The problem is as follows: At the end of the show, a player is shown three doors. Behind one of them, there's a prize for him to keep, while the other two contain goats (signifying no prize to be won). Although the show host knows what is behind each door, of course the player does not. After the player makes a first choice, Monty opens one of the two other doors, revealing a goat. He then offers the player the option to either stick with the initial choice or switch to the other closed door. Should the player switch?
The classical answer to this problem is yes, because the chances of winning the prize are twice as high when the player switches to another door than they are when the player sticks with their original choice. This is because upon the original choice, the player has only a 1/3 chance of choosing the door with the prize; this probability does not change when Monty opens a door with a goat. Hence the chances of winning the prize are 1/3 if the player sticks to their original choice, and thus 2/3 if the player switches.
Instead of one door being opened and thus eliminated from the game, it may equivalently be regarded as combining two doors into one, as a door containing a goat is essentially the same as a door with nothing behind it. In essence, this means the player has the choice of either sticking with their original choice of door, or choosing the sum of the contents of the two other doors. Clearly, the chances of the prize being in the other two doors is twice as high.
It may also be easier for the reader to appreciate the result by considering a hundred doors instead of just three, with one prize behind only one of the doors. After the player picks a door, Monty opens 98 doors with goats behind them. Clearly, there's now a very high chance (precisely 99/100) that the prize is in the other door Monty did not open.
See Empirical proof of the Monty Hall problem for a Perl program which demonstrates the result.
The classical answer presented above makes two assumptions that are rarely made explicit:
If one of these assumptions is violated, the answer is different. In the first case, it could be that Monty does not always open a losing door. Maybe in some shows he does, and on other occasions he does not, he simply gives the contestant whatever is behind his first choice door. If that is the case, it all depends on Monty's character:
In the second case, Monty may open an unpicked door at random, rather than always opening a door with a goat behind it. In this case, if Monty happens to open a door with a goat behind it, then both switcher and sticker have a 50% chance of winning, so it doesn't matter what you do. This is because a correct initial guess means that Monty is certain to open a door with a goat behind it, whereas if the initial guess was incorrect there is only a 50% chance that Monty will open a door with a goat behind it.
In the original game show, there were in fact two contestants. Both of them chose a door; they were not allowed to choose the same one. Monty then eliminated a player with a goat behind their door (if both players had a goat, one was eliminated randomly, without letting the players know about it), opened the door and then offered the remaining player a chance to switch. Should the remaining player switch?
The answer is no. The reason: a switcher in this game will lose if and only if either of two initial choices of the two contestants was correct. How likely is that? Twothirds. A sticker will win in those 2/3 of the cases. So stickers will win twice as often as switchers.
There is a generalization of the original problem to n doors: in the first step, you choose a door. Monty then opens some other door that's a loser. If you want, you may then switch your allegiance to another door. Monty will then open an as yet unopened losing door, different from your current preference. Then you may switch again, and so on. This continues until there are only two unopened doors left: your current choice and another one. How many times should you switch, and when, if at all?
The answer is: stick all the way through with your first choice but then switch at the very end. This was proved by Bapeswara Rao and Rao.
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