Let this stand as a testament to how the popularly and improperly taught truth about the Monty Hall Problem is false, for a variety of reasons.

If you’re unfamiliar with it, check out either this video (5min.com) or that video (YouTube). They’re two seperate explanations.

To set an official designation of the rules:

You are a contestant on a gameshow. The host gives you the option of selecting one door from an available 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 at random. The host now reveals one of the doors you have not selected to be a Goat Door. What is the probability of selecting the Car Door?

The disturbingly frequent misinformed explanation is that switching gives you the greater advantage.

Here is how exactly that line of reasoning is mortally flawed.

The greatest of logical errors is the perception of switching, when in fact you are not switching anything. The first choice is not a genuine choice because the revelation that one of the three doors you are permitted to pick from will be removed. You are in effect simply playing the game. Once you’ve made your misperceived selection, you are given the choice to choose between two options. This is your final decision. The original selection when there were three unknowns does not influence the final decision, because you would have had to choose between two new options regardless of whether you had picked any of the three.

The original choice was one out of three (choose the first door, choose the second door, choose the third door), but the second choice is one out of two (choose one door, choose the other door). The very question itself posed, “do you want to swap (option 1), or do you want to stay (option 2)” in *itself* defines the *new* set — the *new* probability, of 2 choices. Whether or not there had been 2.38 fillion dillion doors is irrespective of whether or not the new question is “stay or swap” at present.

The option between “switch or not switch” are the new options, and now has nothing to do with doors or original previous options.

The oft-cited, poorly-reasoned proof (which Wikipedia does here, is that:

The fact is, is that when you are asked to select a door in the beginning, *you have made a selection that only applies to the original terms*. The flawed reasoning operates under the circumstance of the original question imposed upon the second question — but the new question is what sets the standard for calculation, not the original. The reasoning fails to include the fact that *the announcer will not open the Car Door* AND *the announcer will not open the door you have selected*, which creates an actual probability of 50/50, because the 3rd door he reveals was never an actual option. Any version of the Monty Hall Problem that includes BOTH of these stipulations AND declares there to be a 66% chance of win by switching bears goat-like intelligence.

Rephrased, the most glaring problem is that there is actually only one chance, 100%, that the Car door is the Car Door. There is a 0% chance that Goat Door A is a Car Door, and a 0% chance that Goat Door B is a Car Door. Of the available options, three doors, you have a chance of 100 divided by 3 options. When one option is eliminated, there are now only two options. That makes the probability of 100% that the Car door is the Car door, and 0% that the Goat Door is the Car door. Now that you only have two options, the probability is 50%.

Even if there were 100 doors, as if often used to rationalize that switching makes better sense, there would STILL only be a 100% probability that the Car Door is the Car Door, and 0% for each of the 99 other Goat Doors. By the purported proof offered by Wikipedia, switching would therefore offer you a 99% chance of getting the door correct by switching, when in fact you are actually only choosing between one door or one other. Regardless of how many past choices you had before, the current question between “switch or swap” (2 options) is what sets the stakes.

Take again for instance if you were asked to select 2 of the 3 doors:

1. You pick the Car Door and a Goat Door, a Goat Door is revealed to be the third. You have a choice between your original selections, a Car or a Goat — a 50/50 shot.

2. You pick a Goat Door and the Car Door, and a Goat Door is revealed as one of YOUR doors. You have a choice between your remaining unopened door or the unselected door — a 50/50 shot.

3. You pick a Goat Door and the other Goat Door. The announcer will unfailingly reveal one of YOUR doors as a goat door. You still have a choice between your remaining unopened door or the unselected door — a 50/50 shot.

Take again for instance if you were to select *one* door and the host were to reveal that YOUR door is the Goat Door. Your probability is still a 50/50 chance of your *new choice* correct. The fact here is that you must make a *new choice*, which is identically probable to the dilemma of whether the host had NOT opened the door you first selected — because the choices were between STAY or SWAP.

Take yet again for instance if you were to select *one* door and the host were to reveal that one of the doors you had not selected to be a Goat Door. Now, you are turned around and blindfolded, and the prizes behind the doors are rotated behind the two available doors, so that you’re unsure whether the door you had selected is actually still corresponds to your original Car Door guess. The probability does not change, because there are still only two options. The chances are still 100% that one door will be correct and 0% that the other door will be incorrect, resulting in a 50% chance.

Take yet again a visual illustration of your choices: To pick between three doors is like unto throwing a dart at a rotating (a plane rotation) circle that has three equal sections defined as section 1, 2, and 3. It is revealed that the corresponding door the dart hits is your first selection, and following is revealed that one of the unhit doors is a Goat Door. The *new circumstance* is like unto throwing the dart at a new circle divided half and half into “stay” and “swap” respectively. You are not throwing a dart at a plane-rotation circle with all three options, because you are clearly not going to select the already-opened goat door. The new selection is the identical odds as whether you had not even selected one of them originally.There is no possible way to discern whether the door you have already selected is the Car Door, and LIKEWISE is there no possible way to discern whether the unselected door is the Car Door. However, it IS possible to know whether the third door is a Car Door, because it has already been revealed as a Goat Door. The options are now only between one unknown and another unknown — 50/50.

The reason the purportedly true answer gains acceptance is because it stands to reason only in one case, and not the rest. This is akin to the assertion that the scientific method is the *only way to determine truth*. In order to determine *that very statement*, you would have to establish a fact fact without using the scientific method. It’s the same “How do you know the bible is true? God says so! Where? In the bible!” (even though the bible does not say that) argument.

Wikipedia reasons, *This difference can be demonstrated by contrasting the original problem with a variation that appeared in vos Savant’s column in November 2006. In this version, Monty Hall forgets which door hides the car. He opens one of the doors at random and is relieved when a goat is revealed. Asked whether the contestant should switch, vos Savant correctly replied, “If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch” (vos Savant, 2006).*

This is also incorrect, because the actual probability of one door being 100% the Car Door and the other door having a 0% chance of being the Car Door remains true, regardless of whether the host knows or not. To Switch (option one) or not to Switch (option two) is what is being asked. You are being asked to choose between two actions, the original probability now being completely irrelevant.

Wikipedia also cites a graph showing three doors, where the prizes are actually revealed. It lists the Car Door under a 33% bracket, and brackets the two remaining doors under a collective 66% deliniation (a Venn diagram). The problem here is twofold, at a minimum. *Firstly*, when you divide 66% by two, you do not have 66% remaining. The two doors do not represent a 66% chance each that averages ([66+66]/2) to 66%, because each door is purportedly a 1-in-3 chance each, or 33% each. *Secondly*, there is not a 33% chance that the Car Door is the Car Door — it is 100% the Car Door. The other two options are 0% and 0% respectively. If you were to eliminate one of the 0% probability doors, a Goat Door, you’re left with either the 100% option or the 0% option, making it a 50% chance.

The diagram found here uses inverted teacups concealing a diamond. The diagram makes the assertion that switching your original choice when one is revealed. However, the actual choice being made is between one cup or one other cup, a 50/50 chance. The original formula does not apply since one cup has been eliminated. It’s a false positive.

Playing the game yourself and compiling the results is completely arbirtrary because at what point do you cease the experiment? If you were to select the Car Door in your original 33 selections without switching, could you simply just quit, sufficed that staying with your original guess is the 100% sure strategy? It’s the same question of asking whether a coin flipped 50 times resulting in 50 heads will next flip tails.

I hope you have begun to see reason more clearly if you had been or still are a “66-percent-chancer” in regard the Monty Hall Problem. Please feel free to discuss below =)

**EDITOR’s NOTE, 2013-APR-09** – After about 6 years of this debate in comments, I’ve revised and re-addressed my firm belief of the 50/50 position in a new blog post, 50/50 Is King: Classic Monty Hall Problem Re-Addressed and Re-Debunked, for your perusal (and a fresh slate to declare my utter buffoonery).

there is a 100% chance that your statistics are Just Plain Wrong.

If you still don’t get it after wikipedia explained it thoroughly, write a simulation, and do 1000 trials without switch, and 1000 with switch. You’ll easily see that 33/66 is correct, and 50/50 isn’t. I just did. The results of my random test:

Not switching won: 326

Switching won: 659

Your testing is flawed because you’re looking at it from the context of prior choices affecting the final choice — which isn’t what I’m saying. You have a 100% chance of getting the correct answer when you pick the correct door, and a 0% chance of getting the correct answer when you pick the incorrect door. The results of my random test that you’re a cult follower: 95% yes, 0% no, 5% probably. Anyone can quote numbers.

If you had 100 doors and did it, you’d come up with a ~99% chance of getting the correct door by switching, presumably. That demonstrates that the tactic of employing former choices into a present choice is a flawed strategy, because the present choice has the identical 1-of-2 options in both cases.

You throw a dice. You need a six to win the goose game.

Either it’s a six, in which case you have 100% chance that it’s a six. Or it’s not a six, in which case you have a 0% chance that it’s six. On average: 50%

It’s 1-of-2 options. Either you win, or you don’t win. Either you throw a six, or you don’t throw a six. 50% chance. 1-of-2 options.

Give me a yell when you are in Vegas, and we’ll have a ball! 🙂

With dice, you don’t choose, and options aren’t whittled down to two.

To see that you are wrong, all you have to do is consider that one option is to stick with your original choice. Your original choice has a 1/3 chance of being right. If you stick with it, you’ll be sticking with a 1/3 chance of being right, and a 2/3 chance of being wrong. Eliminating one of the doors can’t change the probability of your initial choice.

Eliminating one of the doors doesn’t change the probability of the original choice, correct. However, you are given a new choice between “stay” and “switch” therefore the previous choice is a completely different probability altogether, and can’t be a variable influencing the new choice.

As demonstrated in Jan-Pieter example, the empirical evidence has long been established and accepted in regards to the “Monty Hall Problem.” I hear what you’re saying though – on one hand it makes no sense, and it does appear to be a 50/50 proposition…but the numbers don’t lie. It is what it is, and regardless of our acceptance of the theory or explanation, over repeated trials in the long run you will win 66% of the time by changing.

It has long been established by those who insist that numbers don’t lie, agreed. The reason it appears to be a matter of 50/50, is that it truly is one.

Ok…I’m open minded. Help me out. Can you explain why, over repeated trials, does one win 66% of the time when changing their choice in this example?

It’s not my intent to be argumentative – I’m enjoying the discussion. There must be an explanation that I haven’t been able to wrap my mind around yet. I only ran it a few hundred times, but I’m getting the same results that Jan-Pieter did…

I just thought of something else. Perhaps this problem only creates the illusion of a 50/50 proposition afrer discounting the presence of the 3rd and invalid option.

The 3rd and invalid option represented a 1 in 3 chance originally, yet in the Monty Hall problem the invalid option is correctly identified 3 out of 3 times. When this invalid option is selected 3 out of 3 times, the other 2 options cannot represent a 1 in 3 chance each. The invalid option in effect transfers its 1 in 3 chance to the other unselected option, as the selected option has its 1 in 3 chance locked up as a result of its selection when there were still 3 supposed equal valued choices.

The structure of the Monty Hall Program creates the illusion of a 50/50 proposition when it’s pre-destined to be a 1 in 3 vs. 2 in 3 choice from the start. The remaining 2 options can’t hold a 50/50 value if a 1 in 3 option is identified 3 our of 3 times.

The reason is because there are two seperate choices being made — it’s not one single equation. For instance, if you were to have 100 doors, and were to remove one non-prize door each time, it would still boil down to a choice between staying with your choice or switching your choice, once you got down to two again. Whether you start from 100 doors (making it a 1-in-100 chance, according to the “empirical” proof) or start out with 3 doors making it 1-in-3, you’re still making the same final choice at the end, which is between staying or switching.

There’s a 66% chance you will pick a goat first.

If you pick a goat first, then switching will get you the car.

If you pick a car first, then switching will get you a goat.

Therefore, because you are most likely to choose a goat, switching will most likely get you a car.

It’s only a different equation if the prizes are shuffled after one door is opened.

There’s no “chance” you’ll pick anything. Either you do or you don’t. Summaries of accumulated previous attempts are only suggestive of those prior attempts — not future attempts. There is a zero percent chance you’ve selected a car if you pick a goat. There’s a 100% chance you’ve selected a car if you pick a car. There’s no in-between — either you have or you haven’t.

I too was skeptical of the Monty Hall theory but it does turn out to be true. You can run my Silverlight tester in your browser that runs through the trials and read the explanation for why it is true here:

http://www.blendblog.net/Blendblognet/tabid/36/EntryID/61/Default.aspx

Sean

But it’s asking you the probability of the selection being a car after one of the goat doors has been revealed, not before.

The problem with all of the “simulations” is that they are based/written using an assumption that there are still 3 choices after the first door is open.

Logically (human reasoning, not binary logic), nobody would choose the open door, so it is not a logical choice… leaving only 2 choices after the door is open.

Mathimatically, as it all of the “simulations”, the 66% proves itself because the computer doesn’t have human reasoning eliminating that third mathematical choice.

Look at it this way… if each choice has a 33.3% chance of being chosen during the first “make a choice”… or let’s look at it as each weighing 33lbs… if one door has a goat (which eliminates it as being the door with the car), the 33.3% chance that door weighed is no longer available. So now it is 33% vs 33.3%. The third door (still there but not going to be chosen holds the other 33.3%) does not give up it’s 33% to the other goat door just because it is open. So if you place 2 doors that are possible to be chosen on a scale with the same weight, the scale would balance. Balanced scales means 50/50.

PS. You are spot on with your explanation, ablestmage.

Thanks for posting this. You explain beautifully what I failed to describe in a fairly heated discussion yesterday. Forwarding it now…

Your stats is so flawed that you should apply for Randi’s $1,000,000 challenge.

You have no brain, this is a simple concept, there are 3 scenarios: you pick gda and gdb is removed,swap and win; you pick gdb and gdb is removed, swap and win; you pick car door and either gda or b is removes, swap you lose. There is a2/3 chance it is not the last scenario, therefore 2/3 chance swapping will win. I am 14 and I can grasp this

No, there are only two scenarios. There are two doors to pick from, and your option is either stay or switch. The problem still exists if there were 99 doors originally — the new scenario is a stay-or-switch choice, with none of the previous 99 choices having anything to do with the rest.

Loller, this is an argument you aren’t going to win.

I had a lengthy discussion about a year ago with Ablestmage (using a deck of cards and the odds of picking the A of Spades as an example), even did 600 trials (300 always sticking, 300 always switching) of his OWN simulation to prove the 33.3/66.7 solution. All my postings have been deleted. I also had an interesting discussion with another poster, whose name I’ve forgotten, about the boy/girl paradox (similar to MH) – these have been deleted too.

Ablestmage simply doesn’t understand statistics. and that a strategy of ALWAYS switching is 2x better than ALWAYS sticking, and is 16% better than a random stick/switch strategy (which is 50/50)

Hi Marley52. It was me that you had a conversation with about the boy/girl paradox. Funny how this all got deleted. Free speech is great but free history is better.

Hi MartinH.

I agree. Ablestmage obviously doesn’t like being wrong, LOL

There is a 1/1 chance that the posting will be deleted along with the conversation.

It all brings to mind that old arab proverb “It is no use lighting candles in blind men’s houses”

The conversation hasn’t been deleted — I changed that whole thread to “awaiting moderation” (generally used for blogs that wish to approve comments for posting before they are made public) because it became derailed. I realized the problem with “my own test” and modified the moderation status around the time I redacted the “Edit” mention of the test.

I’m not sure what your post means, will my “not deleted” comments be reinstated or are they still “awaiting moderation”?. My posts I thought pretty much categorically proved that the “it makes no difference whether you switch or stick” viewpoint is just plain WRONG! Perhaps that’s where the thread got derailed.

I notice that you’ve now removed the java simulation program that you added last December, why is that? Did you even try it out yourself?

Seeing you haven’t reinstated mine or MartinH’s comments, and you’ve removed the simulation program that you originally thought supported your 50/50 viewpoint (but which in actual fact proved the 33/66 solution). do you now finally admit that your initial assertion that it’s a 50/50 proposition whether you stick or switch is incorrect?

No, I just haven’t had time to get around to it. I’m doing a bit of research into whether it is perhaps a larger philosophical question of whether “chance” exists at all or not.

Well perhaps I can save you a bit of time – it’s just simple maths. If your want to maximize your chances of winning the car then you’re better off switching ( 2 times better than sticking and 16.7% better than guessing), there’s nothing philosophical about it

Ablestmage, you hit the nail on the head, you are trying to approach this from a standpoint of random “chance” but that isn’t the situation here. It isn’t pure chance because Monte Hall has knowledge of the goat -AND- he can’t reveal the prize until the last move. In pure chance Monte Hall wouldn’t know where the prize is either and would be able to reveal it after your intial pick, and he would 50% of the time ending the game before you even got to make a choice to stay or switch. In this scenario you would be correct that it would not matter if you switched or stayed, but that is not the scenario we are talking about on the game show. The knowledge Monte Hall has and the rules he has to follow about what he can reveal based on that knowledge completely change the equation.

So to explain it one more time, the 1 in 3 initial odds are reversed if you switch since if you pick a goat the first time (66% likely this will happen) Monte Hall HAS to pick the other goat when he does his reveal. Now if you switch you are switching to the prize guaranteed. When you switch the only time you lose is when you pick the prize at the beginning which only happens 33% of the time. Hence you win 66% of the time by switching and 33% of the time by staying.

I was skeptical too. I wrote my simulation program to prove that it was 50% odds because I was so skeptical (I posted the link above). But once I put together the simulation it became clear that the equation is not pure chance, it’s a rigged game and by switching the player rigs it to his favor.

Perhaps a better way of saying it, is the philosophical question of whether anything is “likely” as a valid concept is more at play, with an attempt to use math to explain how “likelihood” does not exist, under the assumtion that it is possible to calculate a “probability” (that is, setting aside the belief that “probability does not exist”).

Good luck with trying to use math to explain how “likelihood” (or probability) doesn’t exist, especially as probability theory is a branch of mathematics. So, when did you realize that the correct answer to the MHP wasn’t 50/50?

I didn’t understand problem before but now i do, his a clear way of understanding it.

At the beginning there is a car, a goat and a goat. The odds that you pick the car at the beginning is 1/3 while as the odds that you pick a goat is 2/3. When a goat is revealed, odds are 50% but you also need to factor that it more likely that you pick wrong at the beginning then you pick right. This information make you more likely to win if you change.

Changing give you 66.6% because, you pick the wrong answer at the start 2/3 times. So switching counteract it.

Finally someone who uses common sense, people who understand the Monty hall problem aren’t smart there just to simple minded to question it. They accept it for a truth but known subconsciously that it wrong. A lot of math problems can be wrong, because the maker miss interpreting what the variable represent.

Putting a math variable without another separate math variable without taking account of there differences create incorrect formula. Which is what the Monty hall problem is.

The MHP is proven both mathematically and logically and has been for decades. The proofs are trivial, look them up. It isn’t a matter of opinion, there’s no debate, I even think Ablestmage realises he was wrong which is why he’s now going on about the concept of ‘likelihood’ (or probability) and whether it exists or not from a philosophical standpoint.

Your last but one sentence doesn’t make any sense – I don’t think you understand maths as well as you think you do.

The only realization that has come as a result of this discussion is that there are two competing philosophies at work.

The idea that “simple math doesn’t lie” is a foolhardy gesture, in the simple case of whether a glass is half empty or half full, being simple math of halving, whereas application of philosophy when grasping the results that is what is susceptible to untruth. To suggest that results are or are not indicative of this or that, aligns according to the conclusion-drawer’s philosophy of the merit of predictive mathematics.

There are perhaps those who are so keenly astute into discerning that math /is/ capable of prediction but lack talent in detection of the subtly of a propagandized question even if explained in detail, and likewise there are perhaps those who are keenly astute into discerning language’s finer subtly to detect that a question /is/ flawed and lack the detection of predictive math’s merit even if explained in detail.

Is it that hard for someone to admit they are wrong? You were wrong by an order of magnitude, you believed that switching led to a 50/50 outcome but in actual trials the result was a 66/33% outcome. Just like if you flip a coin 10000 times it will come up heads or tails roughly 50% of the time. There is no “philosophy” that dictates this. Pure trial and error simply leads to an objective truth.

Feel free to argue that 2 + 2 = 5 and justify it via some sort of philosophical nonsense. Most of us will laugh at you the way I am right now. Alternatively, grow up and when you are incorrect about something admit it. It’s not the end of the world, in fact its an amazingly positive thing called: learning!

Glad to see that you eventually understood and can also explain the correct solution to the MHP, Congratulations !. Some people are just not prepared to admit they’re wrong.

I believe the method of your testing is what has led you to the false 33/66 solution.

In the body of the original article, I propose an alternate method of testing which has perhaps gone unnoticed, using circles and darts.

Take a circle that can spin on its center, and create three 33.333% areas on it joining at the center, spin it, and throw a dart at it. The area the dart hits will correspond to a door selection, and one goat door you have not selected will be removed. Now on a new circle which is divided in two halves, one is marked stay and the other is marked switch. Throw a dart at one of those, and that will represent your selection. The fact is that the latter circle is the only circle that matters. If you were to ask the capital of Kentucky and remove a door from the available three, you would still just be left with two options. If you were to ask to pick between “beans, understanding, and purple” and chose “airplanes” a door would be removed and you would still have two options. You could ask any conceivable question and offer any conceivable answer before hand, and a goat door would be removed and leave with you with a choice between one or the other. The original choice being “choose one of these three doors” is an ILLUSION of choice, when in fact the original choice doesn’t matter.

(edit–) The first circle of three spaces has no relevance to the second circle, because the options are always stay or switch. You could have a circle with 10 even pie slices that each have a two-door description on them, and a thrown dart indicates which 2 doors to keep in the game, with the dart being closer to one of the two doors to “stay” with, eliminating all other 8 doors, and the option would still be whether to stay or switch.

Do you propose that the very act of asking a specific question changes the probability of getting 1 out of 2 correct?

You are describing a very different problem, getting a different result and then arguing that because of this math is wrong.

Here’s the difference: the darts hit at random. Monty Hall CANNOT pick the door to reveal at random, his choice is based on what you pick initially. He cannot reveal a car until after your second choice. Because of this he HAS to make an informed choice in what he reveals to you. This inadvertantly gives you more knowledge.

Not that I think I’ll do a better job of explaining it but here we go: initially you will be wrong 66% of the time. Because Monty cannot reveal a car after your first choice (THE MOST IMPORTANT PART OF THE EQUATION) that means that in these 66% of occassions he MUST reveal the other goat. So if you switch you are 100% likely to get the car. Since your initial odds of picking a goat are 66% on the initial try and then 100% after switching you will win 66% of the time by switching.

At the time you make the first pick everything is random. But since Monty’s pick is done with not only knowledge but with rules around what he can reveal your second pick is not. In your dart metaphore each throw is random. These are very different situations.

The only evidence that you have to the “fact” that I am describing a different problem, is your word that it is. We are discussing the SAME problem, just with a different arbitrary way of arriving at the problem that you believe matters.

If you were a trickster, and it was your intent to figure out a way to convince people to believe that a 50/50 chance was more than 50/50, what would the best way to go about that? The best way, I think, and whereby this very conversation is evident, would be to give the impression that a calculation slightly related to it, performed leading up to the 50/50 throw, mattered even in the slightest to affect the 50/50 choice following it.

Your choice to stay or switch *is* random, between Stay and Switch. There are no other variables. The question preceding the Stay/Switch question bears no effect on the Stay/Switch question, and could be anything, just as long as a goat door is removed. Regardless of what you do prior to the removal of a goat door, a goat door will be removed. What if there were 4 doors, and Monty removed one door without asking you a question, and then removed another door without asking you a question?

What if you were given a choice between Heads, Tails or a combined Heads/Tail door that would be defined once your selection of the three was made? Regardless of the one you picked, and regardless of their particular name, you are left with either Heads or Tails. You now have a choice between Heads or Tails, and you’re trying to tell me that there is a 33% or 66% chance of getting heads over tails or tails over heads.

I think what you are saying is absent any knowledge of the Monty Hall theory a user would be 50/50 likely to switch versus stay. It is proven ad nauseum that if they switch they win 66% of the time if they stay they only lose 33% of the time (I think even you get that but I’m not sure, if not I honestly give up and would direct you to buy 3 cups, a toy car and two goats and have a friend run through the game with you 50 times where you switch and 50 times where you stay). So for a user who doesn’t know that it is better to switch they will have 50/50 odds of winning because 50% of the time they pick the 66% option and 50% of the time they would pick 33% odds.

But that’s missing the whole point of this discussion. Your choice to switch or stay IS NOT RANDOM once you’ve read this post and followed the numerous trials people have done (myself included). Why would you ever make the choice to stay when you know you are more likely to win when you switch? Why would you ever choose “random” when a choice made with the knowledge of switch versus stay will win more?

If a guys has a 2 headed coin and I randomly pick heads or tails I will win 50% of the time. But once I know the coin has heads on both sides I would never choose to randomly pick heads or tails. I would choose to pick heads and win 100% of the time.

It kind of comes across like you realize you were initially wrong but rather than take your new knowledge and be happy for it you are saying that if you didn’t have that knowledge in the first place you would’ve been right. So rather than condemn your initial theory you condemn knowledge in general. I guess the saying ignorance is bliss is what you are trying to prove here?

This. The choice in round 1 doesn’t give the subject ANY new information, because it is not a choice at all. The host will remove one door with a goat behind it REGARDLESS OF WHAT YOU PICK, and so the only choice you are actually making is in round 2. Obvious. The math is correct however, but it is assuming that the potentialities in round 2 are conditional on round 1, which they most certainly are not, as round 2 is the exact same choice regardless of what you pick in round 1. Round 1 is totally irrelevant. This is a mathematical paradox.

And you’re still wrong John. Get 3 playing cards and discover it yourself.

What is proven ad nauseum is a flawed calculation based on the idea that the original door-elimination phase affects the final phase of stay/switch, yes, I agree. However, the door-elimination phase is a red herring and has no bearing on the stay/switch phase. All of the 33/66 results REQUIRE the door-elimination phase in order to make sense, but those calculations are based on the idea that the door-elimination phase affects a 50/50 choice. To suggest that the stay/switch choice is 33/66, REQUIRES belief that the door-elimination phase is a genuine variable.

Perhaps “semantic fallacy” is what I’m trying to describe. One such example is this “math” word problem: “Three men rent a hotel room. Each pays $10 for a total of $30 spent on the room. The next day the hotel owner tells the three men that they over paid for the room as it only costs $25. The three men tell the owner to give them each a dollar back and he can keep two dollars. If you do the math, each man paid $9 a piece for the room for a total of $27. The owner kept $2 which brings the total to $29. Where did the other dollar go?”

The question of where “the other dollar” went is the red herring, or a misdirection in literary terms, because there is no “other dollar” missing. The men paid $25 for the room, and tipped the owner $2, for a total of $27, and the $1 they each receive back brings the total to $30. Similarly with the MHP there is no “higher probability of switching” because the choice is the same, since the beginning phase is completely irrelevant, in the same way that “The owner kept $2, for a total of $27” is false, since you need to SUBTRACT $2 from $27 instead of adding it. It presents an illusion of being relevant, but the reverse is true. In order to arrive at a total of $29, you have to falsely believe that the $2 must be added rather than subtracted, in the same way that the MHP presents the door-elimination round as a genuine variable.

.”All of the 33/66 results REQUIRE the door-elimination phase in order to make sense”. Of course they do, because the door-elimination phase is the key part of the problem, didn’t you realize that ?!

You clearly do not know what you are talking about. The stay/switch choice IS 33/66 and the proof is trivial, there is no belief required only a basic understanding of probability theory which you obviously lack.

If you RANDOMLY choose to stay or switch you will have a 50% chance of winning the car, but that would be the case even if you knew which door the car was behind, and is true of any either/or question (“Will the sun rise in the east or west tomorrow?” “Is Elvis alive or dead?” Flip a coin, you’ve got a 50% chance of being right in both cases). But why take 50% when switching gets 66.6%?.

Either A) you’re correct and every mathematician in the world, 100’s of maths textbooks and probability theory is wrong

Or B) every mathematician in the world, 100’s of maths textbooks and probability theory is correct and you’re making a fool of yourself with your refusal to accept the self-evident.

Which is it? If you flip a coin you’ll have a 50% chance of getting the right answer

You should stick to your pretentious philosophical musings, alternatively, how about a blog on whether Pythagoras’ Theory is true for all right-angled triangles or only some of them.

Man I give up. Rather than try reason with you how about this: Let’s do a Skype Video chat and run the game 50 times with the rules used by Monty Hall. We can use cups as doors and I’ll put a toy car under 1 and a toy goat under the other 2. I will be Monty Hall and you will be the contestant. You will switch each time. If you win more than 60% of the time you will come back on here and say I’m right and that you are wrong. If the results come out any closer to 50/50 than that I will do the same. If you want we can do 50 runs where you don’t switch and I will admit defeat if you win more than 40% of the time during those runs. We can come up with other side bets if you’d like to, maybe a prize for the winner or something? Oh and we can record our Skype session and post it too for proof.

If you really believe the nonsense you are spewing here you will take me up on this offer! Besides it will be fun!

There’s 2 chances of that happening though Sean: Fat Chance and No Chance. Now that’s a 50/50 choice! Ablestmage doesn’t WANT any evidence that might refute his ridiculous argument. if he did he wouldn’t have removed the computer simulation that he added about a year ago to (presumably) support his stance of “it’s 50/50”. I ran 600 trials and unsurprisingly got a ratio of approx 1:2 for stick/switch and posted the results here (which have been removed “pending moderation”).

I think he just can’t admit he’s wrong, and presents ever more nonsensical arguments, or tries to change the problem entirely, to avoid doing so. I’m not sure what type of personality that behaviour demonstrates.

I refer to previous comment on that old Arab proverb “It is no use lighting candles in blind men’s houses”

LOL. Very true. Ablestmage is ploughing a very lonely furrow, not many people still persist these days in believing staying or switching is a 50/50 option.

In fact I expect him to start back-peddling from his original position, (statements such as “The disturbingly frequent misinformed explanation is that switching gives you the greater advantage” and “Any version of the Monty Hall Problem that … declares there to be a 66% chance of win by switching bears goat-like intelligence”.)

The usual excuses, from those who refuse to admit they were wrong initially, range from the simply pathetic “He opens the door AFTER you make your choice. Sorry I thought it was BEFORE, I must’ve misread etc.” to the more disingenuous “I never said switching wasn’t advantageous, just that any choice between 2 options is by definition 50/50 “. He’s alluded to the second excuse several times in his recent posts (e.g. “Your choice to stay or switch *is* random”. No it isn’t).

He appears to have given up (sensibly) on any attempt to refute that switching is a 66.7% proposition, and now conjures up different games (involving independent spinning circles, or a door marked heads/tails), asserts they’re the same as the original MHP (which they aren’t), and then proceeds to deduce a 50/50 solution.

Oh well, what can you do, hey?

Tell you what. I’ll meet you any time, anywhere in the world. We’ll both bring $50,000. You be Monty. I’ll switch every time. If I win, you give me $45. If I lose, I’ll give you $50. We’ll play until one of us is broke.

I’ll even pay for your airfare.

I am 100% dead serious, by the way.

There is a 50/50 choice for Ablestmage either he takes you on or he does not but a 100% chance he will not.

I’ve seen some dense blog posts before but never one quite so temerarious in it’s disregard for the obvious truth. Came here via a link from the Dilbert blog comments. *shakes head at utter moron ablestmage* For real?

I looked at the Dilbert blog and saw that “AtlantaDude” appears to have come around to accepting the proven solution. I’m not holding my breath that he’ll admit the same here though. I mean it’s been 6 years since he wrote this blog entry, and it would’ve taken about 6 minutes if he’d experimented with 3 playing cards to realise he was wrong in the 1st place. Go figure.

Ok its time to prove 66%/33% wrong. The mage guy just hasn’t fully grasped how easy his TRUTH can be explained.

3 doors, you pick door one. Door three is revealed as goat. Ok, if u stay it’s33% chance of win,if u change to door 2 it’s66% chance to win, right? Was it by TOTAL RANDOM CHANCE that the car ended up behind one of the doors? Yes! Now u have stated its a 33% chance for door 1 & 66% chance for door 2 and you know this cuz the reveal of door 3 gave you “INFORMATION”…definitely! Now remove the revealed door no. 3 and leave the room. Enter next contestant. Now there are only two doors. LET ME TELL the customer that by TOTAL RANDOM CHANCE one door is a winner, the other is a loser. BUT THEN I’Ll TELL THEM THAT STINKIN LIE, YOU KNOW THE ONE…TELL THEM THERES A 33% CHANCE DOOR 1 IS A CAR AND A 66%CHANCE DOOR 2 IS A CAR. Now if they want proof for this hogwash i’ll just make door number 3 reappear & bring you back, you can explain how you picked door number one. This will give contestant 2 all the necessary “INFORMATION” THAT YOU SAID IS PROOF and will also prove that a 66%CHANCE OF goat WHEN THERE’S 3 DOORS IS ABSOLUTELY THE SAME AS 66% CHANCE OF car(s) WHEN THERE’S TWO DOORS, unless it da door we picked then the 33% chance wemains da same for car!

Yes cp “random” pick games are programmed to attach both the 33%(car) AND 66% (goat) to the ONLY the car In the 2nd part of game. Why don’t both spots have a 66%of car? What irks me is your FAKE statistics add up to OVER 100%! What you say is there’s 66%chance of a CAR door number TWO and a 66% chance for a GOAT door THREE! That adds up to 132%! Or look at switch the doors for a (not) probability 33%/33% & shux your still wrong!

All of the xplanations for the problem that I’ve heard use the words “if” & “but” extensively as “if” these words are mathematical. You lose case closed.

Btw, in my home town a man used to practical joke a lot, he’s dead now but with research I might be able to cite some stories of his. One such story, he captured a bunch of birds and airbrushed their wings various colors, then called the bird watchers society. Once a few were Verified and Word spread hundreds came to see these new species. He laughed at such so called “experts” This was but one of his trix. Yes people do these things and id bet a million there’s someone out there laughing at Monty Hall “experts”, some of them even MIT “geniuses”! Lol bwahaha thhhpppt!

Are you the same JonathanDavid who recently posted a similar bunch of gibberish on a YouTube vid of the Monty Hall Problem? You have an even poorer understanding of probability than Ablestmage has, which is saying something. Anyway as I said in my previous post, Ablestmage has, after 6 years, come to realise he was wrong and that 33%/66% is indeed the correct solution.

Why don’t YOU get 3 playing cards (2 red, 1 black) and do the experiment yourself?

I have made an updated and 50/50 re-affirming new Monty Hall post here..

Pingback: 50/50 Is King: Classic Monty Hall Problem Re-Addressed and Re-Debunked | the ablestmage press

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?

The host will randomly eliminate a door in the elimination round from among the available doors which are (a) not the car, and (b) not the door the contestant has assigned as their elimination-round door, which varies in number depending on whether the contestant has chosen the car door, but the host does not eliminate a door randomly among the total 3. The contestant’s choice of the elimination-round door is random, however, and does not impact/affect the choices of the final round.

If the contestant has chosen the prize door, then the number of available doors is 2, so randomly between either goat door.

If the contestant has chosen a goat door, then the available number of doors is 1, since the host will not eliminate the car door, nor the specific goat door the contestant has chosen.

If you mean, “does the host nonverbally indicate or imply which door the host knows to be the winning door based on his method of selecting the door to eliminate whereas one could develop a strategy based on the host’s method of selecting the door to eliminate,” I would say no, the host doesn’t do this.