View Full Version : 5 handed, 5 pocket pairs dealt-- odds of no flopped sets?
djk123
11-17-2005, 10:59 PM
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%
BruceZ
11-17-2005, 11:39 PM
[ 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%
djk123
11-17-2005, 11:48 PM
o crap, i do not know why i did (32,2). thanks, stupid error on my part.
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