Re: Heads up: % of the time high card wins?
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Without some more assumptions about the 2 players hands it becomes a pain to calculate, and not too useful either. So let's assume that neither player has a pocket pair, and that players share no cards. That is, between the two players there are a total of 4 different card ranks. In that case, the chance that neither player pairs by the river is:
(36 choose 5) / (48 choose 5) = 0.220166512
gm
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exactly what i was looking for.
22% seems like a low # though? That means 39% of the time one person will pair up. Thus the person with the high card has a 22+39 = 61% chance of winning?
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