C(50,2) = 1225
Link to BruceZ's post on inclusion/exclusion:
http://archiveserver.twoplustwo.com/...;o=&fpart=
Excerpt from that post (which oddly enough answers your exact question):
This is the simplest non-trivial poker example. If you hold KK before the flop, what is the probability that at least one of your 9 opponents holds AA?
Note that at most 2 opponents can have AA. The probability that any particular player holds AA is 6/C(50,2). So the first term in inclusion-exclusion is simply 9*6/C(50,2). That is the sum of the 9 probabilities of each player having AA. This alone is very close to the exact answer, but it double counts the times that 2 players hold AA. For the second term, we take the probability of 2 particular players having aces, 1/C(50,4), and multiply by the number of ways to pick the two players, which is C(9,2). So the final answer is:
P(KK vs. AA) = 9*6/C(50,2) - C(9,2)/C(50,4)
or about 4.39%