6 max probabilities

[ActionFreak]

I would like to figure out some 6 max probability problems. I know this will be a decent amount of work. Can someone point me in the right direction on any of these problems?

1. I would like to figure out the probability of someone having a bigger pocket pair at a 6 max table for each pair that I may get.

2. I would also like to figure out the probability of someone having an ace with a kicker that is higher than my pair for each pair I get that is 99 or better.

3.For an Ax hand, how do I figure out the probability of someone having a kicker higher than mine or a pair that is higher than my kicker?

[ActionFreak]

For question 1, could I take the number of hand combinations that make a higher pair, divide by the number of total hand combinations (1326) and then multiply by the number of other players at the table?

For example, I have KK. (6 / 1326) * 5 = 0.0226
I have QQ. (12 / 1326) * 5 = 0.0452

EDIT:
I just realized that the total hand combinations is no longer 1326 because I have two known cards in my hand. I guess that makes the total hand combinations (50*49)/2 = 1225

[ActionFreak]

I can tell from this post that I am not doing this right but it seems to be getting number that are close.


http://forumserver.twoplustwo.com/sh...Number=3446464


AA vs KK (6 handed):

5*6/C(50,2) - C(5,2)/C(50,4) =~ 39.9-to-1

AA or KK vs. QQ (6 handed):

5*12/C(50,2) –
C(5,2)*12*7/C(50,2)/C(48,2) +
C(5,3)*12*(6*2 + 1*6)/C(50,2)/C(48,2)/C(46,2) –
C(5,4)*12*(6*2*1 + 1*6*1)/C(50,2)/C(48,2)/C(46,2)/C(44,2)

=~ 19.7-to-1

[Cobra]

All three of these are solved by using inclusion-exclusion. If you do a search of BruceZ threads he shows you how to figure these out. They are relatively complicated. I did an excel spread sheet for the overpair question awhile ago. It is up to three terms so is relatively accurate. Here is the probability that one or more of the 5 remaining players was dealt an overpair if you have a pair.

2 - 26.1%
3 - 24.2%
4 - 22.2%
5 - 20.2%
6 - 18.2%
7 - 16.1%
8 - 13.9%
9 - 11.7%
10 - 9.5%
JA - 7.2%
QU - 4.8%
KI - 2.4%

I will look at the other questions if I have time.

Cobra

[Cobra]

I solved question #2 using two terms of inclusion exclusion, this should give an answer accurate to within .1%. If you have a pair the probability of one or more of your 5 opponents having an AX that is higher than your pair is:

99 - 24.09%
10/10 - 18.45%
JJ - 12.57%
QQ - 6.43%

The third question has a lot of terms and would be easy to solve for one hand, but time consuming for many hands.

Cobra

[BruceZ]

[ QUOTE ]
For question 1, could I take the number of hand combinations that make a higher pair, divide by the number of total hand combinations (1326) and then multiply by the number of other players at the table?

For example, I have KK. (6 / 1326) * 5 = 0.0226
I have QQ. (12 / 1326) * 5 = 0.0452

EDIT:
I just realized that the total hand combinations is no longer 1326 because I have two known cards in my hand. I guess that makes the total hand combinations (50*49)/2 = 1225


[/ QUOTE ]

Using 1225, this is the first term of inclusion-exclusion. Using this one term will be very accurate in cases like this one where the probability that 2 or more opponents hold one of the hands in question is small, which occurs when there are relatively few hands in question. Usually only 2 or 3 terms are required for an accuracy better than 0.1%. In the series of articles by Brian Alspach on inclusion-exclusion titled "I'm In...No I'm Out", he gives tables with some of the numbers that you are interested in.

In cases where there are many hands in question, such as when you are computing the probability that an opponent holds one of many possible pairs, the approximation of independence is often accurate. That is, if the probability that a single opponent holds one of these hands is p, then the probability that he does not hold one is 1 - p, and the probability that one of 5 opponents does not hold one is approximately (1 - p)^5, so the probability that one of 5 opponents holds one is approximately 1 - (1 - p)^5.