Pr(9s8s7s5s4h) = C(3,1)
2 / C(52,7) =~ 14,864,950 : 1
There is only 1 combination of the five-card hand since you specified all exact suits.
Then the remaining two cards must be 3 & 2, both non-spades. There are 3 non-spade deuces and 3 non-spade 3's.
Quote:
Originally Posted by SrslySirius
1/C(52,2) * 9/C(47,2)/C(50,5)
I was about to say I partly see what you were thinking, but now I'm not sure.
1/C(52,2) is the chance of an exact preflop hand. But you don't need an exact preflop hand.
9/C(47,2) presumably represents a 3x an 2x on the Turn and River. But they don't have to be on the turn and river.
C(50,5) is the board cards but you already accounted for some of the board cards by saying C(47,2).
Furthermore, breaking the cards into sections e.g. multiplying C(52,2)*C(50,5) makes for a larger number than C(52,7) and all you actually care about is the 7-card combo, not where the cards occur (preflop, flop, etc).
You could do it that way, but then you'd need to count more possibilities in the numerator since you've counted more in the denominator. You'd also need to fix your denominator. But it's an unnecessarily masochistic approach when order of cards doesn't matter.
For that matter you could go a step further and use permutations everywhere, which count order of each individual card (e.g. order of flop cards would count) as opposed to just sectioning them into 4 ordered groups (preflop, flop, etc). The order in the denominator would again cancel out with the order in the numerator as long as you were consistent. This would be easier than counting partial orders like you were trying to do.
Last edited by heehaww; 04-19-2014 at 05:28 PM.