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Old 08-24-2005, 02:00 PM
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Default 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
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Old 08-25-2005, 04:49 AM
mtrubo mtrubo is offline
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Default 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)
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Old 08-25-2005, 08:44 AM
LetYouDown LetYouDown is offline
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Default 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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Old 08-25-2005, 12:59 PM
aloiz aloiz is offline
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Default 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
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Old 08-25-2005, 06:13 PM
BruceZ BruceZ is offline
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Default 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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