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I have a doubt concerning a question about the Monty Hall Three-Door Puzzle, in probability. I found this problem in Rosen's "Discrete Mathematics and Its Applications".

The Monty Hall Three-Door Puzzle: Suppose you are a game show contestant. You have a chance to win a large prize. You are asked to select one of three doors to open; the large prize is behind one of the three doors and the other two doors are losers. Once you select a door, the game show host, who knows what is behind each door, does the following. First, whether or not you selected the winning door, he opens one of the other two doors that he knows is a losing door (selecting at random if both are losing doors). Then he asks you whether you would like to switch doors. Which strategy should you use? Should you change doors or keep your original selection, or does it not matter?

First of all, before I ask my specific doubt:

I understand that the best strategy is switching doors, because the probability that the initially chosen door is incorrect is high (2/3); therefore, it is most probably not the winning door. So, after the host opens a door (which he knows is a losing door), the probability that the prize is in the other closed door (and not in the initially chosen one) is higher (2/3).

Now, the specific question which I want to ask (found in Rosen's book):

Explain what is wrong with the statement that in the Monty Hall Three-Door Puzzle the probability that the prize is behind the first door you select and the probability that the prize is behind the other of the two doors that Monty does not open are both 1/2, because there are two doors left.

When the contestant chooses one door (before the host opens a door), the probability that it has the prize is 1/3. But, when the host opens one door (that he knows is a loosing door), the possibilities of where the prize can be are reduced by one, because now the contestant knows that the prize can only be either on the chosen door, or on the closed door. So, it seems reasonable to think that the probability that the prize is in any one of the two remaining doors is 1/2.

But this reasoning is apparently wrong. Can any one help me understand why?

Thank you in advance.

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A new contestant that arrives at the game with no information about what door was chosen by the first contestant would indeed have 1/2 chance of choosing the right door. But the initial contestant is not picking between two doors; he is really picking between the original door, and "the best prize behind the other two doors" (if he switches). – Arturo Magidin Feb 5 '12 at 23:40
See also this, and this. – Arturo Magidin Feb 5 '12 at 23:41
You are in good company. There is a story, possibly even true, that when Albert Einstein was asked the question, he first gave the answer $\frac{1}{2}$ for the probability. – André Nicolas Feb 6 '12 at 6:15
@AndréNicolas: Most certainly not true; Einstein died 20 years before the first appearance of the problem in print. The story that Erdős didn't trust the resolution until he saw simulation evidence is more chronologically plausible. – Henning Makholm Jan 3 '15 at 18:24
up vote 5 down vote accepted

The reasoning "because there are two doors, the probability that the prize is behind either is $1/2$" depends on an unspoken premise that there is no reason to distinguish between the two doors.

But the situation is not really that symmetric. One of the two doors were chosen by the contestant without any special information. The other was chosen by the host, under the particular restriction that it and the contestant's choice must not both be non-winning.

Because the two doors were not chosen under the same conditions, the situation is not symmetric between them, and hence the argument that they should be ascribed the same chance of winning is not valid.

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This is true, I think I understand this better now. The problem was that dividing the favorable outcomes by the possible outcomes only works if all outcomes have the same chance of happening. – anonymous Feb 7 '12 at 17:25

Here's the way I like to think about it. When you choose the door, you have 1/3 chance of being right. If you have the opportunity to choose 2 doors, you have 2/3 chance of being right. By offering you to switch, Monty effectively offers you to choose 2 doors rather than 1 (the two doors are the closed one, and the one he has already opened). Thus, you have 2/3 chance of being right if you switch.

Extra: Many people like to give the multiple doors example. Let's say that you have 100 doors instead of 3. You choose one of them and then Monty opens the other 98 doors, leaving one closed and giving you the chance to switch. Most people would feel that there's something fishy going on with that door — why didn't he open that specific door? So most people would now intuitively switch.

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could someone clarify or correct my reasoning?

i've found the concept behind mh problem is clear when changing the scale. for example, instead of three doors, say, use $100$ doors. $99$ goats are behind 99 doors and the car is behind a single door. obviously you have a $\frac{1}{100}$ chance of picking the car. in other words, the odds are slim you picked the winning door. now monty, who knows where the car is located, opens the $98$ doors that all hide goats, leaving your door, which you choose with $\frac{1}{100}$ chance of succeeding, and one other door. he asks you to pick again. clearly your odds on the second attempt are much better than $\frac{1}{100}$.

what is at fault with my reasoning? the crux of my argument depends upon the $98$ doors being revealed before your second choice.


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The naive answer to original Monty Hall is you go from $\frac 1n$ to $\frac 12$ when $n=3,$ so of course for $n=100$ you don't still have $\frac 1n$ after the $98$ doors open--you have $\frac 12$ chance, which is much better than $\frac 1{100}$ (but is still wrong). – David K Nov 7 '14 at 13:57

I like to think of the host combining the two boxes that the contestant has not picked and saying that the contestant can walk away with the contents of both the boxes that were not picked. Almost equivalent if the losing boxes are empty and with negligible difference in the goat variation (though I daresay it matters to the goat).

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