[Cyrus]

"Forget the scraps of paper. Throw them in the trash."

More PC to put them in the recycle bin.

"If you were to just add these 9 6/1225's together to get 9*6/1225, that would be extremely close to what we are looking for, the probability of at least one more AA."

I fully agree, of course. I was only counting on you for the exact answer which I got in the WPT thread.

"...to get the exact answer. At the bottom of my original post, there are combinatorics for both of these terms, and the first will evaluate to exactly 6*9/1225."

This is what I don't understand. How can [(probability of one player having AA)*(number of opponents)] be approximate and exact at the same time? As usual, I must be missing something.