Rephrasing the question
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The math here is off. The probability of 22 hitting a set on the 2nd card (which implies that it didn't hit the set on the 1st card) is
( (1 - p(hitting set on 1st card) ) * (2/47)), or ( (46/48) * (2/47) = about .04078
3rd card = (46/48) * (45/47) * (2/46) = .039989
4th card = (46/48) * (45/47) * (44/46) * (2/45) = .03900
River card = (46/48) * (45/47) * (44/46) * (43/45) * (2/44) = .03812
What you are really asking is: Given that one player has AA and the other has 22, what is the probability that the first of the remaining 2's comes at the 1st, 2nd... card.
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Thanks for the reply. I know that my math was off, if you look at it in the way you took it. But I think I'm asking a different question than what you suggest. That's why I didn't took in calculation the probablity of the occurance happening (or not happening) on the former streets. I was trying to look at each street for its own, only I'm not sure I did it right, and how one should actually do it. I didn't try to solve it, or to get the right probablity, but only to see, in a way, what is "more probable".
I will rephrase my question (or actually phrase it for the first time):
Player A had AA, player B had 22. They were both all-in pre-flop. You *know* player B outdrew player A by hitting his set on one of the streets. What is the probablity he hit his set on any specific street?
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