Probability of an overpair with # opponents -- Whats the Formula?

[etgryphon]

Hey all I try to do a search on this, but I don't know what the title would be...

Here is my stab at it and tell me if it is off...
P = probablity
x = Rank of your PP

This is finding the probability that your opponents DO NOT have a higher PP that you...

P = product of [((13-x)*4/(50-2n)*(((50-2n)-3)/((50-2n)-1))) + (((x-1)*4)/(50-2n))] where n = 0...# of opponents

I hope this is clear and makes sense to you all...or if someone has already written this then please point me in the right direction...

Thanks,

-Gryph

[gaming_mouse]

Gryph,

It's not clear at all what you are asking. Does PP mean pocket pair? Or just the highest card in your hand? Please clarify.

gm

[etgryphon]

Sorry PP stands for Pocket Pair...

I'm look to answer the question if you are dealt JJ what is the probability that one of your opponents has QQ-AA if you have 3 opponents or 4 or n-number of opponents

-Gryph

[gaming_mouse]

Okay.

There are 6 ways to make QQ. Same for KK-AA. So 18 hands total that you are worried about. There are 50 cards total left. So against 3 opponents, the chance that at least one person has QQ-AA given that you have JJ is:

Note:
50 choose 2 = 1225
48 choose 2 = 1128
46 choose 2 = 1035
44 choose 3 = 946

1 - ((1225-18)/1225)*((1128-18)/1128)*((1035-18)/1035)= 0.047

So about 5% with 3 oppos. For 4 oppos, it's about 6.5%:

1 - ((1225-18)/1225)*((1128-18)/1128)*((1035-18)/1035)*((946-18)/946)= .065

The same method can be used to solve for any number of players.

gm

gm

[etgryphon]

GM:

Thanks a lot...I think I need to simplify my thinking in these math problems...

-Gryph

[Cobra]

I did figure out an excel computation for at least one person holding a higher pocket pair than you have. I used inclusion-exclusion with three terms to get an approximation.

N = Number of opponents
X = Number of pair above yours(You have JJ, X = 3)

First term - 6X/C(50,2)
Second term - (36*Power(X,2)-30x)/c(50,2)/c(48/2)
Third term -

(216*Power(X,3)-540*Power(X,2)+324x)/c(50,2)/c(48,2)/c(46,2)

Take the first term and multiply by # opponent combo's, etc.

Term 1 * c(N,1) - Term 2 * c(N,2) + Term 3 * c(N,3)

This will give you the proboability that at least one person will hold a pocket pair higher than you with N number of opponents.

[memphis57]

Seems to me you should base the calc on the number of players who saw the deal times the probability that they would still be in the hand, rather than just the number of current opponents. Thus, in the case where you hold JJ, say its on the turn, and say it's a full 10-man table, you would want to use 9.9 or 10 for your number of opponents since it's extremely unlikely anybody who got that deal would have dropped out of the hand.

[BruceZ]

[ QUOTE ]
Okay.

There are 6 ways to make QQ. Same for KK-AA. So 18 hands total that you are worried about. There are 50 cards total left. So against 3 opponents, the chance that at least one person has QQ-AA given that you have JJ is:

Note:
50 choose 2 = 1225
48 choose 2 = 1128
46 choose 2 = 1035
44 choose 3 = 946

1 - ((1225-18)/1225)*((1128-18)/1128)*((1035-18)/1035)= 0.047

So about 5% with 3 oppos. For 4 oppos, it's about 6.5%:

1 - ((1225-18)/1225)*((1128-18)/1128)*((1035-18)/1035)*((946-18)/946)= .065

The same method can be used to solve for any number of players.

gm

gm

[/ QUOTE ]

This is incorrect. The second term is not (1128-18)/1128 because that assumes that the first player does not have an A,K, or Q. The same is true for the remaining terms. The problem cannot be done easily this way. Cobra has done this problem exactly right with inclusion-exclusion. The answer comes out to 12.6%.

This is the same as the mistake you made in this post.

[gaming_mouse]

Bruce,

Yep. You're exactly right. It was a bad error too, as the answer doesn't even provide a good approximation.

Thanks for the correction,
gm

[BruceZ]

[ QUOTE ]
Bruce,

Yep. You're exactly right. It was a bad error too, as the answer doesn't even provide a good approximation.

Thanks for the correction,
gm

[/ QUOTE ]

The 12.6% was for 9 opponents. For 3 opponents it is 4.36%, and for 4 opponents it is 5.77%. So your results of 5% and 6.5% weren't that far off, but it would actually be more accurate to just use independence since 1 - (1207/1225)^N gives 4.34% and 5.75%.

The advantage of inclusion-exclusion is that you always have a bound on the error. Although the terms can sometimes become complicated to compute for multiple opponents, it converges so fast that it is rarely necessary to compute more than 2 or 3 terms for accuracy to within 0.1% regardless of the number of opponents.