[elitegimp]

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kingstalker is of course right.

The two aces remaining in the other 50 cards do not care about you holding the other two aces. So there are still '50 choose 2 = 1225' possible combinations of which one combination (AA) creates the (more or less) desired situation.

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This is only correct if you are only dealing 2 hands, one of which is already AA. I'm not 100% sure on how to include additional hands, and I've already met my quota for incorrect posts today, but if you search the probability forum for 'AA vs AA' or '2 AAs' or something, you should get approximately a gazillion hits.

edit: for example, for 3 players -- there is a 1/1225 chance that player 2 gets AA if player 1 doesn't have it. There is a 48*47/2450 chance that player 2 is not deal an A at all, in which case there is a 1/1128 chance that player 3 has AA.

So prob that either player 2 or player 3 has the last AA = 1/1225 + (48*47/2450)*(1/1128) = 2/1225.

Since these events can not both happen (players 2 and 3 cannot both have AA if player 1 does), there is no correction to be made.

Anyhoo, it looks like prob(2 players in n-person table both have AA given one player has AA) = (n-1)/1225. So in a 6-person game, once every 245 hands (and about once every 136 hands in a 10-person game).

I'd like to go on record saying I don't see AA vs AA every 136 times online, so party must be rigged (or I made another mistake).

edit 2: D'oh! every 136 given that one person already has AA! I'm a moron.