game in 3. If you make a random choice you’ll win 1 game in 2, but the problem asks ‘Is it advantageous to switch doors?’, and the answer is demonstrably (and mathematically) YES.

No brad. Statistically, there are only two games. One with three doors, and one with two. (Monty Hall removed the third door for the second game.)

It’s not Brad, it’s Marley52. Statistically there are as many games as you want to play to define your population size. In any 1 single game the PROBABILITY of winning by switching doors is 2/3, and the PROBABILITY of winning by staying is 1/3. There are plenty of mathematical proofs out there – why don’t you look one up (Google ‘Monty Hall and Bayes Rule’) and educate yourself.

Marley52 you only get to choose (bet on) one door. You do not get to pick two doors together. You forego your first choice when you switch. I’ve offered a scenario where your odds would be 2/3, and that’s if you were to say “NOT a car”. If three doors remained in play, then you would have the advantage. But here’s one more scenario where you would have a 2/3 advantage. And that is if you were allowed to pick two of the three doors. Go to Las Vegas some time and play Roulette. You might learn something about odds.”I’m going to take my chips off of this row, and bet red vs black instead.” Now, with two outcomes available, your odds are 50/50. (I’m ignoring the green 0 and 00, of course, but you see my point.)

“you only get to choose (bet on) one door.”. Correct , and you should bet on the door that you didn’t pick initially and Monty didn’t open – as that gives you a 2/3 chance of winning the car (mathematically, logically and STATISTICALLY).

And if I were you I’d stay away from Las Vegas, anyone who’s as ignorant of probability as you obviously are would lose the lot.

Marley52, Do you think you can you bet on the 3rd door that Monty just opened? No. The 2nd goat has been removed, Now, with only two doors, you no longer have two losing doors. Just one. Chances of either door losing = 1 in 2. Chances of either door winning = 1 in 2.

“Chances of either door losing = 1 in 2. Chances of either door winning = 1 in 2.”

Not true. Probability the door not picked (by you or Monty) hides the car is 2/3, so if you switch to that door you’ll have a 2/3 chance of winning the car. Why don’t you try it out STATISTICALLY, play 300 games and see how many you win by always staying , and how many you win by always switching. You keep repeating the same nonsense, it’s getting tedious..

Marley52, If I win, do I get two cars? No. If I lose, do I get two goats? No. So neither outcome is 2/3.

There’s little sense in trying to engage Marley in any meaningful discussion, which I learned years ago — he will not even agree to disagree, but rests like a secret corpse among the comments, awaiting anyone to poke him to see whether he is still alive so he can grab their hand and hold them down and whisper “66 percent” over and over until they give up or manage to flee in terror..

There’s NO sense in trying to engage Ablestmage in a meaningful discussion about the MHP. He just wants to talk about spinning dartboards and what’s the capital of Kentucky. Still waiting for your YouTube video on the MHP Ablestmage, the trailers were a hoot.

I had forgotten all about the Switchmen Repent concept.. I had put it on hiatus pending stylistic reasons, which is still on hold indefinitely. I think these threads do a sufficient job of describing my argument so far. I’ve since had another discussion with a friend of mine from high school (who recently got his doctorate in philosophy and mathematics, and he went into the more complex forms of his position with the greek symbols to explain his argument, but couldn’t (or wouldn’t) answer why the two rounds needed to be combined..

“I think these threads do a sufficient job of describing my argument so far.”

Ablestmage, so far you haven’t described your thoughts in regards to those 67 out of 100 contestants who already have a goat behind their door before the host reveals one to them.

Here (it’s even called a trailer) https://www.youtube.com/watch?v=7MZ7-NXyduw

I was looking at the old original account that was a string of numbers -_-

https://www.youtube.com/channel/UC0oP9ZAuQbA3zmYrVCjaJYw

Marley52. Let me give you three scenarios. 1) There is a car behind one of these three doors. 2) There is a car behind one of these two doors. 3) There are cars behind two of these three doors. If you only get one try, what are the odds of winning game #1? What are the winning odds for #2? What are the odds of winning game #3?

2) Assuming the car is randomly placed behind one of these TWO doors, the answer is 1/2.

In the MHP the car is randomly placed behind one of THREE doors. What’s your point?

Let me give you one scenario: There is a car behind one of two doors. You’ve been told (by someone who knows) the car is behind Door1. What are your chances of winning the game?

Marley52. Let me rephrase my three questions make it easier to understand.. …..Question 1) There are three doors on stage. There is a car behind one of these three doors. If you only get one try, what are the odds of winning? …..Question 2) There are two doors on stage. There is a car behind one of these two doors. If you only get one try, what are the odds of winning? ….. Question 3) There are three doors on stage. There are cars behind two of these three doors. If you only get one try, what are the odds of winning game #3?

I understood 1st time round. Here are your answers

1) (Assuming the car has been randomly placed behind one of the THREE doors) 1/3.

2) (Assuming the car has been randomly placed behind one of the TWO doors) 1/2

3) (Assuming the 2 cars have been randomly placed behind two of the THREE doors) 2/3

Now answer mine: There is a car behind one of two doors. You’ve been told (by someone who knows) the car is behind Door1. What are your chances of winning the game?

Marley52, I have no reason to trust the person to tell me the right door to choose. So, statistically, my chances of picking the right door is 50/50. (It’s either behind door 1 or behind door 2).

Marley52, You and three of your friends are invited on stage. There are four doors in front of you, and Monty says that behind one of those doors is a car. One of you will go home a winner. So he starts by letting you pick a door. Next, contestant #2 is allowed to pick one of the three remaining doors. Then, contestant #3 is allowed to pick from one of the two remaining doors. Finally, contestant #4 is given the last door. Are the contestant’s odds of winning equally 1/4? or does each sequential contestant gain increasing advantage due to the fewer number of doors available? Should you keep your door, or switch with the guy who got stuck with the last, unopened door?

“I have no reason to trust the person to tell me the right door to choose”.

Cop out…..OK, then suppose YOU saw which door the car was behind (you took a sneaky peek backstage before the show started), what are your chances of winning? According to you it’s still 1 in 2.

” Are the contestant’s odds of winning equally 1/4?”

Of course. And this is related to the MHP how?.

Marley52, But Monty (or a stage hand) might have seen me peek, and switched prizes before I started the game. When we are discussing the chances when there is 1 car and 2 doors, the statistics are still 1/2. To calculate the odds, you divide the number of prizes (1) by the number of possible spots (3). When there are 3 doors and 1 car, the odds of winning is 1/3. If any parameter changes, so do the statistics. (Add a second car, and play the three doors, the chances of winning is 2/3.) (Play one car, and three doors, the chances of winning is 1/3). (Remove one of the doors, and play with one car, and the chances of winning is 1/2.) That is exactly what happens in the Monty Hall Problem. One of the three doors is removed. So it’s 50/50, and no advantage gained by switching.

“But Monty (or a stage hand) might have seen me peek, and switched prizes before I started the game.”

Really, that’s just lame.

The PROBABILITY (you keep confusing probability with statistics) is only 1/2 (or 1/3 or 1/4) if each possible outcome is equally likely of occurring. I take a dice and roll it , if it lands 1,2,3 or 4 I put the car behind Door1, if it lands 5 or 6 I put the car behind Door2. There are still 2 doors and 1 car, what are your chances of winning the car , you still think it’s 1/2?

Really you should give up now, you’re just embarrassing yourself. .

I’ve been trying to a make a point, but Roy keeps mixing up terms – recently referred to games when I was talking about enumeration of possibilities/outcomes/cases.

I originally wanted to engage ablestmage , so will try once more. Are you familiar with enumerating possible outcomes when solving statistics problems? If so, can you please do so with flip of a coin as a very basic, simple example – assuming that it cannot land on an edge as all of my university professors did but Roy cannot seem to…

“I originally wanted to engage ablestmage”

Brad, I can tell you from past experience that is a complete waste of time …… unless you want to discuss ‘What is the capital of Kentucky?’

The problem you guys seem to have is that you are only looking at 3 of the 4 possible outcomes. For example, when the car is behind Door #3, and the contestant picks Door#1, Monty will open Door #2. If the contestant picks Door #2, Monty will open Door #1. Then you say if the contestant picks Door #3, Monty opens “another door”. And you stop applying statistics. (“Monty opens another door” does not qualify as statistical data.) You have two more arrays instead of one. [1] If the contestant picks Door #3, Monty opens Door #1. [2] If the contestant picks Door #3, Monty opens Door #2. So there are FOUR situations to consider, not just three. (Repeat this exercise with the car behind other doors, and you will find a total of TWELVE arrays, not just nine.) The true statistics will reveal six times out of twelve you should stay, and six times out of twelve you should switch. 6/12 simplified = 1/2. Don’t let the con men presenting the MHP fool you into believing in an imaginary advantage. it’s 50/50.

I am now convinced that we don’t have a math problem here. We have a reading problem. The 50/50 people can’t read well enough to understand the problem. They have to solve a problem that they have re-stated incorrectly.

Charles, the MHP is easy. You forego your first choice if you switch. So 1-1+1=1. Not 2. You do not gain a 2/3 advantage by switching. Maybe my problem isn’t reading. Maybe your problem is comprehension.

I’ll explain this to you one more time. if there are 5 doors, then each door has a 1/5 probability of winning. If there are 7 doors, then each door has a 1/7 probability of winning. If there are 3 doors, then each door has a 1/3 probability of winning. Let’s simplify this, and call the probability of winning, based on the number of doors, “1/n”. So in the MHP, each door has a probability of 1/n. You pick door #1. It has a probability of 1/n. Door #2=1/n, and Door #3=1/n. Monty removes Door #3. Now, Door #1=1/n, and Door #2=1/n. (The new value of “n” represents the available options.) This means any probability factor you want to attribute to Door #1 is EQUAL to the probability factor of door #2. So you started the game, and chose a door with a 1/3 chance of winning. Then Monty Hall removed a door. Instantly, the door you chose had a 1/2 chance of winning. I’ve stated this before, the door you didn’t chose had a 1/3 possibility of winning, and when Monty removed a door, it instantly had a 1/2 possibility of winning as well. The two remaining doors are EQUAL. If you still believe that, magically, one of the last two doors would gain a 2/3 advantage, then you need to make a sound argument as to why it would go to the door you didn’t choose, instead of the door you did choose.

For those of you who want to see your computer calculate the probability factor for each door, then open your Excel Spreadsheet, and do the following: [1] Highlight cells A1-A-15. [2] Right-click your mouse, and select “Format Cells”. [3] Under “Category”, select “Fraction” and under “Type”, select “Up to two digits”. [4] Now click cell A15, and enter the following formula: =1/COUNTIF(A1:A14,1) [5] Now place random 1’s along the column between A1 and A14 [6] Now, cell A15 will display the probability of winning for each cell in play (It will count all cells marked with a 1). This means if there are 7 doors in play, A15 will show 1/7. If there are 3 doors in play, then A15 will show 1/3. And if there are 2 doors in play, then A15 will show 1/2. The MHP starts with 3 doors in play (A15 shows 1/3), and then Monty Hall removes one door (Now, you delete one of your 1’s from A1, A2, or A3) Cell A15 will show 1/2. Now even your computer agrees with me.

Roy, you don’t understand probability, stop pretending to do so. You said “The problem you guys seem to have is that you are only looking at 3 of the 4 possible outcomes”

And those 4 possible outcomes have probabilities of 1/3, 1/3, 1/6 and 1/6. Now get 3 playing cards and go do some statistical trials yourself. Don’t come back until you’ve done at least 300,000 (that’s a reasonable sample size). Bye bye.

Marley52, if you remove Door#3, then 1/6 is added to Door #1, and 1/6 is added to Door #2. Now add 1/6 to 1/3 and you get 1/2 for each door. Why would you add 1/3 to door #2, but nothing to Door#1? Is it because you’re standing in front of it? Please give a logical reason for doing that. Then, if you’re still so inclined, give a logical reason for not adding it the the door you aren’t standing in front of. I’ll be waiting….

I you want to unevenly distribute the odds, then why not do so at the start of the game? Why not say that Door #1 has a 1/3 chance of winning, Door #2 has a 1/2 chance of winning, and Door #3 has a 1/6 chance of winning? You wouldn’t do that in the beginning, because you know it’s stupid to think that way. You shouldn’t do that during the game, either. Because it’s stupid to think that way.

Roy, you are right – the problem is not a reading problem (we all can read the words) it is a comprehension problem. Here is how I comprehend the problem and from your last post, I think you are getting close to comprehending it.

There are three doors (n = 3) – one with a car behind it and two with goats.

I choose one of the doors. There is a 1/n chance that the car is behind that door. That probability is 1/3 no matter what else happens.

There is a 2/n chance that the car is behind one of those two doors. That probability is 2/3 no matter what else happens.

Monty is now going to help me out by giving me some additional information. There is still a 1/3 chance that the car is behind the door that I first picked. Nothing happened (nor could anything have happened) to change that probability from 1/3. There is still a 2/3 chance that the car is behind one of the doors that I didn’t choose. We know that the chances are 1/3 (the car is behind the door that I first chose) and 2/3 that the car is behind one of the other two doors. Monty now helps me out by telling me which of the other two doors does not have the car behind it. There is still a 1/3 chance that the car is behind my first choice, there is 0 chance that it is behind the door that Monty has just revealed and since the probabilities must add to 1, I should switch to the other door with a 2/3 probability. Another way to look at the problem – what would you do if Monty had said, before revealing the door with a goat. Do you want to stay with your first choice 1/3) or do you want the car if it is behind either of these two other doors 2/3? 1/3 + 2/3 = 1. The probabilities don’t change. Do you want a 1/3 or 2/3 chance of winning the car? Do you want to cover 1 door or 2 doors with your bet?

Charles, you say there is a 2/n chance that the car is behind one of those two doors. Don’t you realize what that means? One of the two doors has a 1/n chance, and the other door has a 1/n chance. You can’t “fuse” two doors into one, making it 2/n for just one door. I don’t even need to read the rest of your post, because I think it’s more nonsense based on your failed premise.

Charles, why not “fuse” the two doors that Monty didn’t open? Now, the door you picked, and the door you didn’t pick have a combined value of 2/n. Now, factor that value into the two doors, and each one is 1/n. (Remember that when there were 3 doors, 1/n meant 1/3. And now that there are just 2 doors, 1/n means 1/2). But you still think one of the two doors has a value of 2/n? Then defend your position as to WHY it is not your door. I bet you can’t refute that whatever argument you make can’t be applied TO your door. (Started as 1/3, so now it’s 2/3…because…???) Because you picked it? Because you didn’t pick it? Because it’s the wrong color? Give me something. Anything that is unique to your position.

Roy, Lets take one step at a time. You have 3 doors A, B, and C. One of the doors has a car behind it and each of the other two door has a goat behind it. What is the probability of a car behind door A? What is the probability of a car behind door B? What is the probability of a car behind door C?

Charles, I’m glad you asked. Initially, the probability of a car behind Door #1 is 1/3. Initially, the probability of a car behind Door #2 is 1/3. Initially, the probability of a car behind door #3 is 1/3. Now it’s my turn to ask a question. Once Monty Hall removes one of the three doors, he asks you if you would like to stay or switch. So tell me, Charles, Do you want to keep your door (which had an initial probability of 1/3) or switch to the other door (which also had an initial probability of 1/3)? And what is the advantage of switching from one 1/3 probability to another 1/3 probability?

RIP Monty Hall. 8/25/1921 – 9/30/2017.

Roy, I indicated that I wanted to take one step at a time. I will be glad to answer your questions after we have gone through my questions. If you want to do that, let me know. Otherwise, I’m done.

Charles, I answered your question. I’ll post it here again, so you can read it slowly. [[Charles, I’m glad you asked. Initially, the probability of a car behind Door #1 is 1/3. Initially, the probability of a car behind Door #2 is 1/3. Initially, the probability of a car behind door #3 is 1/3.]]

Charles, Now it’s my turn to ask a question. Once Monty Hall removes one of the three doors, he asks you if you would like to stay or switch. So tell me, Charles, Do you want to keep your door (which had an initial probability of 1/3) or switch to the other door (which also had an initial probability of 1/3)? And what is the advantage of switching from one 1/3 probability to another 1/3 probability?

Roy, the question that you answered was step 1. I have several more and then I will answer all or yours. My next question is to ask you to pick a door (A, B, or C) and tell me which door and what is the probability that the car is behind that door? If you are not interested in finding the correct solution to the problem, we can quit now. Perhaps, you will be able to enlighten me if I am wrong.

Charles. Okay. I’ll play along. I’ll pick A. (It has an equal probability as the other two doors.)

Charles, Mathematically that would be: A=B=C

Charles, Note that is A=B=C. Not A=B+C and Not A=B-C. Do you follow me so far?

Correct. What is the probability that the car is behind door B or door C?

Charles, the probability that the car is behind door B is currently 1/3. The probability that the car is behind door C is currently 1/3. (Therefore, the probability that the car is behind door A is currently 1/3)

Exactly. I assume that you agree that the probability that the car is behind door B or C is 2/3. Probability (A) = 1/3. Probability (B or C) = 2/3.

You’re mistake is already showing. It’s not (B or C), it’s (B+C). Please continue…

But, I may have misunderstood your question. So allow me to answer this way: The combined probability that car is behind B+C is 2/3. However, the combined probability that the car is behind A+B is also 2/3. And the combined probability that the car is behind A+C is also 2/3. [ An easier way to put this would be to say the probability that the car is NOT behind A is 2/3. And the probability that the car is NOT behind B is 2/3. And the probability that the car is NOT behind C is 2/3.]

Exactly. You agree that the probability that the car is not behind A is 2/3 and the probability that the car is behind A is 1/3. The next step in the Monty Hall problem is for Monty to reveal which of the doors that you have not chosen (B or C) hides the goat. Lets re-word the Monty Hall problem slightly but not substantively to eliminate revealing which door hides the goat. Monty now simply asks – do you want to stay with your choice of door A or do you want to win the car is it is behind either door B or C? What would your answer be?

Chaarles, Let’s say that Monty revealed a goat behind C, just to keep this conversation simple. So Monty asks do I want to stay with A or switch to B. I decide to stay with A.

That was not my question, but if you decide to stay with A, you have a 1/3 chance of winning the car. If you decide to switch, you will have a 2/3 chance of winning the car because you have already agreed that the probability that the car is not behind door A is 2/3. Nothing that Monty has done changes that.

Charles, Wrong again. If I decide to stay with A, I have a 1/2 chance of winning the car. If I decide to switch, I have a 1/2 chance of winning the car. Monty changed the fraction’s denominator from 3 to 2 when he took away the 3rd door.

Roy, I think that’s all I have to say for my position. I will be glad to answer your questions now, as I promised.

Charles, Question time: Door B only had a value of 1/3. Why would it increase to 2/3? Door A only had a value of 1/3. Why wouldn’t it increase to 2/3? Just because C was removed? That only changes the statement “A=B=C” to “A=B”.

Roy, I agree that Door A has a value of 1/3 and Door B has a value of 1/3 and Door C has a value of 1/3. The person that chooses Door A and does not switch is going to win 1/3 or the time (if the car is behind Door A). The person who chooses Door A and then switches wins if the car is not behind Door A that is if the car is behind either Door B or Door C because the value of Door B is 1/3 and the value of Door C is 1/3. The union of those two is 2/3. The point that the 50/50 person misses is that the person who switches is covering a bet on 2 of the 3 doors.

Charles, why do keep insisting that C’s value has to go to door B? Why not add it to door A? Switching from A to B is not betting on B+C any more than A becomes A+C. Don’t forget, C was removed. The new value of C is ZERO. So even if you believe that C has been added to B, then A=(B+0).

And by the way, I was originally a 50/50 person who finally saw the light.

Charles. I started as a 50/50 person. But I was intrigued by the 1/3 2/3 argument. So I looked at it in more depth. 50/50 is the right answer. I don’t claim 50/50 lightly. I have made several points debunking the 33 percenters.

Charles, picture a river. It divides equally into three smaller streams before flowing to the bay. The amount of water dumped into the bay is 300 gallons per minute. Stream #1 handles 100 gallons. Stream #2 handles 100 gallons, and Stream #3 handles 100 gallons. Monty Hall builds a dam, and blocks one of the streams. But 300 gallons still flow into the bay each minute. Monty’s stream dumps 0 gallons. But now, the 300 gallon flow is divided between the two remaining streams. Do you think it’s equal? Do you think one stream dumps 100 gallons while the other dumps 200 gallons? If so, which one has a heavier flow, and why? It can’t be because you “picked it”. Think clearly and tell me your opinion.

Probability (A) = 1/3. Probability (B) = 1/3. Probability (C) = 1/3. Probability (not A) =2/3. Probability of (B U C) is 2/3. Once you discover that the probability of either B or C = 0, the probability of the other (B or C) becomes 2/3. Discovering that the probability of B or C becoming 0 does not change the probability of A which remains 1/3.

Charles, why wouldn’t A increase to 2/3? You still haven’t explained why B would change and why A wouldn’t. That’s the magic argument right there.

Original probabilities:

A = 1/3

B = 1/3

C = 1/3

You choose A

not A = (B U C) = 2/3

Now Monty looking at B and C (not A because you have chosen A) reveals that C = 0

What does B equal

I contend that B = 2/3

(B U 0) = 2/3

Charles, I contend that A = 2/3. It’s just as good an argument as you make. Not B = (A U C) = 2/3. How about that? [I’m just showing you that you’re making a nonsensical argument, based on flawed logic.]

How are those 300,000 trials coming along Roy. They’ll take you forever if you waste your time posting gibberish here. Stick to the task at hand!!

Marley52 wrote: “How are those 300,000 trials coming along Roy. They’ll take you forever if you waste your time posting gibberish here. Stick to the task at hand!!” Marley, I win 50% of the time.

Roy, that’s odd. When I did it I won 67% of the time. I guess my strategy is better than yours.

Let P represent your first Pick.

Let X represent Monty Hall reveals a goat.

Let S represent Switching

Car is behind Door #1 Car is behind Door #2 Car is behind Door #3

1 2 3 1 2 3 1 2 3

Car Goat Goat Goat Car Goat Goat Goat Car

P S X P S X P X S

P X S X P S X P S

S P X S P X X S P

S X P X S P S X P

Impossible scenarios have been eliminated. (Monty can not open door with car.)

How many P’s are under a car? (6)

How many S’s are under a car? (6)

Odds of winning with first Pick is 6/12.

Odds of winning by Switching is 6/12.

There’s your proof that it is 50/50.

I hope that is the magic argument.

I am going to give up for today – but not until you see the light.. In the mean time i suggest that you take a look at the following web site:

http://marilynvossavant.com/game-show-problem/

It may help you to understand the the problem.

Talk to you later.

Charles, I’ve heard that website’s argument before. It’s the old “Open a bunch of doors, to give support to a different door” ploy. What if you opened a hundred other doors? But after you open each door, say “It’s a good thing you didn’t switch to this door, because you would have lost!” Eventually, you will start to feel that you’re much safer by staying. This is an appeal to emotion no matter which way you approach it. Statistics don’t care about your emotions. But nice try.

By the way, you still haven’t given any solid evidence as to why 1/3 is added to B instead of A. I have shown that any argument you make for B can also apply to A. All I want is for someone to prove why B is so unique, then I will happily accept your conclusion. But so far, nobody has been able to do that.

My first chart didn’t print correctly, so here it is again:

Let P represent your first Pick.

Let X represent Monty Hall reveals a goat.

Let S represent Switching

Car is behind Door #1……….Car is behind Door #2……….Car is behind Door #3

1……… 2……….3…………………1……….2………3………………..1……….2………..3

Car…..Goat…..Goat…………..Goat…..Car…..Goat…………..Goat…..Goat…..Car

P………S……….X…………………P………S……….X……………….P……….X……….S

P………X……….S…………………X……….P………S……………….X……….P……….S

S………P……….X…………………S……….P………X……………….X……….S……….P

S………X……….P…………………X……….S………P……………….S……….X……….P

Impossible scenarios have been eliminated. (Monty can not open door with car.)

How many P’s are under a car? (6)

How many S’s are under a car? (6)

Odds of winning with first Pick is 6/12.

Odds of winning by Switching is 6/12.

Roy, how are you picking the car door 50% of the time with your first pick (P)? Back to the drawing board for you.

Marley52, [1] If I pick the car a door 1, and Monty opens door 2. (Stay=win, switch=lose) [2] If I pick the car at door 1, and Monty opens Door 3. (Stay=win, switch=lose). [3] If I pick the goat at door 2, and Monty opens door 3.(Stay=lose, switch=win) [4] If I pick the goat at door 3, and Monty opens door 2. (Stay=lose, switch=win).

Marly52, if you want to see it another way, then imagine a black goat and a white goat. [1] Monty reveals a black goat. My door is either the car or the white goat. Also, the door I didn’t pick is either the car or the white goat. (Stay or switch odds = 50/50.) [2] Monty reveals a white goat. My door is either the car or the black goat. Also, the door I didn’t pick is either the car or the black goat. (Stay or switch odds = 50/50) That covers everything. Simple enough?

You didn’t answer my question. On your 1st pick (P) you’ve picked the car 6 times out of 12. That’s never going to happen, you’ll only pick the car 4 times out of 12. Back to the drawing board Roy.

Marley 52. You said “You’ll only pick the car 4 times out of 12.”

No. I’ll pick Door #1 four times out of twelve.

That doesn’t guarantee I’ll pick the car 4 times out of 12.

It means if I pick door #1, I will get the car 2 of those 4 times.

If I include all three doors, I will have gotten the car 6 times out of 12.

“I’ll pick Door #1 four times out of twelve… if I pick door #1, I will get the car 2 of those 4 times”.

Only if the car is behind Door1 50% of the time….which it isn’t.

Roy, in 12 games how is it possible to pick, from THREE doors, the car door 6 times (50%) as your diagram depicts? Until you can answer that, don’t bother me further with your nonsense.

Marley52. The car is behind Door #1 four times out of twelve (or 33%).

Of those four times, here are the possible situations:

[1] I pick Door #1, and Monty reveals a goat behind Door #2.

[2] I pick Door #1, and Monty reveals A goat behind Door #3.

[3] I pick Door #2, and Monty reveals the goat behind Door #3.

[4] I pick Door #3, and Monty reveals the goat behind Door #2.

That is why I say I have a 50% chance of picking the car first. Because in 2 of the 4 possible situations, I picked Door #1 with the car. The other two times, I didn’t.

“The car is behind Door #1 four times out of twelve (or 33%).”_

Correct! Well done!!

“in 2 of the 4 possible situations, I picked Door1 with the car”

There are 12 possible situations and you only picked the car 4 times (33% of the time)

You still don’t know what’you’re talking about Roy. You’re not being wilfully ignorant are you?

Why did you pick Door1 twice but Door2 and Door3 only once each. You’re not trying to cheat are you Roy?

After reading that exchange I give up. There is no hope that Roy will get it.

Charles, I don’t blame you. I only keep going because it’s amusing watching Roy repeatedly tie himself in knots.

“[1] I pick Door #1, and Monty reveals a goat behind Door #2.

[2] I pick Door #1, and Monty reveals A goat behind Door #3.

[3] I pick Door #2, and Monty reveals the goat behind Door #3.

[4] I pick Door #3, and Monty reveals the goat behind Door #2.”

Roy, use your head. The car door is still picked 1/3 of the time, as is Goat A and Goat B. Half of the time when the car is picked, host shows Goat A, and the other half Goat B. But you’re DOUBLE counting the car being picked!!

And here’s the other thing you halfers ALWAYS miss….the contestant’s chance of what he picks EQUALS the chances of what doors are left over. If it’s a 1/3 chance to pick the car then it’s a 1/3 chance two goat doors remain for the host. Therefore there’s a 1/3 chance the host leaves a goat to switch to AND NOT 1/2!!

If the chances of the contestant to pick a goat is 2/3, then it’s a 2/3 chance that a goat door and a car door remain for the host. Therefore there’s a 2/3 chance the host leaves a car to switch to AND NOT 1/2!!

Get that through your head once and for all.

“Charles, Wrong again. If I decide to stay with A, I have a 1/2 chance of winning the car. If I decide to switch, I have a 1/2 chance of winning the car. Monty changed the fraction’s denominator from 3 to 2 when he took away the 3rd door.”

Roy, Monty didn’t change the fraction’s denominator….YOU DID!! You determined the average probabilities of the doors, not the probabilities of each door. Knowing the average is totally useless in the MHP…it’s like saying that the average human being has only one testicle.

“it’s like saying that the average human being has only one testicle.”

But that IS true (unlike Roy’s comments)

Lol, thanks Marley. When I said the average probability of the doors is totally useless in the MHP I should have referred that knowing the average human being having one testicle is useless in understanding anatomy.

P.S. I always enjoy reading your comments and respect/admire your knowledge in mathematics.

Thanks Klaus. I read your comments on YouTube, always well argued and often quite funny, along with HDO (who I feel a close affinity with) and Freddie of course (who’s been quiet recently) and serious Ted. We need people like Roy and Ablestmage and Richard Buxton of course to keep the merriment going.

Klaus, Monty changed the denominator when he changed the number of doors. First, there were three doors. (The probability of any door hiding the car was 1 in 3). Now, there are two doors. (The probability of one of those two doors hiding the car is 1 in 2).

There are only two prizes you can end up with. [1] The goat that Monty didn’t reveal. [2] The car. Just two possible outcomes. So the odds of either outcome is 50/50.

Roy, I’ll give this one last shot. In an earlier post I asked you to answer this question, which you did not do. Could you answer it now?

Lets re-word the Monty Hall problem slightly but not substantively to eliminate revealing which door hides the goat. Monty,without revealing a door with a goat, now simply asks – do you want to stay with your choice of door A or do you want to win the car if it is behind either door B or C? What would your answer be?

Charles, you are confusing offering the contestant to switch “Picks” (Door1, or Door2, or Door3) with offering the contestant to switch answers (“Yes, I think the car is behind Door1”, or “No, I think the car is NOT behind Door1”)

When you allow the contestant to change his position, (Yes/No) then the odds will go from 33% correct to 67% correct. (That is, of course, if all three doors are still in play.)

But if you only give the contestant the chance to switch which door he picks, then his odds always remain the same. (33% right if all three doors are still in play, and 50% right if only two doors are in play.)

Klaus and the gang can’t seem to differentiate between these two propositions. And any time I call them out, they want to switch the subject and talk about testicles.

So I offer them the challenge to take the 33/67 argument (as it relates to the door they picked), and apply it to the door they didn’t pick.

Every definition and every argument applies to both doors in exactly the same way. (That shouldn’t happen if their position was unique) But it does happen, and therefore it does not make their case at all.

Roy answer this ONE question…….How can half of the contestants end up with a car behind their door when 67 out of 100 had goats BEFORE the host revealed a goat?????????????

Klaus 74 asked “Roy answer this ONE question…….How can half of the contestants end up with a car behind their door when 67 out of 100 had goats BEFORE the host revealed a goat?????????????”

No Klaus. Half of the contestants do not end up with a car BEFORE the host revealed a goat.

You’re starting off wrong again.

The contestant has a 1/3 chance of picking the car BEFORE the host revealed a goat.

33/67 are the odds BEFORE

BEFORE

BEFORE

BEFORE

Understand?

Now let’s look at the set-up AFTER the host revealed a goat.

There are only 2 doors. AFTER Monty removes a door.

There are only 2 possible outcomes. AFTER Monty removes the 2nd goat.

So when there are only a choice left between two options, the new statistics are:

50% Chance for the Car and 50% Chance for the remaining Goat.

“There are only two prizes you can end up with. [1] The goat that Monty didn’t reveal. [2] The car. Just two possible outcomes. So the odds of either outcome is 50/50.”

Nobody cares about that argument BECAUSE IT DOESN’T WORK!!!!

1 Car 2 Goat 3 Goat

Pick Door 1… host shows either goat, stay wins. Switching loses.

Pick Door 2. host shows Door 3, THERE ARE NOW 2 DOORS LEFT, stay loses, switch wins

Pick Door 3…host shows Door 2, THERE ARE NOW 2 DOORS LEFT, stay loses, switch wins.

You win 1/3 of the games by staying whether the host shows one, two, or NO doors!!

Since you are too %^$#&*@ lazy to do the experiment yourself I showed you. So STFU idiot!!!

Charles, or Marley52, are either of you around? I have a probability question that I cannot answer myself and I need an explanation as to how I am approaching the problem incorrectly. Let me know if you would like to help and I will describe the problem. Thanks!

Hi Klaus. I’m in Thailand at the moment on holiday but tell me the problem and I’ll see if I can help

I’m here. I’ll try to help.

Roy, just humor me and answer the question that I asked. Before Monty reveals the goat, would your strategy be to stay with Door A, or switch?

“So I offer them the challenge to take the 33/67 argument (as it relates to the door they picked), and apply it to the door they didn’t pick.Every definition and every argument applies to both doors in exactly the same way. (That shouldn’t happen if their position was unique) But it does happen, and therefore it does not make their case at all.”

Easily disputed.

1 Car 2 Goat 3 Goat

Contestant picks Door 2, host shows Door 3. Now you are comparing the chances of Door 2 with Door 1 that was not picked. You are playing the ‘What if’ game….’what if contestant picks Door 1?’ what if contestant picks Door 2?’ But this is what you ignore….WHAT IF THE CONTESTANT PICKED DOOR 3?????

“Klaus, Monty changed the denominator when he changed the number of doors. First, there were three doors. (The probability of any door hiding the car was 1 in 3). Now, there are two doors. (The probability of one of those two doors hiding the car is 1 in 2).”

That’s the average!! Chances to win by switching are calculated by multiplying the probabilities of both the contestant and the host picking goats, 2/3×1=2/3.

The most stupid person I have ever dealt with in my life is pattystomper, you sound just like him!!

“So when there are only a choice left between two options, the new statistics are:

50% Chance for the Car and 50% Chance for the remaining Goat.”

And I showed you it DOESN’T work!!

Okay guys, here are your statistics one last time.

Please pay attention and try to learn something.

Behind Door #1 is either THE CAR or THE GOAT MONTY DOESN’T REVEAL

(Odds for Door #1 = 50/50)

Behind Door #2 is either THE CAR or THE GOAT MONTY DOESN’T REVEAL

(Odds for Door #2 = 50/50)

Behind Door #3 is either THE CAR or THE GOAT MONTY DOESN’T REVEAL

(Odds for Door #3 = 50/50)

So it doesn’t matter how many doors there are, the possible outcome is 50/50.

It’s like flipping a coin.

It doesn’t matter how many times you flip, the possible outcome is 50/50.

Quit describing stupid arguments and just show us….

1Car 2Goat 3Goat

Which door wins by staying and which doors win by switching.

JUST SHOW US!!

Hey are you actually pattystomper? There’s a way of finding out. He never could answer this question, and if you can’t then you must be him!!!

1Car 2Goat 3Goat

Which door wins by staying and which two win by switching?

Klaus 74, “Hey are you actually pattystomper? There’s a way of finding out. He never could answer this question, and if you can’t then you must be him!!!”

That’s like saying “My dog can’t breathe underwater, so if you can’t breathe underwater either, then that makes you my dog.”

More flawed logic from Klaus 74, ladies and gentlemen (applause-applause)

Can’t answer this can you?

1Car 2Goat 3Goat

Which door wins by staying, which ones win by switching?

Very hard question isn’t it?

Ladies and gentlemen, Roy cannot find the winning door even with them open!!!

1Car 2Goat 3Goat

Can’t find the door that wins by staying?

Klaus 74,

1 Car…….2 Goat…….3 Goat Revealed

So your options are

1 Car…….2 Goat

Door#1, Stay and win / Switch and lose.

Door #2, Stay and lose / Switch and win.

There’s your answer.

You can’t factor in Door #3 BECAUSE IT’S GONE

IDIOT!

“You can’t factor in Door #3 BECAUSE IT’S GONE IDIOT!”

You ARE pattystomper!!

1Car 2Goat 3Goat

So the contestant cannot choose Door 3 because the host needs it for revealing the goat?

Klaus 74,

We know that if we pick a goat, Monty will reveal the other goat.

So let’s call one goat “Our Goat” and the other one “Monty’s Goat”

We know it is impossible for Monty to reveal the car.

We also know it is impossible for Monty to reveal “Our Goat”

So,

Hidden behind Door#1 is either The Car, or Our Goat (Stay/Switch = 50% win)

Hidden behind Door#2 is either The Car, or Our Goat (Stay/Switch = 50% win)

Hidden behind Door#3 is either The Car, or Our Goat (Stay/Switch = 50% win)

Flip a coin attempt#1 (Heads/Tails = 50% win)

Flip a coin attempt#2 (Heads/Tails = 50% win)

Flip a coin attempt#3 (Heads/Tails = 50% win)

The number of attempts doesn’t affect the statistics.

Klaus 74 says “Quit describing stupid arguments and just show us….

1Car 2Goat 3Goat

Which door wins by staying and which doors win by switching.

JUST SHOW US!!”

The answer depends on which door it taken out of the available options.

If Monty removes Door #3,

then Door#1 has a 50% chance of winning/losing and Door #2 has a 50% chance of winning/losing.

I SHOWED YOU!!

I SHOWED YOU!!

Still don’t know how to calculate probability can you patty?

1Car 2Goat 3Goat

So the contestant cannot choose Door 3 because the host needs it for revealing the goat?

My oh my……

Klaus 74, now you’re just spouting sass and back-talk instead of addressing the statistical problem. That’s a sign of a weak-minded person. But maybe you are really looking to find a concrete argument to stand against the 33/67’ers, and you haven’t been able to explain it yourself, so you’re playing “Devil’s Advocate”, hoping my logic and reasoning will help you. Either way, you’re annoying.

Roy, forget about starting your argument AFTER the host reveals a goat. Your logic failed BEFORE that. If the contestant picks the car (1/3 chance), then the host WILL leave a goat. If the contestant picks a goat, (2/3 chance) the host WILL leave a car.

Get this through that thick skull of yours.

When Monty offers you the option to stay/switch, What are your chances of picking a goat by switching?

“When Monty offers you the option to stay/switch, What are your chances of picking a goat by switching?”

1/3

Now answer this…if there’s a 1/3 chance to pick the car when there are three doors, what are the chances that two doors with goats remain for the host? 1/3 or 1/2?

Klaus 74 “When Monty offers you the option to stay/switch, What are your chances of picking a goat by switching?”

1/3″“

Wrong, Klaus. Guess again.

Think carefully Klaus,

Monty doesn’t give you the option to stay/switch until after he has removed a goat (and the 3rd door)

So now that there is one less goat, and Monty offers you the option to stay/switch, What are your chances of picking a goat by switching?

“Think carefully Klaus, Monty doesn’t give you the option to stay/switch until after he has removed a goat (and the 3rd door)”

Think carefully Roy, if the contestant picks the car 1/3 of the time, then 1/3 of the time it leaves two doors with goats for the host. So the host reveals one and leaves a goat to switch to 1/3 of the time.

If the contestant picks a goat 2/3 of the time, then 2/3 of the time it leaves a car door and a goat door for the host. So the host reveals the goat and leaves the car to switch to 2/3 of the time.

The host CAN’T leave a goat to switch to 1/2 of the time when only 1/3 of the time it’s available IDIOT!!!!!

Roy, let me ask once more. If you have chosen Door A. Before Monty reveals the goat, what is the probability of a car behind Door A? Door B? and Door C? Just respond with three fractions.

Charles 1/3, 1/3, and 1/3

Correct Roy. Now let me ask you – before Monty reveals the goat, what is the probability that the car is behind either Door B or Door C? Just respond with one fraction.

Charles, Let me understand your question

Are you asking me what is the probability that the car is Not behind Door A, Door B, Door C?

(Respond with three fractions as I did for the first question?)

2/3, 2/3, 2/3

(It’s kind of funny, but when you look at those three fractions together, you know you can’t add them up to make 3/3 or 1)

It’s more like 6/3 or 2.

(Maybe that isn’t relevant, but it might be, once I’m asked to add something involving these fractions.)

“Wrong, Klaus. Guess again.”

The chances to get the goat by switching is 1/3, which is the SAME chances of the host having two goat doors remaining from which to reveal one, and the SAME chances as the contestant’s in picking the car!

No Klaus. Monty removed 1/3 from the setup. Now the hidden door either hides the car, or the remaining goat. So the answer is 1/2. Think about that for a while.

It’s impossible to still hide two goats once one has been revealed.

“No Klaus. Monty removed 1/3 from the setup. Now the hidden door either hides the car, or the remaining goat. So the answer is 1/2. Think about that for a while.”

I am right, you are wrong. Think about this. BEFORE the host reveals a goat, 2/3 of the contestants ALREADY have a goat behind their door!!

Klaus 74 said: “BEFORE the host reveals a goat, 2/3 of the contestants ALREADY have a goat behind their door!!”

Klaus, I agree with you there.

But when Monty removes a door, then 1/3 of the contestants are gone, too.

So if there were 150 contestants playing, then 50 of them got sent home.

So now there are 100 contestants choosing between two doors.

50% of them have a door with a car behind it, and 50% of them have a door with a goat behind it.

No, Roy. I am asking what is the probability that the car is behind Door B or Door C before Monty reveals a goat. One fraction is that answer.

Sorry Charles, I shouldn’t interfere right now as Roy is confused enough as it is.

“Sorry Charles, I shouldn’t interfere right now as Roy is confused enough as it is.”

Thanks, Klaus. I’m willing to talk with you again later.

Roy, One fraction.

Charles B+C = 2/3

Also A+B = 2/3

And A+C = 2/3

Charles, if, however, you are combining B and C as 1 door, then [Door A]=1/2 and [Door BC] = 1/2

Roy, you’re still here!!! Is Ablestmage paying you to post this drivel to bump up his blog count? I must say you’re going above and beyond the call of duty. I haven’t read so much rubbish in such a short time since PattyStomper was around. Outstanding!

Did you ever explain why you picked Door1 twice but Door2 and Door3 only once each in your “proof” (I use the word loosely) from yesterday?

Thank you Roy, The fraction that I asked for is B+C = 2/3. We now have the following state: The contestant has chosen Door A which has a 1/3 probability that there is a car behind it and, as you have admitted, there is a 2/3 chance that there is a car behind Door B or Door C. Monty has not yet revealed which Door B or C has the goat. If the car is behind Door A, the contestant will win by staying. If the car is not behind Door A, the contestant will win by switching to the Door that Monty does not reveal, because Monty has told him that the probability for that Door = 0. Since you admitted that the probability of B+C is 2/3 if C = 0 then B = 2/3 or if B = 0 then C = 2/3 and the contestant switches to the Door that Monty does not reveal and gets a 2/3 chance to win. Nothing that Monty does changes the 1/3 chance that the car is behind Door A because that Door is not part of the 2/3 probability for B+C.

Charles, We also have the following state:

( I had to re-word your statement a little bit, but the argument is just as valid as yours.)

The door which the contestant has not chosen has a 1/3 probability that there is a car behind it. So there is a 2/3 chance that there is a car behind {Door A combined with the door Monty did not open}.

Monty has not yet revealed which Door B or C has the goat. If the car is behind the door Monty didn’t open, the contestant will lose by staying with A.

If the car is behind Door A, the contestant will win by keeping Door A.

If C = 0 then A = 2/3 or if B = 0 then A = 2/3. So when the contestant switches to the Door that Monty does not reveal, he gets a 2/3 chance to lose. Nothing that Monty does changes the 1/3 chance that the car is behind the door he didn’t open, because that Door is not part of the 2/3 probability for A+C.

“But when Monty removes a door, then 1/3 of the contestants are gone, too.

So if there were 150 contestants playing, then 50 of them got sent home.

So now there are 100 contestants choosing between two doors.

50% of them have a door with a car behind it, and 50% of them have a door with a goat behind it.”

Oh my, lol.

Roy, I am not combining it as one door. It is obviously 2 doors.

With a total probability of 2/3 as you said.

The statistics are a mathematical process used to estimate the probability of things which are hidden from you.

So in the beginning of the game, you were dealing with three hidden objects. (Probability per choice = 1/3)

After Monty reveals a goat, you are dealing with just two hidden objects. (Probability per choice = 1/2)

With three doors, you can be offered Stay, Switch, Switch. (33% chance of winning)

With two doors, you are offered Stay, Switch. (50% chance of winning)

So I’ll ask my question again:

After Monty removes the 2nd goat, what are your chances of getting the one remaining goat by switching?

Roy, you are Pattystomper and I claim my $10.

LMAO

Marely52, I should claim Pattystomper as my alter-ego. I make the case for a 50/50 split, and Pattystomper would demonstrate the flaws on the 1/3 vs 2/3 argument. lol

Roy/Patty . The only case you’re making is one showing the unfathomable depths of your ignorance.

Why did you pick Door1 twice but Door2 and Door3 only once each?

Roy, After Monty reveals the goat, your chances are 2/3 for getting the car if you switch because switching gives you the 2/3 probability assigned to the two doors that you did not originally choose and Monty has told you which one of those doors has a probability of 0. If you stay you have the 1/3 probability because that is the probability assigned to each door. I don’t know why you don’t understand that the switch strategy gives you the two possibilities (2/3) that you did not originally choose and Monty tells you which one to pick by revealing a goat behind one of the two doors that you are covering. If you switch and the car is behind your original door you win 1/3 of the time. If you stay with your original pick you don’t get the advantage of covering two doors. When you decide to switch you are deciding to win a car if it is not behind the door that you originally chose. This is not rocket science, it is really a quite simple probability problem with a trick by making you miscalculate the probability when Monty reveals the goat. Your options are to stay and cover one door with a 1/3 probability or switch and cover the other two with a 2/3 probability.

I made one misstatement. Should be if you switch and the car is behind the original door you lose 1/3 of the time..

I bet you guys like to play beer pong.

So I will set 3 cups on the table, marked A,B,and C.

(Let’s assume you’re not a ringer, and you land your shots randomly.)

If you don’t miss, and your ping pong ball lands in a cup,

What are the odds that it will land in cup A? (Answer: 1/3)

What are the odds that it will land in cup B? (Answer: 1/3)

What are the odds that it will land in cup C? (Answer: 1/3)

Now, I take away one of the cups. (I remove C, leaving just A and B)

What are the odds that it will land in cup A? (Answer: 1/2)

What are the odds that it will land in cup B? (Answer: 1/2)

See how the 1/3 argument disappeared when one of the 3 cups was removed?

Charles, do you play Roulette?

(Fast forward to 3:12)

So Roy/Patty, tell me again why you picked Door1 twice?

Marley, I didn’t pick door 1 twice. But, a “Door 1 pick” appears twice in two of the 4 scenarios.

“Marley, I didn’t pick door 1 twice”

Yes you did Roy, you labelled it (P) where (P) you denoted as ‘first pick’, and you did it again here:

“[1] If I PICK Door #1, and Monty reveals a goat behind Door #2”

Your English comprehension skills are as bad as your mathematical ones.

So why PICK (P) Door1 twice Patty?

Marley52 asked

“Did you ever explain why you picked Door1 twice but Door2 and Door3 only once each in your “proof” (I use the word loosely) from yesterday?”

No, I didn’t “pick” Door 1 twice. I only showed that Door 1 had four possible scenarios.

Assuming the Car is behind Door #1, so Monty can not open it, they are:

[1] If I pick Door #1, and Monty reveals a goat behind Door #2.

[2] If I pick Door #1, and Monty reveals A goat behind Door #3.

[3] If I pick Door #2, and Monty reveals the goat behind Door #3.

[4] If I pick Door #3, and Monty reveals the goat behind Door #2.”

“No, I didn’t “pick” Door 1 twice”

Yes you did, you labelled it ‘(P) – First Pick’, remember?

“[1] If I pick Door #1, and Monty reveals a goat behind Door #2.

[2] If I pick Door #1, and Monty reveals A goat behind Door #3.

[3] If I pick Door #2, and Monty reveals the goat behind Door #3.

[4] If I pick Door #3, and Monty reveals the goat behind Door #2.”

So if you pick out the door with the car two times out of four they why even bother having the host reveal a goat? Then they all can play instead of 1/3 of them going home without even getting a goat.

So cowpie, 1/3 of the contestants go home without finishing the game do they? You tested the game thoroughly and those were the results?

Yes cowpie tested it 300,000 times just yesterday. 100,000 contestants never got to finish the game, and of the remaining 200,000 who did get to complete their game, 100,000 (or 50%) won the car. Isn’t that right Roy?

Lol, Marly

One thing you have to admire about ‘Cowpie Roy’ is, he doesn’t mind making a fool of himself over and over again.

OK, now I know you guys are just trolling.

Sorry about the spelling, I’m laughing that much!!

By the way, I didn’t “pick” Door#1 twice, anymore than Monty opened Door #3 twice. But it does appear twice in the array.

“I didn’t “pick” Door#1 twice”

Yes you did Roy. In 2 out of the 4 scenarios you PICKED Door1. Don’t you even read your own comments before posting?

Marley, so you agree there are 4 scenarios? Then base your probability statistics on that, and not the 3 doors.

“Then base your probability statistics on that, and not the 3 doors”

Certainly Roy

[1] If I pick Door #1, and Monty reveals a goat behind Door #2. Probability = 1/3 * 1/2 = 1/6

[2] If I pick Door #1, and Monty reveals A goat behind Door #3. Probability = 1/3 * 1/2 = 1/6

[3] If I pick Door #2, and Monty reveals the goat behind Door #3 .Probability = 1/3 * 1 = 1/3

[4] If I pick Door #3, and Monty reveals the goat behind Door #2. Probability = 1/3 * 1 = 1/3

There you go Patty, knock yourself out

Cowpie, why are 1/3 of the contestants sent home early anyways? And how is that determined?

The rules include that the host knows where the car is, must reveal a goat, and offer the contestant a chance to switch to the remaining door or keep his original one. It doesn’t say anything about 1/3 of them can’t finish the game. Is this something new?

Klaus 74, when you create a scenario of 100 people playing the game, I just respond to it. (Unless something happened earlier, and the problem was demonstrated by pretending there were 100 people.) But that didn’t seem to register with you either, so why bother to bring it up? I can only assume you believe that throwing a bunch of unrelated arguments to the table will somehow sway the debate in your favor.

So instead of doing all of that, I’ll let you begin with a simple argument.

List everything that could be behind the door after Monty reveals a goat.

“I can only assume you believe that throwing a bunch of unrelated arguments to the table will somehow sway the debate in your favor.”

Really? So there is no relation between the chances of the contestant picking the car and the chances of what doors that leaves for the host?

So you don’t think that picking the car 1/3 of the time leaves two goat doors 1/3 of the time for the host?

Klaus. Yes. picking the car 1/3 of the time leaves two goats for the host. But removing a goat leaves just one goat for the contestant.

You tested it out 300,000 times yesterday and never noticed that picking the car door leaves two goat doors each time?

“You tested it out 300,000 times yesterday and never noticed that picking the car door leaves two goat doors each time?”

When your ‘strategy’ for picking between the 2 remaining closed doors is : close your eyes and go “eeny, meeny, miney, mo……”, I very much doubt Roy noticed anything.

“Yes. picking the car 1/3 of the time leaves two goats for the host. But removing a goat leaves just one goat for the contestant.”

So if the host has two goat doors 1/3 of the time, then how often will he leave a goat to switch to?

““Yes. picking the car 1/3 of the time leaves two goats for the host. But removing a goat leaves just one goat for the contestant.”

So if the host has two goat doors 1/3 of the time, then how often will he leave a goat to switch to?”

Monty will leave one of the goats for the contestant 100% of the time.

The question cowpie was: “how often will the host leave the car to SWITCH TO?”

Naughty Roy, answering a different question to the one asked, but fairly typical for a troll like you.

“Yes. picking the car 1/3 of the time leaves two goats for the host. But removing a goat leaves just one goat for the contestant.”

So if picking a goat 2/3 of the time leaves a goat door and a car door for the host, then how often will the host leave the car to switch to?

“So if picking a goat 2/3 of the time leaves a goat door and a car door for the host, then how often will the host leave the car to switch to?”

Monty will leave the car for the contestant 100% of the time.

So if picking a goat 2/3 of the time leaves a goat door and a car door for the host, then how often will the host leave the car to switch to?

So Cowpie, did your hundreds of thousands of tests conclude that even though the host has a goat door and a car door 2/3 of the time, revealing the goat leaves the car available 1/2 of the time, right?

Is it safe to assume that Marilyn vos Savant, Dr. Selvin, all the greatest mathematicians of the world, and us mere mortals that accompany you on this subject, have not tested the problem thoroughly enough and therefore we are all wrong?

Klaus, “revealing the goat leaves the car available 1/2 of the time, right?”

No, Revealing the goat leaves the car available 100% of the time, and the other goat available 100% of the time.

Now that you understand that there is 1 goat, and 1 car, and that there are just 2 doors in play, you can ignore the initial odds of 3 doors and 2 goats.

They no longer apply after Monty makes the changes.

Neither does your assumption that your door would keep it’s 1-in-3 odds.

Roy, back to the 4 scenarios you so helpfully enumerated yesterday. As you correctly stated you win by staying in scenarios 1 and 2, and win by switching in scenarios 3 and 4. I’m sure you can add up the probabilities for for each pair of scenarios to get the overall probability of winning by staying and winning by switching. (let me know what answers you get as I’m not 100% certain you can manage basic addition of fractions)

Marley52, “you win by staying in scenarios 1 and 2, and win by switching in scenarios 3 and 4.”

Right. That means you win in 2 of 4 scenarios when you stay, and you win in 2 of 4 scenarios when you switch.

“Right. That means you win in 2 of 4 scenarios when you stay, and you win in 2 of 4 scenarios when you switch.”

At last Roy you agree with me!! The PROBABILITY of the 2 scenarios occurring whereby you win by staying is 1/6+1/6 = 1/3, and the PROBABILITY of the 2 scenarios occurring whereby you win by switching is 1/3+1/3 = 2/3.

It took a long time but you got there in the end Patty. Well done!

Marley52, when does the 1/3 chance of a car being behind a door become 1/6? (or is it the goat?) Sorry, but you lost me.

“Sorry, but you lost me.”

Roy, you were lost long before I got here.

[1] If I pick Door #1, AND Monty reveals a goat behind Door #2. Staying wins. Probability of this event = 1/3 * 1/2 = 1/6

[2] If I pick Door #1, AND Monty reveals A goat behind Door #3. Staying wins Probability of this event = 1/3 * 1/2 = 1/6

[3] If I pick Door #2, AND Monty reveals the goat behind Door #3 .Switching wins. Probability of this event = 1/3 * 1 = 1/3

[4] If I pick Door #3, AND Monty reveals the goat behind Door #2. Switching wins Probability of this event = 1/3 * 1 = 1/3

You’ve got the attention span of goldfish Patty. There’s probably some medication you can take for it

“Now that you understand that there is 1 goat, and 1 car, and that there are just 2 doors in play, you can ignore the initial odds of 3 doors and 2 goats.”

Ignore the initial odds? You are having such a difficult time plugging up the holes in your logical dyke that you have to send 1/3 of the contestants home as a result!!

“You mean how often does Monty have a (1-car / 1-goat) situation once the contestant picks his first door.”

2/3 of the time because the contestant picks a goat 2/3 of the time. So the host leaves the car 2/3 of the time to switch to.

The contestant picks the car 1/3 of the time, so the host is left with two goat doors 1/3 of the time, so he leaves a goat available to switch to 1/3 of the time.

Your chart and your 1/3 of the contestants go home early argument are completely bullshit. Even your chart contradicts the ‘going home’ argument, lol.

Chances to win by switching is 2/3×1=2/3. Your calculating AFTER the reveal which is the same as the host picking first….1×1/2=1/2 which is a wrong answer on ANY MATH TEST!!!!

“Monty will leave the car for the contestant 100% of the time.”

No. He will leave the car for the contestant 2/3 of the time.

“He will leave the car for the contestant 2/3 of the time.”

Are you saying that 1/3 of the time Monty removes the car?

“Are you saying that 1/3 of the time Monty removes the car?”

No idiot. You said the host leaves the car for the contestant 100% of the time. I said he leaves the car