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JNash
11-16-2005, 02:13 AM
Can someone help me calculate the probability that there is an AA-AA matchup at a ten-handed table?

I know that if there are exactly 2 players, the prob of player 1 having an AA is 6/1326, and prob of player 2 also having AA is 1/1225, so the prob of both having AA is the product of these two. But with 10 players, it gets more complicated...

Thanks in advance for your help!
BruceZ
11-16-2005, 02:20 AM
[ QUOTE ]
Can someone help me calculate the probability that there is an AA-AA matchup at a ten-handed table?

I know that if there are exactly 2 players, the prob of player 1 having an AA is 6/1326, and prob of player 2 also having AA is 1/1225, so the prob of both having AA is the product of these two. But with 10 players, it gets more complicated...

Thanks in advance for your help!

[/ QUOTE ]

Multiply the product by C(10,2). This gives the exact answer, since no more than 2 can have it. C(10,2)*(6/1326)*(1/1225) =~ 1 in 6016.
JNash
11-16-2005, 02:32 AM
That was quick! Thanks a lot!