Likelihood AA vs KK
 

Hello,

we were talking at lunch about the likelihood of having AA vs KK.
As far as I remember it is sth. like 1:210 to get Aces, so it is the same likelihood to get Kings.

We were not sure about the following solution:

(1/210 x 1/210) x (number of players) = likelihood to have AA vs KK.

Re: Likelihood AA vs KK
 

have a look here for those kind of questions:

WikiPedia - Poker Probability In the chapter: "Dominated Hands - Pocket Pairs"

The answer to that would be:
0.0391, which gives 4% you have KK, you are facing AA. Even though on FT it's about 50%, 'cause FT is RIGGED [img]/images/graemlins/laugh.gif[/img]

Re: Likelihood AA vs KK
 

thx!

Re: Likelihood AA vs KK
 

[ QUOTE ]
Hello,

we were talking at lunch about the likelihood of having AA vs KK.
As far as I remember it is sth. like 1:210 to get Aces, so it is the same likelihood to get Kings.

We were not sure about the following solution:

(1/210 x 1/210) x (number of players) = likelihood to have AA vs KK.

[/ QUOTE ]

Instead of multiplying by the number of players, you must multiply by the number of pairs of players, such as C(10,2) = 10*9/2 = 45 for 10 players. Then being dealt a particular pair is 1/221, not 1/210, and instead of (1/221)^2 this should be (12/1326)*(6/1225), where the first player can have one of 12 pairs (AA or KK) out of C(52,2) = 52*51/2 = 1326 total cards, and the second player can have one of 6 pairs of the other denomination out of C(50,2) = 50*49/2 = 1225 total cards. Note that this is roughly double (1/221)^2 since either player could have the AA. This is a very close approximation, though it over counts the hands where 3 or 4 players have AA or KK, and the exact answer would require 2 more terms to account for this by the inclusion-exclusion principle, but these terms are so small that this isn't worth doing. So for a 10 player table we have

C(10,2)*(12/1326)*(6/1225) =~ 0.2% or about 1 in 500 hands.

Note that the answer from Wikipedia assumes that you already have the KK and want the probability of running into AA, so this probability is much greater. You can find solutions to this and similar problems in this post.