[ QUOTE ]
What are the odds of 2 players to have the same pocket pair?
ie. both with pocket 7s?
[/ QUOTE ]
For 10 players and any duplicate pocket pair (not a specific pair) by the
inclusion-exclusion principle:
C(10,2)*13*6/C(52,2)/C(50,2) -
C(10,2)*C(8,2)/2*13*6*12*6/C(52,2)/C(50,2)/C(48,2)/C(46,2) +
C(10,2)*C(8,2)*C(6,2)/3!*13*6*12*6*11*6/C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2) -
C(10,2)*C(8,2)*C(6,2)*C(4,2)/4!*13*6*12*6*11*6*10*6/C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)/C(40,2)/C(38,2) +
C(10,2)*C(8,2)*C(6,2)*C(4,2)*C(2,2)/5!*13*6*12*6*11*6*10*6*9*6/C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)/C(40,2)/C(38,2)/C(36,2)/C(34,2)
=~ 462-to-1.