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Title Author Date
3 Doors (the Monty Hall show) Simon David, Nesa Nov 06, 2002

I still had doubts, so i ran 2 more simulations... one where first choice was -always- kept.. and the other simulation where the choice was -always- changed. In the first simulation, the chance of winning was just 33% approx, and in the second simulation, the chance of winning was 66% approx. So i guess you're right... the data has spoken..

After that i've been thinking about the problem, and now understand why it is so... it's MUCH easier to understand this problem, if instead of 3 doors, we extrapolate and use say... 10 doors. (Participant first chooses a door, then the compere opens 8 doors.... the remaining door is almost sure to be the winning door, therefore participant has almost 100% chance of winning if the first choice is changed.) This 10 door game better helps to understand the odds involved in this type of game. If you use this example, i'm sure ppl who doubt will see why the game works the way it does...

Thanks & best regards... :)
Simon David

Related Articles: Improbable Probabilities

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

Related Articles: Improbable Probabilities