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Stork
11-07-2004, 03:06 PM
Can someone please tell me the odds, or show me how to get them, when holding KQo, of being up against AK, AQ, AA, KK, or QQ in a 10-handed game? What about for 6 max?
Thanks.

zaxx19
11-07-2004, 10:21 PM
After a raise pretty damn high thats all that yu really need to know.........and for a called raise somewhere in between.........KQ off so darn overatted imho for the same reason kx suited is weak...

Stork
11-07-2004, 11:20 PM
I'm not talking about being up against a raise, or using information from the way the hand is played out, I just mean statistically.

uuDevil
11-08-2004, 12:43 AM
Finding exact answers can get kind of messy. Here is a long thread on a similar question, with input from pzhon, BruceZ, and David Sklansky, amoung others:

Odds of KK running into AA at a full table? (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=842374&page=&view=&sb =5&o=&vc=1)

Precision1C
11-08-2004, 03:08 AM
There are 1326 combinations of Texas hold'em hole cards, 50!/(48!*2!)= C(50,2).

AK= 4*3 combinations
AQ= 4*3
AA= 6
KK= 3
QQ= 3

for a total of 36 combinations. So the odds someone dominating your hand is 36/1326= 2.71%

At a 10 person table the odds that no one has a dominating hand is (1-.0271)^9= 78.1% At a 6 person table (1-.0271)^5= 87.1%.

Hope this helps.

pzhon
11-08-2004, 03:14 AM
When you have KQ, there are 6 ways to have AA, 3 ways to have KK and QQ, and 12 ways to have AK and AQ, for a total of 36 hands out of (50 choose 2)=1225. The probability that any particular opponent has one of these hands is 36/1225, so the average number of dominating hands you face is 9*36/1225 = .2645 at a full table and 5*36/1225 = .1469 at a 6-handed table. These are overestimates for the probability you are dominated because they overcount the times you face multiple opponents with dominating hands. It is quite messy to determine the exact answer.

Precision1C
11-08-2004, 11:10 AM
Oops, pzhon has the answer. C(50,2) is 1225 not 1326.

pzhon
11-08-2004, 12:46 PM
[ QUOTE ]
Oops, pzhon has the answer.

[/ QUOTE ]
No, I posted a way of getting an estimate. My estimate (.2645) is good if the dominating hands repel each other perfectly. You posted a way of getting an estimate, too. Your estimate (.2354) is good if the dominating hands don't repel each other at all. The actual value is between the two, and it is messy to work out.

To refine my estimate, we can consider the probability that at least two players have dominating hands. After you have KQ, a pair of players can have (50 choose 2)(48 choose 2) cards. If the first player has one of the 6 AA hands, The second player can have a dominating hand 19 ways (8 choose 2 - 9 ways to have KQ). If the first player has one of the 3 KK or QQ hands, the second player can have a dominating hand 25 ways. If the first player has one of the 12 AK or KQ hands, the second player can have a dominating hand 22 ways. These total 6*19+3*25+3*25+12*22+12*22 = 792. The probability that a pair of players has dominating hands is at most 45*792/(50 choose 2 * 48 choose 2) = .0258.

So, the probability that KQ is dominated at a full table is between .2649 and .2645-.0258 = .2387. It should be much closer to the latter.

Stork
11-08-2004, 09:52 PM
Thanks to everyone who replied /images/graemlins/cool.gif. I'm really just trying to get a feel for the worth of KQo in general.