[BruceZ]

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Here is a simple explanation .

If you're at a 10-handed table and you hold A-x , then the approximate probability that someone has pocket aces is :

1-[(1225-3)/1225]^9 ~ 0.021826135

The probability player i doesn't have pocket aces is (1225-3)/1225 . We may approximate the solution by assuming independence . So the probability neither of the 9 players holds pocket aces is [(1225-3)/1225]^9 . Now the last step is to take the complement which is 1 - [(1225-3)/1225]^9 .

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You don't neeed to approximate in this case since at most 1 player can have AA, so the exact probability is 9*3/1225 =~ 0.0220. However, your approximation method works in the more general case where more than 1 player can hold the hand in question.

EDIT: Fixed typo 0.220 -> 0.0220