[BruceZ]

The original question was "Can anyone tell me what the odds are of running into AA against your KK in a ten handed hold 'em game?" I took that to mean the poster wanted to find out abt the odds of running into any AA, which means one or two opponents having pocket Aces.

I AM answering the question of the probability of running into one or two AA. The approximation is not exact, but it's within a cat's whisker. It gives 4.4% which is the exact answer rounded off, and it gives 1 in 22.7 compared to 1 in 22.77 for the exact answer. This is always close enough for all practical poker purposes. This approximation does not assume that only one player gets AA, that would be a different number (though equally close in the other direction). By ignoring this possibility it actually double counts these cases.

This is why I objected that the example with the "1,225 pieces of paper" on which each of the C(50,2) combinations is written, cannot aply in the "real world". The probability is indeed appoximated with [9*(6/1225)] but it is not exact.

The slip of paper analogy really has nothing to do with this approximation, and I really have no use for it, because that would allow multiple players to get AA, and it would give the wrong answer in that case. I don't think in terms of that for this approximation. All that is required is to notice that the probability of any player getting AA is 6/1225, and it's the same for every player, so summing them gives the first order approximation. If you subtract the probability of two people getting AA from this, you get the remaining 0.01% to the exact answer (4.39% or 1 in 22.77). That is what is answered at the bottom of the original post.