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

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)

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.

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

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.