The Odds Of Pocket Aces

#1 19th January 2008, 11:25 PM

Does anyone know how to claculate the odds of 2 players in a 10 handed game getting pocket aces in the same hand? I saw it happen 2 nites in a row recently and got curious. I know the odds of getting pocket aces are 4/52*3/51=1/221 and the odds of 2 players getting pocket aces playing heads up is 4/52*3/51*2/50*1/49=1/270,725. Since there are 45 2 player combinations possible in a 10 handed game, I'm wondering if the answer is maybe 270,725/45 or about once every 6017 hands. If you know the correct calculation or even think you know it, I'd appreciate your telling me.

#2 19th January 2008, 11:28 PM

I only seen it happen twice and one of those times i was one of the ones with AA, sry but i cant do math

#3 19th January 2008, 11:41 PM

The odds of one player getting aces is 1/221, so 1/22.1 in ten hands. Now you have to look at the odds of a second player getting aces. That would be 2/50*1/49 or 1/1225. But there are 9 other hands, so the second player is 9/1225 or 1/136+ So, 1/22.1*1/136= 1/3,008. You must play a lot of hands to see that twice.

#4 19th January 2008, 11:48 PM

Many thanks for your answer. I just use to play at a weird site called Jungle Poker.

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