Odds of higher PP in 6-max game...

What are the odds of at least 1 higher PP being dealt in a 6-handed game for the following hands...

KK,QQ,JJ,

If this has discussed here before then can someone please point me in the right direction to the thread -- I came up empty...

[BruceZ]

[ QUOTE ]
What are the odds of at least 1 higher PP being dealt in a 6-handed game for the following hands...

KK,QQ,JJ,

If this has discussed here before then can someone please point me in the right direction to the thread -- I came up empty...

[/ QUOTE ]

If you had just searched for KK or QQ, you would have found this post of a few days ago that answers those two cases.

Now for AA or KK or QQ vs. JJ (6 handed):

5*18/C(50,2) -
C(5,2)*18*13/C(50,2)/C(48,2) +
C(5,3)*18*(12*8 + 1*12)/C(50,2)/C(48,2)/C(46,2)

=~ 7.18% or 12.9-to-1.


The exact expression would have 5 terms since all 5 opponents could have these hands. These 3 terms are accurate to better than 0.00001%.

Wow people must really drive you nuts on here [img]/images/graemlins/smile.gif[/img]

Thanks you so much for your help, im new here, but its seem you are a great asset to the forums. I'll work on my searching...

[KJL]

What does the term "C(50,2)/C(48,2)" represent, and what does "12*8 + 1*12" represent?
thanks

[BruceZ]

[ QUOTE ]
What does the term "C(50,2)/C(48,2)" represent, and what does "12*8 + 1*12" represent?
thanks

[/ QUOTE ]

The total number of ways to deal hands to 2 opponents with 50 cards, including order, is C(50,2)*C(48,2), so we divide by this amount. The number of ways that 3 opponents can hold 3 pairs AA-QQ is 18*(12*8 + 1*12). After the first opponent gets one of the 18 pairs, the second opponent has 12 ways to pick one of the other 2 pairs, or 1 of the same pair. If he gets one of the 12, that would leave 8 ways for the third player to choose a pair (6+1+1). If he chooses the 1 pair, that leaves 12 ways for the the third opponent to choose a pair (6 + 6).