AA v KK?

[vkotlyar]

you are in a 10 handed game and pick up KK UTG. what is the prob that some1 has AA? how would i figure this out?
V

[bocablkr]

[ QUOTE ]
you are in a 10 handed game and pick up KK UTG. what is the prob that some1 has AA? how would i figure this out?
V

[/ QUOTE ]

I saw this same question last night on the WPT except it was for a 6 handed game. They claimed the answer was 44-1. I do not understand how that is correct. Can anyone explain it. Thanks.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
you are in a 10 handed game and pick up KK UTG. what is the prob that some1 has AA? how would i figure this out?
V

[/ QUOTE ]

I saw this same question last night on the WPT except it was for a 6 handed game. They claimed the answer was 44-1. I do not understand how that is correct. Can anyone explain it. Thanks.

[/ QUOTE ]

They're wrong then, it's 40-1.

5*6/C(50,2) - C(5,2)/C(50,4) = 40-1.

For 10 players it's 22-1. See inclusion-exclusion principle (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Board=&Number=417383&page=&v iew=&sb=5&o=&fpart=).

Come on, this gets asked every week. Next time type into the search engine. That's what it's there for.

[LetYouDown]

What I'm wondering, is how the hell did they came up with 44 to 1? I saw it too, so I know he's not misquoting. I've tried "screwing up" the math with every oversight I can think of. I can't get to 44-to-1.

[BruceZ]

[ QUOTE ]
What I'm wondering, is how the hell did they came up with 44 to 1? I saw it too, so I know he's not misquoting. I've tried "screwing up" the math with every oversight I can think of. I can't get to 44-to-1.

[/ QUOTE ]

Replace C(50,2) with C(52,2). In other words, they didn't take the KK into account. That gives 1 in 44 which they could misquote as 44-1.

[LetYouDown]

Indeed, I missed the most blatent possible mistake. Thanks for that. They forgot to exclude the K-K, and don't seem to know the difference between odds and probability, yet they figured out how to handle Inclusion/Exclusion, as you call it? Bizarre.

Edit: Scratch that...answer still rounds to the same odds.

[BruceZ]

[ QUOTE ]
Indeed, I missed the most blatent possible mistake. Thanks for that. They forgot to exclude the K-K, and don't seem to know the difference between odds and probability, yet they figured out how to handle Inclusion/Exclusion, as you call it? Bizarre.

[/ QUOTE ]

They don't really need the second term to get close.

I once gave the WPT the answers for 9 and 4 handed here (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Number=212797&page=&view=&sb =5&o=), along with a simple way to estimate the answer (which doesn't give 44-1).

[bocablkr]

Thanks for the answer guys.

[BruceZ]

Note that this isn't the first time they made this error (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=1913645&page=&view=&s b=5&o=).

[LetYouDown]

I believe this was a repeat of that episode. It was the Legends of Poker. Had to leave, but Doyle had Watkinson outchipped like 4 to 1. Not sure what year that was, don't watch much WPT.

[LetYouDown]

Just doing this for my own benefit...have a feeling this is wrong.

You have A-Q at a 10 person table. What are the odds that someone has A-K?

(9 * 12)/50 C 2 - C(9, 2)/C(50,4) - C(9,3)/C(50,6) ?

[BruceZ]

[ QUOTE ]
Just doing this for my own benefit...have a feeling this is wrong.

You have A-Q at a 10 person table. What are the odds that someone has A-K?

(9 * 12)/50 C 2 - C(9, 2)/C(50,4) - C(9,3)/C(50,6) ?

[/ QUOTE ]

9*12/C(50,2) -
C(9,2)*12*6/C(50,2)/C(48,2) +
C(9,3)*12*6*2/C(50,2)/C(48,2)/C(46,2)

= 10.6-1.

The C(50,4) trick only works when all 4 card are the same, like all aces.

[LetYouDown]

Yeah, I had that thought but it was too late to edit my post. Thanks.