A question about AA v. KK for the prob. forum

[27offsuit]

Can someone please give me the math behind this probability listed below. I'm certain I've heard this correctly somewhere, but I have multiple people telling me no way. I will post what I think I've heard in white below.

Thanks for any help.


In a 10 handed game, if you look down at KK, what is the probability that someone is holding AA?


What I could swear I've heard before: <font color="white"> 1 in 44 </font>

[Cobra]

Your UTG and look down at KK's, the probability that one or more of your nine remaining opponents has aces are as follows.

First term

=6/(50c2)*9 = .04408

Second term = (6*3/(50c2)/(48c2))*(9c2) = .00047

First term - second term = .043613 or 1 in 22.93 times

This is how often one or more of your opponents will have aces. This was solved using inclusion/exclusion. If you do a search on BruceZ and use the above words he has an excellent description on how to use this technique.

Cobra

[27offsuit]

Thanks Cobra.

Could you explain the 'first term' and 'second term' thing? I'll also look up the BruceZ info. Could you also explain what/where the numbers in your equation are/come from?

Also, does the equation change if I am not specifically UTG, but just saying "regardless of position, what are the odds any other of the 9 have an AA?

[27offsuit]

No need for further explanation. I've found all I need.

Now, to make some big monies from some suckers with my newfound info I got from the internets. Yayz.

[BruceZ]

[ QUOTE ]
Your UTG and look down at KK's, the probability that one or more of your nine remaining opponents has aces are as follows.

First term

=6/(50c2)*9 = .04408

Second term = (6*3/(50c2)/(48c2))*(9c2) = .00047

First term - second term = .043613 or 1 in 22.93 times

[/ QUOTE ]

The second term should be C(9,2)*6*1/C(50,2)/C(48,2). There are 6 ways for the first player to have AA, and 1 way for the second. This gives 4.39% or 1 in 22.8.