Probability question (KK vs AA)

[willybeer22]

Given: You have been dealt KK.

Question: What are the odds that someone else is holding AA at a 9 man table?

I guess it would be something close to the odds of one guy getting dealt AA knowing the deck is short 2 kings, then multiplied by 8 choose 2 spots for where it could be or something? :O I'm a little rusty on my probability hope that Cobra is out there and reading /images/graemlins/smile.gif. My ballpark guess: 1/24

thanks,

will

[BruceZ]

[ QUOTE ]
Given: You have been dealt KK.

Question: What are the odds that someone else is holding AA at a 9 man table?

I guess it would be something close to the odds of one guy getting dealt AA knowing the deck is short 2 kings, then multiplied by 8 choose 2 spots for where it could be or something? :O I'm a little rusty on my probability hope that Cobra is out there and reading /images/graemlins/smile.gif. My ballpark guess: 1/24

thanks,

will

[/ QUOTE ]

Welcome to the forum.

There are 6 ways to have AA, so the probability of 1 particular player having it is 6/C(50,2).
Multiply this by 8 players to get 8*6/C(50,2). This alone gets you very close, but it will double count all cases where 2 players have AA, so for the exact answer we have to subtract this off. The probability that 2 specific players have AA is 1/C(50,4) since there is only one way to deal 4 aces to 2 players. Since no more than 2 players can have AA, we can just multiply this by the number of 2 player pairs C(8,2). So the exact answer for the whole thing is:

8*6/C(50,2) - C(8,2)/C(50,4) = 1 in 25.6 = 24.6-to-1.

For more discussion, see this thread (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=1457813&page=&view=&s b=5&o=), (for a 10 person table) and the link to the inclusion-exclusion principle explanation in that thread.