RE: The role of probability in solving the Monty Hall problem
March 13, 2016 at 8:19 am
(This post was last modified: March 13, 2016 at 8:21 am by robvalue.)
Er... what?
I don't understand you. Yes, this whole thread has been explaining why you change.
The only assumption I make is that the prize has initially been selected randomly to be behind one of the doors.
If I chose wrong initially, with 2/3 chance, I will guarantee to win if I change.
If I chose right initially, with 1/3 chance, I will lose.
Chances I win = (2/3) * 1 + (1/3) * 0 = 2/3
It's how you calculate the mean of a probability distribution.
Sum(i=1..n)[i*Prob(X=i)]
I don't understand you. Yes, this whole thread has been explaining why you change.
The only assumption I make is that the prize has initially been selected randomly to be behind one of the doors.
If I chose wrong initially, with 2/3 chance, I will guarantee to win if I change.
If I chose right initially, with 1/3 chance, I will lose.
Chances I win = (2/3) * 1 + (1/3) * 0 = 2/3
It's how you calculate the mean of a probability distribution.
Sum(i=1..n)[i*Prob(X=i)]
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Index of useful threads and discussions
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Quickstart guide to the forum