Re: Probability of Flush Given 3 clubs on flop
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Seems right:
P(first player has flush) = C(10,2)/C(49,2) = 45/1176
P(1st and 2nd player has flush) = [C(10,2)/C(49,2)][C(8,2)/C(47,2)] ~ 0.0009911457645 (call this p)
P(1st, 2nd and 3rd player has flush) = p x C(6,2)/C(45,2) = p/66
P(1st, 2nd, 3rd and 4th player has flush) = p/66 x C(4,2)/C(43,2) = p/9933
Using inclusion-exclusion, the probability is approximately
7(45/1176) - C(7,2)p + C(7,3)p/66 - C(7,4)p/9933
~0.2475652.
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Very cool, I didn't even know what inclusion-exclusion was prior to your post (or if I did, I had forgotten about it in the many years since I last did stuff like this). I did this the hard way, calculating the probability that no one had a flush (it would be more effort than it's worth to try to post the formula I came up with) and subtracting from 1.
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