[AaronBrown]

I would state this a little differently.

There are 52*51/2 = 1,326 possible two-card combinations. A specific pair can be formed 4*3/2 = 6 ways. A specific suited combination can be formed 4 ways. A specific unsuited combination can be formed 4*3 = 12 ways. Just to check, there are 13 possible pairs and 13*12/2 = 78 possible non-pair combinations. So there are 13*6 + 78*4 + 78*12 = 1,326 possible hands.

There is one chance in 17 of getting a pair, so out of 10 players the expected number of pairs is 10/17. However, that is not the probability that at least one player has a pair, because there will be some deals with more than one pair among the 10 players. The exact computation is not simple, but a good approximation is to say the probability of no one getting a pair is exp(-10/17) = 56%. Under that assumption, the probability of getting exactly one pair is 33%, two pairs is 10% and three pairs is 2%.