What are the odds on this happening?

Igetp2s · May 4, 2005

2 people at the same 10 person table both getting pocket aces in HE?

This happened last night at my limit table, and needless to say they were both raising furiously building up a monster pot, all for next to nothing after splitting and the rake.

I could probably figure this out myself but I'm just too lazy.

Thanks

QuickLearner · May 4, 2005

Well, you've got 2 of the 4 aces. The odds of another person getting dealt the third ace would be 2/50; the odds that the same person would get the fourth one would be 1/49. Multiply 2/50 x 1/49 = 1/1225. That's the chance that one of the players (in particular) will be dealt the other AA. You asked about the chances of any of the other 9 players though. Multiply the 1/1225 chance by the 9 other players = 9/1225 or 1/136 or about 0.73%. Rare, but you should see it happen about once every 136 hands.

Igetp2s · May 5, 2005

Quick, you are claculating the odds of someone else getting pocket aces after I already have them. I am asking about before cards are dealt, what about the odds of the first person getting them. I think the answer would be (4/52)* (3/51)*(2/50)*(1/49)*9. If this is correct the answer would be a little less than 1/30,000.

QuickLearner · May 5, 2005

I guess I misunderstood the situation/question.

I believe that the odds of a particular player getting dealt AA is about 1/220. The odds of just anybody getting the first pair of aces is 1/220 x 10 = 1/22. The odds of a second person getting AA is as I said 1/136. So if I calculate the chances of both occuring I get 1/22 x 1/136 = 2/2992 = 1/1496.

So, once every 1500 hands all four aces are dealt to two out of 10 players. I think you were doing too much multiplying. Kind of like the human race. :toast:

Igetp2s · May 6, 2005

Actually 1/22 * 1/136 = 1/2992 rather than 2/2992. I left off a 10 in my multiplication for any of the 10 people getting the first pocket aces, rather than just me. OK its not nearly as rare as I thought it would be.