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Odds of loosing AA vs AA
I was reading a post here about the KJ vs KJ with the turned straight. It made me think about this hand.
Seeing that I'm still re-learning how to use combinations, I have a couple questions. I was in a hand, about 2 months ago, in a S&G at PS. It was the 5th hand in the tourney, I had already gotten AA (second hand). I got AA again UTG. I raised about 3x the pot and the BB goes all in. I call. It's AA vs AA. Now the odds of getting dealt AA is: C(4,2)/C(52,2) = 1 in 221 This is pretty straight forward. Now how do you calculate the odds of two players being dealt AA on the same hand? If I would hazard a guess I would say: C(4,2)/C(52,2) * 1/C(50,2) = 1 in 270,725? Is this correct? Now, with this being said, and assuming that I have a chance in being correct thus far.... I lost the hand to a board 4 flush. The probability of that is: C(12,4)/C(48,5) = 1 in 3459? So the probability of this happening is 1 in 936,491,919? Is this correct? If so, I guess I can stop worrying about this happening to me again. |
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