Odds two stud players rolled up in same hand

[BruceZ]

That's like the guy who always carries his own bomb aboard an airplane, since the chance of 2 people bringing a bomb are miniscule.

Once you are rolled up, it becomes over 100 times more likely that 2 players will be rolled up.

Here is the exact probability of 1 or more players being rolled up with you when you are rolled up:

C(7,1)*P(12,1)*4^1/C(49,3) –
C(7,2)*P(12,2)*4^2/C(49,3)/C(46,3) +
C(7,3)*P(12,3)*4^3/C(49,3)/C(46,3)/C(43,3) -
C(7,4)*P(12,4)*4^4/C(49,3)/C(46,3)/C(43,3)/C(40,3) +
C(7,5)*P(12,5)*4^5/C(49,3)/C(46,3)/C(43,3)/C(40,3)/C(37,3) -
C(7,6)*P(12,6)*4^6/C(49,3)/C(46,3)/C(43,3)/C(40,3)/C(37,3)/C(34,3) +
C(7,7)*P(12,7)*4^7/C(49,3)/C(46,3)/C(43,3)/C(40,3)/C(37,3)/C(34,3)/C(31,3)

= 1 in 55.3

It isn't necessary to go through all this to get close, you can just use the first term or two:

7*12*4/C(49,3) = 1 in 54.8