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3

Frisbee's answer is correct. But it only answers for eight players (although the other numbers are hidden in that table). Since the OP asked for 8-10, I will add the answers for nine and ten players. The odds of an ace not occurring in the next x cards dealt (after you're dealt two aces) is ( 48!/(48-x)! )/( 50!/(50-x)! ). That's the probability that no one ...


1

I have been struggling over the same example for the same reason. Let P(A) = Prob A has nuts = 0.2, P(A') = Prob A has dead hand = 0.8, P(B) = Prob A bets, P(A'|B) = Prob A has dead hand given A bets P(A|B) = Prob A has nuts given A bets Then P(A|B) = [P(A)P(B|A)]/[P(A)P(B|A)+P(A')P(B|A')] = (0.2*1)/(0.2*1 + 0.8x) and P(A'|B) = ...



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