[ QUOTE ]
[ QUOTE ]
with 50 unknown cards you can have a posiblity of 1225 hands 6 are AA. Against 1 opponent then (6/1225).

Against X number of opponents then aprox:
1 - (1219/1225)^x

If x approx.
2 then .00977
3 then .01462
4 then .01945
6 then .02903
9 then .04323

Matt

[/ QUOTE ]

x*6/1225 is more accurate for any number of players. The error is C(x,2)/C(50,4).

Your independence approximation works better for some problems where several players can have a hand, such as the probability that any pair is out.

[/ QUOTE ]

I would have guessed that the independence assumption would work better than the x*6/1225 that you suggest, but of course I haven't calculated it.

The precise calculation would get quite messy, I think. It would require calculating the probability of having a total of zero, one, or two Aces in four of other five hands (assuming none of these four hands contain AA), and then calculating the probability of AA in the 6th hand, and multiplying by five. Right?

Could you please explain how you calculated the error here, C(x,2)/C(50,4)?