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 (N1) times. Cheers,
Mark
Best wishes, Mark
