The problem is as follows: at the end of the show, a player is shown three doors. Behind one of them, there's a car for him to keep, behind the other two there are goats. Of course, the player does not know where the car is, but Monty knows. The player chooses one door. Before that door is opened however, Monty opens one of the two other doors with a goat behind it. He then allows the player to switch to the other closed door. Should the player switch?
The answer is yes. Suppose you are a "sticker". This means you point to a door, ignore what Monty does, and stick with your choice. When you chose, all three doors were equally likely, so you win the car with probability 1/3. Now consider a switcher. He chooses a door, waits for Monty to expose a goat and then switches to the other remaining door. The switcher wins if and only if his initial door had a goat behind it (think it through). How likely is it that his initial choice had a goat? Two thirds, of course. So switchers are twice as likely as stickers to win the car.
Not convinced? How about this. Suppose there are a hundred doors instead of just three. Only one has a car. You get to pick a door, then Monty opens 98 loser doors, then he allows you to switch to the other remaining closed door. Would you? Of course, it's almost sure that the car is behind the other door, and very unlikely that it is behind your original choice.
Still not convinced? Here's another try. Suppose Monty has three boxes, one of them contains a hundred dollar bill, the other two contain nothing. You have to pick. But before you do, Monty agrees to help you out: you can point to two boxes, and he will gladly combine their contents into a new box. Now, after that's done, would you go for the new box, or for the old untouched one? Of course for the new one, it is twice as likely to have the bill. Now to make it more like the original problem, suppose that Monty doesn't combine the contents into a new box, but instead throws away an empty box from among the two you pointed out to him. That's effectively the same thing, and has the same result: the other box is twice as likely to have the bill as the untouched one.
Finally, here's an /Empirical Proof.
This was the classical analysis, and it is based on an assumption that has not been fully spelled out.
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. If he is a helpful person and wants you to win, then you should always switch, because the fact that Monty offered you a second chance means that your first choice was a goat. But if Monty is mean or received pressure from his tv station, you should always stick. Because the fact that Monty offered you a second chance only means that he wants to lure you away from your correct first choice. If you don't know anything about Monty's character, it is safest to stick, because that at least guarantees a probability of 1/3 of winning the car, while switchers may end up with a zero probability of winning the car in the face of an adversary Monty. If Monty is truly devious, he could on very rare occasions open a door even if the contestant's initial choice was incorrect, just to fool future contestants. The story gets interesting...
There is one additional twist. 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? Two-thirds. 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 M. Bhaskara Rao.
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