[aloiz]

If you want the odds that you have AA v KK v QQ v XX v XX where XX does not include QQ-AA then the following will work:

I split the equation up into two halves. P(3 players have AA, KK, QQ) * P(remaining 2 don't have AA, KK, QQ|first 3 do)

5*C(4,2) * 4*C(4,2) * 3*C(4,2) / C(52,2)/C(50,2)/C(48,2)
Choose what player has AA, and which suits. Choose what player has KK and which suits. Choose what player has QQ and which suits.


P(2 remaing players don't have AA,KK,QQ|first three do) = 1 - P(either or both players have AA,KK,QQ|first three have AA,KK,QQ). So we have two cases. The first being that only one of the two has QQ-AA, the second being that both have QQ-AA.

First case:
2*3*(C(44,2)-2) / C(46,2)/C(44,2)
Choose what player has QQ-AA, choose QQ, KK, or AA. Choose hand for second player.

Second case:
C(3,2)*2 / C(46,2)/C(44,2)
Choose the two pairs, then the division of those pairs between the two players.

So we get:
5*C(4,2) * 4*C(4,2) * 3*C(4,2) / C(52,2)/C(50,2)/C(48,2) *
(1 - 2*3*(C(44,2)-2)/C(46,2)/C(44,2) + C(3,2)*2/C(46,2)/C(44,2)) =~ 7.032 * 10^-06 or about 142201:1


If you want to include the possiblity that the other two players could have QQ-AA then you need to use inclusion exclusion principle, and the problem becomes pretty difficult. I'd actually be surprised if an exact answer has been given before on this forum.

aloiz