[BugsBunny]

For a 10 player table:

1st term:
C(9,1) * [11 * C(4,2)]/C(50,2)

- 2nd term: 2 players hold pairs. The combinations can be in the format xxxx (same pair held by 2 players) or xxyy (2 players hold 2 different pairs)

C(9,2) * {[C(4,4) * 11] + [C(4,2) * C(4,2) * 11 * 10]/2}/C(50,4)

+ 3rd term: 3 players hold pairs. We have xxxxyy and xxyyzz as possibilities.

C(9,3) *{[C(4,4) * 11 * (C(4,2) * 10] + [C(4,2) * C(4,2) * 11 * 10 * 9 ]/3}/C(50,6)

- 4th term: 4 players hold pairs. Possibilities are xxxxyyyy, xxxxyyzz, and wwxxyyzz (now things start getting a bit hairy)
C(9,4) * {[C(4,4) * C(4,4) * 11 * 10]/2 + [C(4,4) * 11] * [C(4,2) * C(4,2) * 10 * 9]/2 + [C(4,2) * C(4,2) * C(4,2) * C(4,2) * 11 * 10 * 9 * 8]/4}/C(50,8)

I'm pretty sure that's right, as far as it goes. I'll leave the remainder of the terms for someone else.