Post deleted by Mat Sklansky

[jakeoneil]

[bigpooch]

From the twodimes.net poker hand analyzer:

1) Same suited:

cards win %win lose %lose tie %tie EV
As Ah 1410336 82.36 292660 17.09 9308 0.54 0.826
Ks Kh 292660 17.09 1410336 82.36 9308 0.54 0.174

Equity for KK: 297314/1712304 or about 0.173633887

2) One suit different:

cards win %win lose %lose tie %tie EV
As Ah 1399204 81.71 305177 17.82 7923 0.46 0.819
Ks Kd 305177 17.82 1399204 81.71 7923 0.46 0.181

Equity for KK: 309138.5/1712304 or about 0.180539495

3) Both suits different:

cards win %win lose %lose tie %tie EV
As Ah 1388072 81.06 317694 18.55 6538 0.38 0.813
Kc Kd 317694 18.55 1388072 81.06 6538 0.38 0.187

Equity for KK: 320963/1712304 or about 0.187445103


Since the combinations above have a ratio of 1:4:1, the
average EV is approximately 0.180539495323. If you know the
exact suits, then use one of the other numbers above. The
equity for AA is just 1-(equity for KK).

[cold_cash]

I don't know about Texas, but in Oregon KK is a big underdog. /images/graemlins/grin.gif

[Nottom]

The other posters answered how big an underdog KK is, but I don't think thats what you were asking.

I think you are asking what are the odds of an opponet having AA when you have KK.

If you are heads-up the odds are about 1 in 204.

At a full table each person has about a 1 in 204 chance of having Aces so at a 10 man table you should run into Aces about once every 22 times you have KK.

Note: My answers aren't exact and I'm too lazy and/or incompetent to do the real math.

[bigpooch]

For an individual opponent, the odds of him having AA when
you have KK is 6/C(50,2) or 6/1225 or 203.1667 to 1 against.

If there are more opponents, it will be approximately the
number of opponents multiplied by 6/1225. The exact answer
is n x (6/1225) - C(n,2) x (1/230300) where n is the number
of opponents.

[tpir90036]

is the "- C(n,2) x (1/230300)" term an inclusion-exclusion term to subtract out the times you double-counted when two people have KK?

thanks,
-tpir

[bigpooch]

Correct! Since there is no possibility of more than two of
the opponents holding AA, there are no other terms. BTW,
I believe you intended "AA" instead of "KK" in your post.

The term comes from the probability that two players have
AA: there are C(n,2) choices of players among n opponents,
and the probability that specifically two opponents have AA
is just 6/C(50,2) x 1/C(48,2) = 6/1225 x 1/1128 = 230300.

[tpir90036]

yeah, i meant AA. thank you for the response. you and bozeman are the probability masters!!

[bigpooch]

Have you forgotten other gurus that make their presence felt
on these forums? Clearly, BruceZ comes to mind and there
are quite a few others that have put up some meaningful
posts!

[daryn]

i thought it was something like 1 in 220, but then i read your disclaimer about not being exact.. so nevermind!

[tpir90036]

crap! i forgot to mention BruceZ. sorry bruce if you read this....i have a bunch of your threads bookmarked. i have not been on this forum for very long but you 3 (bruce, bozeman and you) are definitely the ones whose posts i go to first for the straight dope. keep up the good work...

-tpir

[Nottom]

[ QUOTE ]
i thought it was something like 1 in 220, but then i read your disclaimer about not being exact.. so nevermind!

[/ QUOTE ]


Its about 1 in 220 to be dealt AA, but if you take out 2 Ks the odds get better.

[sthief09]

It's 220-1 when there's 52 cards in the deck. In this situation, there's 50, since you hold 2 kings. Hence 203-1.

[tpir90036]

why on earth was this post deleted? it was asking about the odds of someone having AA if he had KK. bueller??