# Re: Monty Hall

From: Georghiu, Christos (cg196@soton.ac.uk)
Date: Fri Feb 21 1997 - 12:08:06 GMT

Think about what the statement "there is a 1/3 probability of the first
door picked containing the prize" actually means, i.e. that in one out
of three occasions it will be the winning door. What do you suggest
happens on the opening of the empty door. There is no way this can
alter the probability of the prize being under the one picked since it
is already either there or not there, the probability of it being under
this door therefore must remain the same as before (1/3) you are
provided with the extra imformation, and as such the probability of it
being under the other door (the one you'd switch to) must be 2/3.

Probability of winning if you stick with original door;

pick 2, prize is under 1, you lose.
pick 2, prize is under 2, you win.
pick 2, prize is under 3, you lose.

pick 1, prize is under 1, you win.
pick 1, prize is under 2, you lose.
pick 1, prize is under 3, you lose.

pick 3, prize is under 1 , you lose.
pick 3, prize is under 2, you lose.
pick 3, prize is under 3, you win.

3/9 (1/3) chance of winning.

Probability of winning if you switch;

pick 1, prize is under 1, you lose.
pick 1, prize is under 2,
door 3 is opened and you are told it's empty, switch to 2, you win.
pick 1, prize is under 3,
door 2 is opened, you are shown it's empty, switch to door 3 you win.

pick 2, prize is under 1, door 3 is opened, you switch to 1, win.
pick 2, prize is under 2, you lose.
pick 2, prize is under 3, door 1 is opened, you switch to 3, win.

pick 3, prize is under 1, door 2 is opened, you switch to 1, win.
pick 3, prize is under 2, door 1 is opened, you switch to 2, win.
pick 3, prize is under 3,
you are shown either of the two empty doors, switch to other and lose.

i.e. 6/9 chance of winning or 2/3.

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