It's really not that tricky. There are 52!/(2!*(52-2)!) = 1326 number of different holdem hands. If you hold AA you can have it 6 different ways. The same goes for any pair. So, AA, KK, QQ, JJ, that sums up to 24 ways.


24 out of 1326 for a probability of 1.81%


Now, instead of calculating the probability that one or more players have such a pair we will calculate the probability that none of them has one. The probability that one player does not hold a pair is 98.19% so the probability that none of the remaining 9 players have one is:


0.9819^9 = 0.8484


So, there is about an 85% chance that the players do not have such a pair. Consequently there is about a 15% chance that one of the players has such a pair.


Hope this didn't cause as many questions as it answered. [img]/images/wink.gif[/img]


Greets, Andreas