We continue figuring the probability of pocket aces and pocket kings among ten players at a hold’em table. The first table (not shown) shows the known cells and the remaining five cells for which we obtain these probabilities.
 
For the unknown cells, (2,4), (3,3), (3,4), (4,2), and (4,3), the integers denote the number of kings and aces which lie among the top twenty cards of a standard deck.
 
We already calculated the probability that exactly six aces and kings lie among the top twenty cards dealt to the players; it equals 0.02555. Those six key cards could lie among the ten players as: 1-1-1-1-1-1, 2-1-1-1-1, 2-2-1-1, or 2-2-2. The third pattern could produce two pairs with six of the key cards dealt, a pattern probability of C(6,4) x 3!! x 14 x 13!! / 19!!, or 0.13003. The fourth pattern could produce two or three pairs, a pattern probability of 5!! x 13!! / 19!!, or 0.00310.
 
We consider now the three six-card cells: KKKKAA, KKKAAA, and KKAAA. The probability that we have KKKKAA equals C(4,4) x C(4,2) / C(8,6), or 0.21429, as does KKAAAA. The probability that we have KKKAAA equals C(4,3) x C(4,3) / C(8,6), or 0.57143.
 
When we have the pattern, 2-2-1-1, then of the fifteen ways to deal out those six cards to the two hands each with two key cards, we looked at each of those ways that aces and kings could fill that pattern and found the probability of having two pairs equals 0.2 for each of the three cases.
 
When we have the pattern 2-2-2 we looked at each of the fifteen ways to deal out the six cards, and found that the probability of having two pairs equals 0.2 for both KKKKAA and KKAAAA, and equals 0.6 for KKKAAA.
 
We now have all the numbers we need to fill in three cells of the probability table by multiplying and summing the probabilities for each case. The table below shows these probabilities, rounded to five decimal places.
 
KKKKAA or KKAAAA . . . . . . . . . . . . . . .0.0001458
KKKAAA . . . . . . . . . . . . . . . . . . . . . . . .0.0004068
 
We know the probability of seven aces and kings dealt out among the ten players; it equals 0.00330.
 
We again figure the probabilities for the patterns, 0.26006 for 2-2-1-1-1 and 0.02167 for 2-2-2-1.
 
The probability of having four kings and three aces, KKKKAAA, equals that of KKKAAAA, 0.5.
 
We looked at each possible distribution of aces and kings for both pattern. For example, KK-KK-A-A-A produces two pairs, as does KK-AA-K-A-K. We found that the probability of two pairs for the pattern, 2-2-1-1-1, equals 0.2. We also looked at each possible distribution of aces and kings for the pattern 2-2-2-1: we found the probability of two or more pairs equals 0.6 if the odd card produces three aces and three kings among the three hands, 0.2 when it doesn’t.
 
Summing all the cases and multiplying produces the probability for the cells (3,4) and (4,3), as shown.
 
We sum all the cells to find the probability of two or more pocket aces and kings among ten hold’em players, which equals 0.00230, about 434-to-1 against. Note that they could hold not only pairs of aces & kings, but also aces & aces, kings & kings, aces & aces & kings, aces & kings & kings, and finally, aces & aces & kings & kings.
 
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