Aces!!!.........to aces

[Mano]

I did some cancelling above, and my answer does indeed reduce to 9/C(50,2). For some reason I thought that just multiplying the probability of getting the aces by nine would only be approximate. Bruces answer is obviously much simpler.

[BruceZ]

[ QUOTE ]
I did some cancelling above, and my answer does indeed reduce to 9/C(50,2). For some reason I thought that just multiplying the probability of getting the aces by nine would only be approximate. Bruces answer is obviously much simpler.

[/ QUOTE ]

The reason it is exact is that only one other player can have AA, so the events of each player having AA are mutually exclusive.

This would not be the case if we wanted the probability of someone having AA when we hold KK, for example. In that case, multiplying by 9 would double count the times that 2 players have AA, so we must subtract that off, to get 9*6/C(50,2) - C(9,2)*1/C(50,4). In this case, the events of 2 players having AA are mutually exclusive, so we can multiply 1/C(50,4) by C(9,2). This is the inclusion-exclusion principle.