We return to figuring the probability that two or more players hold pocket pairs at a ten-player table. In a previous column we figured the probability that two players at a ten-player table would have pocket pairs of aces and/or kings, when four aces and kings lie among the top twenty cards in a standard deck. In this column we continue filling out the probability table shown below (not shown):
 
The probability that exactly five aces and kings lie among the top twenty cards dealt to the players equals C(8,5) x C(44,15) / C(52,20), or 0.10219. Those five key cards could lie scattered among the ten players: as 1-1-1-1-1-0-0-0-0-0, 2-1-1-1-0-0-0-0-0-0, or 2-2-1-0-0-0-0-0-0-0. Only the last pattern could produce two pairs with five of the key cards dealt, a probability of C(5,4) x 3!! x 15!! / 19!!, or 0.04644.
 
We consider now these four cases: KAAAA, KKKKA, KKAAA, and KKKAA. The probability of either KAAAA or AKKKK equals C(4,4) x C(4,1) / C(8,5), or 0.07143 . The other distribution has a probability of C(4,3) x C(4,2) / C(8,5), or 0.42857.
 
Finally we need to compute the probability that the 2-2-1 pattern has two pocket pairs. For either KAAAA or KKKKA, we have two pairs only if the odd card stands alone, a probability of 0.2. For KKAAA or KKKAA any one of the three aces or three kings must stand alone, a probability of 3/5, and the other aces and kings must lie as AK AK (two ways) or as AA KK (one way), a probability of two pairs equal to 1/3, so we have the probability of 3/5 x 1/3, again a probability equal to 0.2.
 
We now have all the numbers we need to fill in four cells of the probability table by multiplying the four probabilities for each case. The table below shows these probabilities, rounded to five decimal places.
 
KKKKA or KAAAA . . . . . . . . . . .0.000068
KKKAA or KKAAA . . . . . . . . . . .0.000407
We have enough space left to consider the case of all eight aces and kings dealt out among the ten players, a probability of C(8,8) x C(44,12) / C(52,20), or 0.00017.
 
We could have two pairs if the cards lie 2-2-1-1-1-1 among the players, a probability of C(8,4) x 3!! x 12 x 10 x 11!! / 19!!. We could have two or three pairs if the cards lie 2-2-2-1-1, a probability of C(8,6) x 5!! x 12 x 11!! / 19!!. We could have two or more pairs if the cards lie 2-2-2-2, a probability of 7!! x 11!! / 19!!.
 
Now it gets tedious: We must consider all the possible combinations of kings and aces, well over 100 cases, and compute the probabilities of each.
 
We examined all 143 cases for all eight aces and kings dealt out. We determined that the probability of two or more pairs equals 0.0000184, as shown in the table below rounded to five decimal places.
 
The remaining five empty spaces must await a later issue of Poker Player newspaper. When completed, we plan to sum all the table’s probabilities to find the probability of two or more players among ten holding at least pocket aces and kings 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