We turn now to the task of figuring the probability that two or more players hold pocket pairs at a ten player table. We expect this proof to take several columns. We start by figuring the probability that from 0 to 8 of aces and kings lie among the twenty cards dealt to the players. Those probabilities are obtained from the expression, C(4,A)*C(4,K)*C(44,20-AK)/ C(52,20), where A and K stand for the number of aces and kings lying among the top twenty cards in the fifty-two card deck. This table (not shown) shows the resulting probabilities.
 
We aim to find the probability of two or more pocket aces and kings among the players, so we exclude all those cells for which the probability of those two pairs equals zero. The next table shows the number of cases we must consider. Thirteen combinations of aces and kings dealt out could form two or more pocket pairs of aces and kings among the ten players. Because of symmetry we need consider further not thirteen but only eight of the table’s cells.
 
Just because they could form pocket pairs doesn’t mean they must; we need to examine their distribution, their patterns, among the hands dealt. Suppose the deck had exactly two aces and two kings among the top twenty cards dealt to the players, a probability of 0.11907. Those four cards could lie in four players’ hands, in three players’ hands, or in two players’ hands. Because we can’t have two pairs unless all four key cards lie in two players’ hands, we need to find the probability of this pattern, 2-2-0-0-0-0-0-0-0-0.
 
Since we care not which two players hold the key cards, that probability equals 3!!*15!!/19!!, which means we can deal the four key cards 3!! ways, times 15!! ways for the other sixteen cards, divided by the 19!! ways we could deal all twenty cards among ten players. This probability equals 0.00929, about one chance in a hundred that those four cards fall into two players’ hands.
 
The four key cards could lie AK AK (two ways) or AA KK (one way), so the probability of two pairs equals 0.11907 x 0.00929 x 1/3. We now know the probability for the cell (2,2): it equals 0.00037.
 
Similarly, all four aces and no kings, or all four kings and no aces, could lie in two hands. Those four key cards make two pairs no matter how they lie, so the probability that two players have pocket aces, or pocket kings, equals 0.00331 x 0.00929 x 1.0. We now know the probabilities for cells (0,4) and (4,0): They equal 0.00003.
 

The table shows the probabilities we calculated for the cells (0,4), (2,2), and (4,0). In subsequent columns we plan to fill in the rest of the cells. Their sum, when complete, equals the probability that two or more players at a ten-player table have pocket pairs before the flop.
 
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