How do you calculate this?

[Jordan Olsommer]

OK, I don't know what you're saying here because I think you're changing the formula I had at the end of my last post for four players when it's only three that remain, so let me start again.

here is what I'm looking at as what I came up with to find the probability:



here you have:

(# of ways to choose 2 people to get KK) * (# of ways we can give them KK) * (# of ways to choose 1 person to get AA) * (# of ways to give them AA) * [ (# of ways we can give the remaining three people hands) - (# of ways we can give the remaining three people hands that involve one of them getting the other AA)]

all divided by the number of ways we can deal out six hands.

So what I'm asking is where in that is the double-count of the AA. I'm subtracting all the possible ways that someone else can get AA too in the numerator.

[BruceZ]

[ QUOTE ]
So what I'm asking is where in that is the double-count of the AA.

[/ QUOTE ]

I explained that in my first paragraph, so you need to reread that one. Think about it carefully until you understand it. I'll try again here:

[ QUOTE ]
(# of ways to give them AA) * [ (# of ways we can give the remaining three people hands)

[/ QUOTE ]

The same 2 pairs of aces can be dealt by these two terms, just in reverse order. The same aces are dealt to the same players. That's where the double counting is coming from. Again, it's OK to double count as long as you subtract off what you are double counting.


[ QUOTE ]
OK, I don't know what you're saying here because I think you're changing the formula I had at the end of my last post for four players when it's only three

[/ QUOTE ]

I'm including the 4th player who has the ace. That's why I've multiplied 4*6 by the first term, same as you did. You need to recognize that this term double counts the cases where 2 players have AA as I explained in the first paragraph. Then in the second term, you have to consider C(4,2) players who could both have AA, times 6 ways they can have it. You tried to consider only 3 players in the second term, but this was incorrect. You actually have to consider both players who have AA in this term. I was showing how to fix it to get the correct answer.

You didn't need to show your whole formula as I already understood it completely and corrected it to agree with mine. Here is the corrected formula, by your method. Note that only the second term is different from yours:

2 KK + at least 1 AA:

C(6,2)*C(4,2)*[
C(4,1)*C(4,2)*C(46,2)*C(44,2)*C(42,2) -
C(4,2)*6*C(44,2)*C(42,2) ] /
C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)

= 849,502-to-1.


2 KK + exactly 1 AA:

C(6,2)*C(4,2)*[
C(4,1)*C(4,2)*C(46,2)*C(44,2)*C(42,2) -
2*C(4,2)*6*C(44,2)*C(42,2) ] /
C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)

= 850,737-to-1.

[paperchamp]

[ QUOTE ]

This actually happened in a big tournament not long ago! (There was also a JJ!).

http://cardplayer.com/poker_magazine...amp;m_id=65553

it's hand #3

[/ QUOTE ]

Thanks for the link! Kind of ironic that they were exactly six-handed as well.