Re: Odds of set-over-set on flop?
Well since the odds of flopping a set for one player is .11 the odds of 2 doing it would be about .11*.11=.012, but this isn't exactly correct because the second player has slightly better odds. So the correct calculation would be:
C(2,1)*C(2,1)*C(46,1)/C(48,3)=.106 or ~1.1%
Note: that calculation includes the chance that one one person flops a set and one flops quads, if you didn't want that you can replace the number 46 with 44, but the answer would still be the same.
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