#1
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5 handed, 5 pocket pairs dealt-- odds of no flopped sets?
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% |
#2
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Re: 5 handed, 5 pocket pairs dealt-- odds of no flopped sets?
[ QUOTE ]
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% [/ QUOTE ] C(32,3) / C(42,3) =~ 42.3% |
#3
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Re: 5 handed, 5 pocket pairs dealt-- odds of no flopped sets?
o crap, i do not know why i did (32,2). thanks, stupid error on my part.
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