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slimpikkens
11-25-2004, 07:14 PM
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....

BruceZ
11-26-2004, 08:20 AM
[ QUOTE ]
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....

[/ QUOTE ]

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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Board=probability&amp;Number=417 383) 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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=1121303&amp;page=&amp;view=&amp;s b=5&amp;o=&amp;vc=1), so the third and fourth terms should be neglible which is typical for these problems.

slimpikkens
11-26-2004, 05:08 PM

Cobra
11-26-2004, 07:58 PM
I am new to the probability area as well but am trying to study it. I am not sure when you can use multiplication and skip inclusion-exclusion.

Why couldn't you say the chance of no one having a dominating hand is

c(50,2)-24/c(50,2) and then raise this to the ninth power for eight other people.

This gives you the percentage no one dominates, subtract 1 and you have the percentage of hands that at least one person dominates.

This comes out to .16312 percent, I am confused at why there is a difference.

Thank you,

Cobra

BruceZ
11-26-2004, 08:21 PM
[ QUOTE ]
I am new to the probability area as well but am trying to study it. I am not sure when you can use multiplication and skip inclusion-exclusion.

[/ QUOTE ]

You can just multiply and skip inclusion-exclusion only when the events are mutually exclusive, meaning that only 1 player can have the hand. An example would be the probability of an opponent holding AA when you also hold AA. Since only one opponent can have AA with you, you can just multiply the number of opponents by the probability that just one opponent has it, N*6/1225. Note that we are actually adding the N probabilities in this case to compute the probability of a union. If more than one opponent can have the hand, then this would double count the times that 2 players have the hand, triple count the times 3 players have it, etc.

[ QUOTE ]
Why couldn't you say the chance of no one having a dominating hand is

c(50,2)-24/c(50,2) and then raise this to the ninth power for eight other people.

This gives you the percentage no one dominates, subtract 1 and you have the percentage of hands that at least one person dominates.

This comes out to .16312 percent, I am confused at why there is a difference.

Thank you,

Cobra

[/ QUOTE ]

Now this method is not exact because the player's hands are not independent. That is, the event of one player having or not having a dominating hand changes the probability of the other players having a dominating hand, so you cannot simply multiply these probabilities together. You must instead multiply the conditional probabilities that each player does not have the hand given that other players do not have the hand. Your (1225-24)/1225 only applies to the first player, while the remaining conditional probabilities will be slightly different.

Cobra
11-26-2004, 08:53 PM
Thank you Bruce, that was a great explanation.

Cobra