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My virgin run in the probability forum.
Is this an acceptable estimate for my AQo being dominated in a 10-handed holdem game?
With 50 cards still out, there are 1225 two-card combinations available. (50*49/2*1) = 1225
AA 3 combinations
KK 6
QQ 3
AK 12
for a total of 24 combinations that dominate me....
24/1225 = .01959
multiplied by 9 = .1763 = 17.63%
Close enough? I've read a couple posts today that mention multiplying by 9 doesn't account for the chance of more than one dominating hand....is there a way to estimate how much I could be off? Or is it negligible?
Thanks for any input....
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Welcome to the forum.
Your approximation is close, but it actually over counts the times that more than one player holds these hands. You can improve this approximation to be as accurate as you like by using the
inclusion-exclusion principle to compute successive terms which converge to the exact result. Since the succeeding terms get smaller, you can see how accurate you are by computing the next term. The second term would correspond to summing the C(9,2) probabilities of 2 players having these hands. The first two terms give:
9*24 / C(50,2) -
C(9,2)*(3*13 + 6*13 + 3*21 + 12*13) / C(50,2) / C(48,2)
=~ 16.76% or 5-to-1.
The terms in the second line are:
1st player AA (3 ways); 2nd player KK (6 ways), QQ(3 ways), AK(4 ways) = 13 ways
1st player KK (6 ways); 2nd player AA (3 ways), KK (1 way), QQ (3 ways), AK (6 ways) = 13 ways
1st player QQ (3 ways); 2nd player AA (3 ways), KK (6 ways), AK (12 ways) = 21 ways
1st player AK (12 ways); 2nd player AA (1 way), KK (3 ways), QQ (3 ways), AK (6 ways) = 13 ways
So now we know the answer lies between 16.76% and the 17.63% that you computed. The exact answer requires 4 terms since a maximum of 4 opponents can hold these hands. The appendix of Ken Warren's book gives this result as 5.0-to-1 once you apply the correction to the error I described
here, so the third and fourth terms should be neglible which is typical for these problems.