Re: The chance someone has AA if you have KK at a 9 person table
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I follow the math on the AA vs. KK example that closer posted, and was thinking about the probabilty someone will have AA or KK if I have QQ on a ten person table. This math becomes another level more complicated, right?
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If you want an exact answer, then you need 4 terms since up to 4 people can have AA or KK, but as a practical matter, 2 terms will still get you to within 0.001%. Here is the exact calculation:
P(AA or KK vs. QQ) =
9*12/C(50,2) -
C(9,2)*(12*7)/C(50,2)/C(48,2) +
C(9,3)*12*(6*2 + 1*6)/C(50,2)/C(48,2)/C(46,2) -
C(9,4)*12*(6*2*1 + 1*6*1)/C(50,2)/C(48,2)/C(46,2)/C(44,2)
=~8.6% or 1 in 11.6 or 10.6-1.
As an explanation of the third term, notice that the first player has a choice of 12 hands, the second player has a choice of either 6 of one hand or 1 of the other, and if the second player chooses one of the 6, then the remaining player has a choice of 2 hands, otherwise he has a choice of 6 hands. As these problems get more complicated, you can draw a tree-like structure to describe the various possibilities.
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