Re: Odds house vs. four of a kind
Okay, let me try again.
If you play 1,000 hands of Hold'em, with the full board dealt, 423 of those hands will have one pair but no more on the board.
Given the board, there are 47*46/2 = 1,081 possible pocket cards. 9 of them make a full house. So you expect 9*423/1,081 = 3.5 full houses. Only one pocket combination makes four of a kind, but with 9 opponents you expect it to happen the same 9*423/1,081 = 3.5 times.
It is slightly more likely that one of your opponents has four of a kind when you have a full house, because the two cards in your hand that don't match the pair mean there are only 990 pocket combinations. So the chance of both things happening on the same deal, your full house losing to someone's quads, is (1,098,240/2,598,960)*(9/1,081)*(9/990) = 0.000032.
The chance of that happening 3 or more times in 1,000 hands is 1/188,394, less likely than being dealt a Jack high straight flush or better in five cards.
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