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View Full Version : Odds of higher PP in 6-max game...

09-23-2005, 09:14 AM
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
09-23-2005, 12:07 PM
[ 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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=3446464&amp;page=&amp;view=&amp;s b=5&amp;o=&amp;vc=1) 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%.

09-23-2005, 04:30 PM
Wow people must really drive you nuts on here /images/graemlins/smile.gif

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
09-24-2005, 10:33 AM
What does the term "C(50,2)/C(48,2)" represent, and what does "12*8 + 1*12" represent?
thanks

BruceZ
09-25-2005, 01:46 AM
[ 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).