Re: Three sets on the flop?
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Whats the probability of 3 sets on the flop given that there are 3 pairs going into the flop? (I get .057%)
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2*2*2/C(46,3) =~ 0.0527%
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Whats the probability of 3 sets on the flop at a 9 handed table before the pockets are dealt? (ie the probability of 3 pocket pairs being dealt AND one of each being dealt on the flop?
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Multiply the above probability by C(9,3)*[13*6/C(52,2) * 12*6/C(50,2) * 11*6/C(48,2)] to give about 1 in 111,666. The expression in [] is the probability of 3 specific players being dealt different pairs. Note that multiplying by C(9,3) ways to choose the 3 players is exact because only a single group of 3 players can all have sets at the same time, so the C(9,3) groups of 3 players are mutually exclusive. Also note that this is only exact for computing the probability of 3 sets, and multiplying by C(9,3) above does NOT give the exact probability of getting 3 pairs since these are not mutually exclusive, and we don't need to compute the exact probability of 3 pairs to compute the exact probability of 3 sets.
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