[BruceZ]

I made a mistake, so I'll fix it here.

C(n,k) is the number of ways to choose k combinations of cards out of n cards, irrespective of order. You can use Excel to compute it, the function is called COMBIN.

C(n,k) = n! / [ (n-k)!*k! ] = n*(n-1)*(n-2)*...*(n-k+1)/ [ k*(k-1)*(k-2)*...*1 ]

So C(52,4) = 52*51*50*49/(4*3*2*1) = 270,725.

and C(10,2) = 10*9/(2*1) = 45.

To get the probability of someone holding AA out of 10 players, first multiply 1/221 by 10 to get 10/221. This is almost the right answer, except that it counts the cases where 2 people get AA twice, so we have to subtract the probability that 2 people get AA. There are C(10,2) ways to choose the 2 people, and there is 1 way to choose the 4 aces for those 2 people out of C(52,4) total ways to deal the 4 cards. So the probability of 2 people having aces is C(10,2)/C(52,4). So the final answer is:

10/221 - C(10,2) / C(52,4) = 4.5% = 1 in 22.2