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.
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).