View Full Version : Flopping Multiple Sets..
sekrah
07-15-2005, 11:54 AM
2 pocket pairs going to the flop.. What are the odds of both of these players flopping a set.
3 pocket pairs going to the flop.. What are the odds of 2 of the 3 players flopping a set? What are the odds of ALL 3 players flopping a set?
Thanks guys, you're great!
LetYouDown
07-15-2005, 12:28 PM
Someone correct me if I'm wrong.
Say one player holds 5-5 the other 10-10. The probability of the flop coming 5-10-X where X is not a 5 or a 10:
(C(3,1) * C(3,1) * C(44,1))/C(48,3)
396/17296 = 2.29%
Say one player holds 5-5, another 10-10 and another Q-Q. The probability that 2/3 will flop a set is:
5-10-X where X is not a 5, 10 or Q = (C(3,1) * C(3,1) * C(40,1))/C(46,3) = 360/15180 = 2.372%
5-Q-X where X is not a 5, 10 or Q = 2.372%
10-Q-X where X is not a 5, 10 or Q = 2.372%
So for 2 out of 3 flopping a set, I get 1080/15180 or about 7.11%.
For all three flopping a set, I get 27/C(46,3) or .178%
Cobra
07-15-2005, 03:01 PM
LetyouDown,
I believe you have a couple of errors.
(C(3,1) * C(3,1) * C(44,1))/C(48,3)
Should be C(2,1)*c(2,1)*44/c(48,3) = 1.02%
The probability of two people getting exactly two sets if three people have pairs and see the flop is:
C(3,2)*c(2,1)*c(2,1)*40/C(46,3) = 3.16% That is three ranks choose two, each rank has two cards choose one, and there are 40 misses.
Cobra
LetYouDown
07-15-2005, 03:04 PM
Yeah, that was about as glaring as it possibly could be. Thanks for posting that. I dunno what the hell I was thinking. I think I was just reading a post about flopping trips with only one of your hole cards and my math went right out the damned window. Thanks though.
LetYouDown
07-15-2005, 03:26 PM
Corrected (I hope):
Say one player holds 5-5 the other 10-10. The probability of the flop coming 5-10-X where X is not a 5 or a 10:
(C(2,1) * C(2,1) * C(44,1))/C(48,3)
176/17296 = 1.02%
Say one player holds 5-5, another 10-10 and another Q-Q. The probability that 2/3 will flop a set is:
5-10-X where X is not a 5, 10 or Q = (C(2,1) * C(2,1) * C(40,1))/C(46,3) = 160/15180 = 1.05%
5-Q-X where X is not a 5, 10 or Q = 1.05%
10-Q-X where X is not a 5, 10 or Q = 1.05%
So for 2 out of 3 flopping a set, I get 480/15180 or about 3.16%%.
For all three flopping a set, I get 8/C(46,3) or .053%.
This seem right now or am I still fubaring it?
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