Thread: odds of two players flopping the same trips, 6 players
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Old 08-03-2005, 08:49 AM
LetYouDown LetYouDown is offline
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Default Re: odds of two players flopping the same trips, 6 players
Well, the way you phrased it...I'm not sure that the number of people dealt in matters. If I hold A-J and you hold J-4...the odds that the flop comes J-J-X where X is not A or 4 is:

[C(2,2) * C(40,1)]/C(48,3) = 0.2312673% or about ~431 to 1
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