What are the odds of pocket aces beating pocket aces?

[jay_shark]

It's really simple actually . If you hold pocket aces , then the probability one of n opponents holds aces is n/50c2 . At most , one other player may hold aces other than yourself and so the events are mutually exclusive .

However , if you didn't hold aces , then the probability two players get dealt aces is 6/52c2*1/50c2*nc2 ~0.00016622

Given that you do hold aces , the probability you will lose to another player with aces is :

2*[n/50c2*12c4*36/48c5]+2*[n/50c2*12c5/48c5] ~ 2n*0.000008873114

So if n=9 , then the answer is ~ 0.000159716