steve968574
05-31-2005, 11:01 PM
This is a rather esoteric question that is interesting nonetheless.
Given a 10-player table, what is the probability that any one or more of my nine opponents holds JJ, QQ, KK, or AA? (Assume I don't hold such a pair.)
I have found this very difficult even to conceptualize. I begin by saying, okay, how many combinations of nine 2-card hands are there in which all nine hands are J+ pocket pairs? A: None. There are only 16 J-Q-K-A cards, so at most only eight of the hands can consist of such a pair.
Okay, in how many ways can the eight hands form eight J+ pairs? Well, there are 3 ways to make two pairs of the same rank. So, 3*3*3*3 = 81. Then 81 * 34c2 = 45,441 ways for nine opponents to hold eight J+ pairs between them.
In how many ways can exactly seven opponents hold such a pair? The seven will consist of one pair of one of the ranks and two pair of each of the remaining three ranks. So, 6*3*3*3 = 162. Great. What do we multiply this by?
Here's where I start to go off the deep end. The 8th player will have one of 36c2 = 630 possible hands; the 9th player will have one of 34c2 = 561 possible hands. Actually 629 and 560, respectively, since neither can hold the 2 remaining J-A cards (all the situations with eight J+ pairs have been counted already).
So how do I find the number of combinations of 4 cards arranged as the 8th and 9th players' 2-card hands? Is it (629*560)/2! ? I have no idea.
And, coming at it from the other end, in how many ways can exactly two of my opponents hold pocket pairs J or better? Exactly one player holding such a pair is easy--24 * the number of combinations for the eight remaining hands. But two players--are the pairs of the same rank? If so, then 3* the number of combinations for the seven remaining hands. If different, then 36 * that number. (Any 1 of the 6 for the first rank * any one of the 6 for the 2nd rank).
The complexity is beyond my grasp. If someone can just point me to the correct chapter of an introductory statistics textbook, that would be huge. I would of course like to reduce it to a general formula--given pocket pairs of <i>x</i> rank, what is the likelihood of any of <i>y</i> opponents holding a pair of rank <i>x + 1</i> or greater?
What is the practical application? I assume it could only matter when you are first to act before the flop, since for players in later position more relevant data will come from the action and the tells. There may in fact be no practical use, since you can just memorize the winning percentage of the hand you draw against all comers. Maybe the likelihood of a particular subgroup of those competing hands (J+ pairs as opposed to AK AQ AKo etc) is never particularly relevant.
I don't care. This has been driving me crazy and I need to figure it out.
Given a 10-player table, what is the probability that any one or more of my nine opponents holds JJ, QQ, KK, or AA? (Assume I don't hold such a pair.)
I have found this very difficult even to conceptualize. I begin by saying, okay, how many combinations of nine 2-card hands are there in which all nine hands are J+ pocket pairs? A: None. There are only 16 J-Q-K-A cards, so at most only eight of the hands can consist of such a pair.
Okay, in how many ways can the eight hands form eight J+ pairs? Well, there are 3 ways to make two pairs of the same rank. So, 3*3*3*3 = 81. Then 81 * 34c2 = 45,441 ways for nine opponents to hold eight J+ pairs between them.
In how many ways can exactly seven opponents hold such a pair? The seven will consist of one pair of one of the ranks and two pair of each of the remaining three ranks. So, 6*3*3*3 = 162. Great. What do we multiply this by?
Here's where I start to go off the deep end. The 8th player will have one of 36c2 = 630 possible hands; the 9th player will have one of 34c2 = 561 possible hands. Actually 629 and 560, respectively, since neither can hold the 2 remaining J-A cards (all the situations with eight J+ pairs have been counted already).
So how do I find the number of combinations of 4 cards arranged as the 8th and 9th players' 2-card hands? Is it (629*560)/2! ? I have no idea.
And, coming at it from the other end, in how many ways can exactly two of my opponents hold pocket pairs J or better? Exactly one player holding such a pair is easy--24 * the number of combinations for the eight remaining hands. But two players--are the pairs of the same rank? If so, then 3* the number of combinations for the seven remaining hands. If different, then 36 * that number. (Any 1 of the 6 for the first rank * any one of the 6 for the 2nd rank).
The complexity is beyond my grasp. If someone can just point me to the correct chapter of an introductory statistics textbook, that would be huge. I would of course like to reduce it to a general formula--given pocket pairs of <i>x</i> rank, what is the likelihood of any of <i>y</i> opponents holding a pair of rank <i>x + 1</i> or greater?
What is the practical application? I assume it could only matter when you are first to act before the flop, since for players in later position more relevant data will come from the action and the tells. There may in fact be no practical use, since you can just memorize the winning percentage of the hand you draw against all comers. Maybe the likelihood of a particular subgroup of those competing hands (J+ pairs as opposed to AK AQ AKo etc) is never particularly relevant.
I don't care. This has been driving me crazy and I need to figure it out.