Classic Cooler

[BruceZ]

[ QUOTE ]
What's the probably of aa-kk in a 6 mx table? This seems to be happening a lot latly.

[/ QUOTE ]

As usual, treating the hands as independent gives a very accurate approximation of about 0.06647% or 1 in 1504, while the exact answer is about 0.06625% or 1 in 1509. The approximation is

1 - [1 - 12*6/C(52,2)/C(50,2)]^C(6,2)

=~ 0.06647% or 1 in 1504


where the expression in [] is 1 minus the probability that 2 particular players have AA and KK, which is the probability that these 2 players do not have it. This is raised to the power of C(6,2), which is the number of pairs of players. This approximates the probability that no 2 players have it, and this is all subtracted from 1 to get the probability that some 2 players have it.

The exact solution from the inclusion-exclusion principle is

C(6,2)*12*6/C(52,2)/C(50,2) -
C(6,3)*12*6*1*3/C(52,2)/C(50,2)/C(48,2) +
C(6,4)*12*6*1*3*1/C(52,2)/C(50,2)/C(48,2)/C(46,2)

=~ 0.06625% or 1 in 1509.


Note that this is the probability that at least 1 AA and at least 1 KK are dealt at a 6 player table. If you want the probability of being up against AA when you have KK, or other similar problems, see this post.