[ QUOTE ]
Sorry if asked a million times. What are the odds you'll have the same hand as another player in a full (10) ring game?
[/ QUOTE ]
P(pair)*P(matching pair) + P(XYs)*P(other XYs) + P(XYo)*P(other XYo)
= (1/17)*P(matching pair) + (4/17)*P(other XYs) + (12/17)*P(other XYo)
From the
inclusion-exclusion principle:
1/17*[9*1/C(50,2)] +
4/17*[9*3/C(50,2) -
C(9,2)*3*2/C(50,2)/C(48,2) +
C(9,3)*3*2*1/C(50,2)/C(48,2)/C(46,2)] +
12/17*[9*7/C(50,2) -
C(9,2)*(4*3 + 2*2 + 1*2)/C(50,2)/C(48,2) +
C(9,3)*(4*3*1 + 2*2*1 + 1*2*1)/C(50,2)/C(48,2)/C(46,2)]
=~ 4.1%.