[Writeup stolen from Bai]
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The Monty Hall Three-Door Puzzle (from an old television
game show called ``Let's Make a Deal''):
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* Suppose you are a game 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. 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?
* Solution: The probability you select the correct door (before the host opens
a door) is 1/3, since the three doors are equally likely to be the corret door.
The probability this is the correct door does not change once the game show
host opens one of the other doors, since he will always open a door that the
prize is not behind. The probability that you selected incorrectly is the
probability the prize is behind one of the two doors you did not select.
Consequently, the probability you selected incorrectly is 2/3.
If you selected incorrectly, when the game host opens a door to show you that
the prize is not behind it, the prize is behind the other door. You will
always win if your initial choice was incorrect and you change doors.
So, by changing doors, the probability you win is 2/3. In other words, you
should always change doors. This doubles the probability that your will win.