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Old 06-15-2005, 11:37 AM
jason1990 jason1990 is offline
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Join Date: Sep 2004
Posts: 205
Default Re: Running It Twice

Call the players A and B. The key observation is this: the probability that player A wins the second run is the same as the probability that he wins the first run. What may confuse some people is this: the conditional probability that he wins the second run, given that he won the first or given that he lost the first, is something different. But before the runs take place, the probability he wins the second run is exactly the same as the probability he wins the first.

So let P be the size of the pot, let X=1 if player A wins the first run and 0 otherwise, and let Y=1 if player A wins the second run and 0 otherwise. Then the amount that player A wins is

X*P/2 + Y*P/2

and his EV is

E[X*P/2] + E[Y*P/2] = [P(X=1)+P(Y=1)]*P/2
= [2*P(X=1)]*P/2
= P(X=1)*P.

On the other hand, if they only run it once, he will win X*P and his EV will also be P(X=1)*P.
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