The Monty Hall Problem is a mathematical game, which is counter-intuitive to our common sense, and I’ve been trying to understand the math behind it. If you pick a door hoping to get a car, and the Gameshow Host opens another door to reveal a goat, and he asks if you would like to switch your choice, the math says switching is 2/3 more likely to win a car than remaining on your first choice. WTF!? Trust the math…

If we use X to represent winning a car, and O to represent winning a goat, then we can use a diagram to show the possibilities dependent on choices.

1 X O O

2 O X O

3 O O X

Say you choose door 1, and he shows door 3 contains a goat. So the possibilities are XOO, or OXO. But if he shows door 2 has a goat, the possibilities are XOO, or OOX. Common sense would tell us that whichever door he chooses, our probability of winning should be 50%. But look carefully,

2 X O O O X O

3 X O O O O X

becomes probabalistically:

O X O

X O O

O O X

Stay and win = 1/3

Switch and win = 2/3

Thus, the possibility XOO (that you have already picked a winning door), cancels out because it is only one possibility. The limiting factors of the game, that the Host cannot open your door, and he cannot show a winning door, mean that you can infer if there was a car in the door he didn’t pick, it limited him to choosing the door he did.

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