Probability of aces in hold\'em
 

I'm too far removed from college statistics class to know how to do this. I'm collecting my thoughts around lower limit hold'em after some serious variance recently.

In my analysis of KK as a starting hand, I'd like to figure out the following data poitns:

In an X handed HE game where X = 2 through 10, what is the probability that Y aces are out where Y = 0 through 4?

Thanks!
Scott C

Re: Probability of aces in hold\'em
 

In an X handed HE game where X = 2 through 10, what is the probability that Y aces are out where Y = 0 through 4?

For exactly Y aces:

C(4,Y)*C(48,2X-Y) / C(52,2X)

For at least Y aces, sum this from Y to 4.

Re: Probability of aces in hold\'em
 

Thanks, Bruce. The results are below if anyone is interested.


Players / Prob of at least Y aces
2 28.13% 2.57% 0.07% 0.00%
3 39.72% 6.08% 0.35% 0.01%
4 49.86% 10.72% 0.94% 0.03%
5 58.66% 16.25% 1.94% 0.08%
6 66.24% 22.45% 3.43% 0.18%
7 72.73% 29.11% 5.48% 0.37%
8 78.24% 36.04% 8.12% 0.67%
9 82.87% 43.08% 11.38% 1.13%
10 86.72% 50.07% 15.26% 1.79%