Simple hold em odds question

[Gavagai]

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]

Yes.

1 - C(42,2)/C(47,2) = 20.35%

[Gavagai]

Cool thanks. Care to explain what C is and the calculation "C(42,2)/C(47,2)" ? [img]/images/graemlins/smile.gif[/img]

[LetYouDown]

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]

Thanks