We finished figuring the probability of pocket aces and pocket kings among ten players at a hold’em table in our last column, finishing with the table shown below. The probabilities of each possible way to make two or more pairs with aces and kings sum to 0.00230. (Charts not shown.)
 
Now we expand our horizon by finding the probability of pocket pairs of any two ranks. We could have aces and sevens, or threes and twos, for examples. With thirteen possible ranks we have C(13,2) ways to have two ranks, so we multiply 78 and 0.00230 to obtain the probability that players at a ten-handed hold’em table hold pairs of any two ranks: 0.17923, about 4.6-to-1 against.
 
Fred reviewed our draft of this column and asked, “What’s to prevent someone else having a pair of a different rank?”
 
Because we didn’t specifically exclude that possibility, it could happen. The cards dealt to the other players could end up as a pocket pair of a different rank in a player’s hand. Strictly speaking we should say the probability of players at a ten player table holding pairs in two or more ranks before the flop equals 0.17923.
 
“So,” Fred said, “to find the probability of pocket pairs in exactly two ranks, you have to subtract the probability of pocket pairs in three ranks, right?”
 
It may come to that. Let’s first explore an approximation using Pascal’s triangle. For a ten-player table, we use the bottom row of the triangle shown nearby to find the coefficients of the probability equation. The probability of not having a pocket pair equals 16/17. We find a probability for no one having a pocket pair as 1*(16/17)10, which equals 0.545. That seems about right.
 
We find a probability for exactly one player having a pocket pair from the second cell in the bottom of the triangle 10*(16/17)9*(1/17)1, which equals 0.34. That seems reasonable enough.
 
Using the third cell of the bottom line in the triangle, we find a probability for exactly two players having pocket pairs as 45*(16/17)8*(1/17)2, which equals 0.096.
 
That comes about half-way to the probability we computed for two or more paired ranks, 0.179, which took us three-and-a-half columns to relate! So, we bid goodbye to many hours of work trying to figure this analytically for ten players. Using Pascal’s triangle (actually the binomial expansion) for figuring out pocket pairs works well, so we must have an error somewhere in our computations.
 
We told Fred that he could pursue this topic of multiple pocket pairs by studying “Multiple Pocket Pairs,” at http://www.math.sfu.ca/~alspach/comp35/. Of course you can too: we  warn that you may find it a bit deep.
 
Before we lurch off to find our error or compute the probability of three or more ranks of pocket pairs among ten hold’em players, (if we ever do), we need to compute the probability of one or more ranks of pocket pairs among ten players. That column, titled “Aces Among Us,” should follow in an issue of Poker Player newspaper coming soon to a poker room near you.
 
Mr. Burke is the author of Flop: The Art of Winning at Low-Limit Hold ’Em, on sale at amazon.com & kokopellipress.com. E-mail your Hold ’Em questions to richardburke@comcast.net