Re: 5 handed, 5 pocket pairs dealt-- odds of no flopped sets?
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If in a 5 handed game 5 different pocket pairs were dealt, would this be the way to go about calculating the odds of no one flopping a set?
( nCr(10,0) * nCr(32,2) )/ ( nCr(42,3) )
which comes out to equal : 4.3206%
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C(32,3) / C(42,3) =~ 42.3%
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