What are the chances that two players will have pocket aces at a nine player Texas Hold 'em table?

Answer 1

Bigsky_52 gave you the correct answer if you you want the odds of two specific players both having pocket aces. That would be correct with only two players, but not for a nine player table.

The odds of one other player having pocket aces when you also hold pocted aces would be eight times higher since any of eight other people could also hold pocket aces. The odds would be

8 in 270,725 or 1 in 33,841

To get the odds of any two players having pocket aces, you have to take into account that there are 36 different pairs of players, making the odds

36 in 270,725 or 1 in 7,520

Answer 2

1 in 270,725

Chances of the first player getting pocket aces = (4/52)*(3/51) = 1/221

Chances of the second player also getting pocket aces = (2/50)*(1/49) = 1/1225

Multiply both probabilities to get the final probability

Edit-
Zman is correct, vote him for best answer. If you'd like to learn more about why you'll find another player with pocket aces 1 in 136 times that you yourself hold pocket aces, take a look at http://www.worldpokertour.com/magazine/index.php?x=detail&aID=359

Answer 3

zman got it. The first answer is only correct for a heads up situation. Your question was about a full table, so you have to factor in the extra players.

Answer 4

I have had that done to me once and the other guy won with four suited with one of his aces....i wanted to get sick