#1
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AA v KK v QQ at a five handed table
I played at a home game last night, and the following hand came up with five players at the table.
P1 AA, P2 KK, P3 QQ, P4-xx, P5-xx I read that the chances of hitting pocket aces are about 1/221? Does this mean that the odds of this particular combination are roughly as follows? (5/221) * (4/221) * (3/221) Thanks |
#2
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Re: AA v KK v QQ at a five handed table
1 in 221 are the odds of hitting a pocket pair of a specific rank, so the odds of getting AA, KK or QQ all have 1 in 221 odds.
so the odds are 221*221*221 = 221^3 However, there are C(10,2) ways of arranging starting hands for 5 players. so the odds are 221^3/C(10,2) = 221^3/45 = 239863.6 1 in 239863.6 (someone please correct me if i'm wrong) |
#3
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Re: AA v KK v QQ at a five handed table
You need the inclusion/exclusion principle for this to get an exact answer. Do a search, you should find it instantly.
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#4
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Re: AA v KK v QQ at a five handed table
If you want the odds that you have AA v KK v QQ v XX v XX where XX does not include QQ-AA then the following will work:
I split the equation up into two halves. P(3 players have AA, KK, QQ) * P(remaining 2 don't have AA, KK, QQ|first 3 do) 5*C(4,2) * 4*C(4,2) * 3*C(4,2) / C(52,2)/C(50,2)/C(48,2) Choose what player has AA, and which suits. Choose what player has KK and which suits. Choose what player has QQ and which suits. P(2 remaing players don't have AA,KK,QQ|first three do) = 1 - P(either or both players have AA,KK,QQ|first three have AA,KK,QQ). So we have two cases. The first being that only one of the two has QQ-AA, the second being that both have QQ-AA. First case: 2*3*(C(44,2)-2) / C(46,2)/C(44,2) Choose what player has QQ-AA, choose QQ, KK, or AA. Choose hand for second player. Second case: C(3,2)*2 / C(46,2)/C(44,2) Choose the two pairs, then the division of those pairs between the two players. So we get: 5*C(4,2) * 4*C(4,2) * 3*C(4,2) / C(52,2)/C(50,2)/C(48,2) * (1 - 2*3*(C(44,2)-2)/C(46,2)/C(44,2) + C(3,2)*2/C(46,2)/C(44,2)) =~ 7.032 * 10^-06 or about 142201:1 If you want to include the possiblity that the other two players could have QQ-AA then you need to use inclusion exclusion principle, and the problem becomes pretty difficult. I'd actually be surprised if an exact answer has been given before on this forum. aloiz |
#5
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Re: AA v KK v QQ at a five handed table
[ QUOTE ]
If you want to include the possiblity that the other two players could have QQ-AA then you need to use inclusion exclusion principle, and the problem becomes pretty difficult. I'd actually be surprised if an exact answer has been given before on this forum. [/ QUOTE ] I solved this for a 10-handed table here. Regular inclusion-exclusion won't do it. You need a generalization of inclusion-exclusion. Also see some of the other posts in that thread for an explanation. |
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