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 Mark Perakh's Web Site

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Title Author Date
3 Doors (the Monty Hall show) Perakh, Mark Nov 06, 2002
Hi, Simon David. Thanks for your message. I am glad that you have figured
it
out. Of course, there is also a rigorous mathematical way to prove it, but
since my article was written for a general audience, I did not use it. Here
is a brief rendition of a formal proof. Denote the doors A, B, and C. P(X)
is probability of X being the winning door. Obviously P(A)=P(B)=P(C)=1/3 and
P(A)+P(B)+P(C)=1. Assume A was chosen. Then P(A)=1/3 and
P(~A)=P(B)+P(C)=2/3; Assume B is opened and found empty. Now P(B)=0, hence
P(A)+P(C)=1. Since P(A)=1/3, P(C)=2/3. QED. Instead of 3, any number N of
doors can be used, all of the above remains valid by replacing 3 with N, so
changing the choice increases probability of winning (N-1) times. Cheers,
Mark

Best wishes, Mark

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