AK against AA or KK

[slickpoppa]

In a ten handed game, what are the chances that any other player has KK or AA if you have AK. No need to post the calculation. I'm just looking for the answer if anyone happens to know it already.

[kyro]

4.41%

someone corroborate this for me please though

[gaming_mouse]

I get 4.28%

(6/1225)*9 - nCr(9,2)*(6/1225)*(3/1128)=.0428

gm

[kyro]

hmm, i did

9 * 2 * (3/50) * (2/49)

I'm sure your answer is correct, but I would like an answer as to why mine isn't.

As for how I got that. 9 players, AA and KK are two possibilities. (3/50)*(2/49) the chance that one player gets dealt those exact two cards...

[gaming_mouse]

kyro,

(3/50) * (2/49) is the correct way to calculate that one particular player is dealt one particular hand, either AA or KK.

I don't what your thought process was when you multiplied that number by 9*2, so I can't tell you where you are going wrong, except to say that it's not how you do it.

Tell me what you were thinking and maybe I can explain.

I solved it using the inclusion-exclusion principle, the formula for solving P(A or B or C ...) where, in this case, the events are:

A = opponent 1 has AA or KK
B = opponent 2 has AA or KK
etc

gm

[slickpoppa]

The reason you cannot multiply by 9*2 is because the prob. of someone being dealt AA or KK is not independent of what the rest of the table is holding. If none of the 8 other players at the table has an A or a K, the prob of the 10th person holding AA or KK is much higher (2*(3/34)*(2/33)).

[BruceZ]

[ QUOTE ]

I don't what your thought process was when you multiplied that number by 9*2, so I can't tell you where you are going wrong, except to say that it's not how you do it.

Tell me what you were thinking and maybe I can explain.


[/ QUOTE ]

His answer is the first term of the inclusion-exclusion solution. It is not exact because it double counts the times that two players hold AA or KK, so this must be subtracted off. It would be exact if no more than one player could hold AA or KK.