This equation is still under construction. I will post a note later when it is complete.
Quote:
Originally Posted by spadebidder
What is the chance that at a 9-player table, two players hold matching hands like AK AK?
You can get as accurate as you like with the
inclusion-exclusion principle. I will compute it to within 0.1%. I have aleady shown in a separate post that the upper bound is 25.1%.
Although this looks like alot of work, there is really alot of repetition of simple terms, and the whole thing can be done quite quickly.
P(at least 2 players with same hand) =
P(at least 2 players with same pair) +
P(at least 2 players with same non-pair) -
P(at least 2 players with same pair AND 2 players with same non-pair)
P(at least 2 players with same pair) =
C(9,2)*P(2 specific players with same pair) -
C(9,4)*P(2 specific pairs of players which each share a pair) +
...
= C(9,2)*13*6 /C(52,2)/C(50,2) -
C(9,4)*13*6*12*6 /C(52,2)/C(50,2)/C(48,2)/C(46,2)
=~ 0.173%
P(at least 2 players with same non-pair) =
C(9,2)*P(2 specific players with same non-pair) -
C(9,3)*P(3 specific players with same non-pair) +
C(9,4)*[ P(4 specific players with same non-pair) + P(2 specific pairs of players which each share a non-pair) ] -
...
= C(9,2)*C(13,2)*4*4*3*3 /C(52,2)/C(50,2) -
C(9,3)*C(13,2)*4*4*3*3*2*2 /C(52,2)/C(50,2)/C(48,2) +
C(9,4)*[ C(13,2)*4*4*3*3*2*2*1*1 +
C(4,2)*C(13,2)*4*4*C(12,2)*4*4 ] /C(52,2)/C(50,2)/C(48,2)/C(46,2)
=~ 24.7%
P(at least 2 players with same pair AND 2 players with same non-pair) =
C(9,4)*P(2 specific players with same pair)*P(2 specific players with same non-pair)
-
...
= C(9,4)*C(4,2)*13*6*C(12,2)*4*4 /C(52,2)/C(50,2)/C(48,2)/C(46,2)
=~ 0.003%
This last term is neglible compared to the others.
Adding the 3 gives about
24.9%.
Last edited by BruceZ; 11-13-2009 at 11:23 AM.