[easypete]

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.