Probability of AA vs AA

JNash · #1 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 · #2 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 · #3 11-16-2005, 02:32 AM

That was quick! Thanks a lot!