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Old 08-12-2003, 01:19 AM
Custodian Custodian is offline
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Default 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
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Old 08-12-2003, 03:18 AM
BruceZ BruceZ is offline
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Default 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.
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Old 08-13-2003, 01:40 PM
Custodian Custodian is offline
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Default 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%
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