AA V KK V QQ 6-Max

by **Kuso** » Sat Feb 04, 2006 12:33 pm

interesting math... i hope someone can get a definitive answer for this.

by **Rhound50** » Sat Feb 04, 2006 5:33 pm

The number I come up with is 1:449774

by **GooperMC** » Sat Feb 04, 2006 9:55 pm

by **Kuso** » Sat Feb 04, 2006 11:04 pm

by **Kuso** » Sat Feb 04, 2006 11:18 pm

i come up with about one in 1.8 million, but i think that this is not right.

218/221*218/221*218/221*3/221*2/221*1/221

this includes instances when there are ONLY these three instances of AA, KK, and QQ. that being said, it's almost the same if you add other instances with multiple occurances of any or all of these PPs.

i feel like you have to divide this by n/(n-3) or something like that..

by **Rhound50** » Sun Feb 05, 2006 12:10 pm

by **GooperMC** » Tue Feb 07, 2006 8:49 am

by **Rhound50** » Tue Feb 07, 2006 2:39 pm

by **GooperMC** » Thu Feb 09, 2006 1:47 pm

by **Rhound50** » Thu Feb 09, 2006 2:35 pm

I think the number you are trying to get at is C(4,2)/ C(52,2) * C(4,2)/C(50,2) * C(4,2)/ C(48,2) since we agree the order doesnt matter this is correct for a game with 3 players in it , but this number doesnt account for the 3 other random hands at the table that arent going to be AA, KK, or QQ.

by **GooperMC** » Sun Feb 12, 2006 9:44 am

by **laynegt** » Mon Feb 13, 2006 5:28 pm

by **Kuso** » Tue Feb 14, 2006 4:47 am

by **laynegt** » Tue Feb 14, 2006 5:23 pm

nah, I went to virginia. rhound and gooper got me started, and this is all normal college-level probability, anyway...also google helps.

AA V KK V QQ 6-Max

AA V KK V QQ 6-Max

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