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Probability of aces in hold'em
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08-12-2003, 03:18 AM
BruceZ
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Join Date: Sep 2002
Posts: 1,636
Re: Probability of aces in hold\'em
In an X handed HE game where X = 2 through 10, what is the probability that Y aces are out where Y = 0 through 4?
For exactly Y aces:
C(4,Y)*C(48,2X-Y) / C(52,2X)
For at least Y aces, sum this from Y to 4.
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