View Full Version : Simple hold em odds question
Gavagai
09-12-2005, 12:19 PM
Am I right in thinking that if you flop 5 outs in hold em, you have about a 4:1 shot of hitting one of them by the river? And is this the correct way to work it out?:
1-((42/47)*(41/46)) = 0.20
Thanks,
Gavagai
LetYouDown
09-12-2005, 12:31 PM
Yes.
1 - C(42,2)/C(47,2) = 20.35%
Gavagai
09-12-2005, 12:38 PM
Cool thanks. Care to explain what C is and the calculation "C(42,2)/C(47,2)" ? /images/graemlins/smile.gif
LetYouDown
09-12-2005, 12:42 PM
Essentially the same thing you did. C = Combinations. Basically determining the number of boards that don't contain any of your outs, over the total number of possible turns/rivers, and then subtracting from 1.
C(47,2) is the number of ways to choose 2 cards from 47. Order is irrelevant.
C(N,R) = N!/[(N-R)! * R!]
C(47,2) = 47!/[(47-2)! * 2!]
C(47,2) = 1081
Gavagai
09-12-2005, 12:46 PM
Thanks
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