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View Full Version : The chance someone has AA if you have KK at a 9 person table

Gavagai
05-04-2005, 08:22 PM
What are the odds of this? Is it right to just do 220:8 since there are 8 others with a 220:1 chance each at the table? Or is this really dumb?

Thanks,

Gavagai

closer2313
05-04-2005, 09:15 PM
This is the simplest non-trivial poker example. If you hold KK before the flop, what is the probability that at least one of your 9 opponents holds AA?

Note that at most 2 opponents can have AA. The probability that any particular player holds AA is 6/C(50,2). So the first term in inclusion-exclusion is simply 9*6/C(50,2). That is the sum of the 9 probabilities of each player having AA. This alone is very close to the exact answer, but it double counts the times that 2 players hold AA. For the second term, we take the probability of 2 particular players having aces, 1/C(50,4), and multiply by the number of ways to pick the two players, which is C(9,2). So the final answer is:

P(KK vs. AA) = 9*6/C(50,2) - C(9,2)/C(50,4)
or about 4.39%

http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&amp;Board=&amp;Number=417383&amp;page=&amp;v iew=&amp;sb=5&amp;o=&amp;fpart=

LINES
05-05-2005, 09:53 AM
44:1

eOXevious
05-05-2005, 01:22 PM
If you have KK, the chances of someone having AA is the same with no matter what hole cards you have. You could have 72o and the chances for AA would be the same. If you hade AK or something to that effect, the chances would change because one of the aces is taken. So the chances of someone haveing aces is 4*3/52*51 Every card in every order = 12 / 2652 = 12:2640 or .45 percent But you gotta remember you got 8 chances (8 other people), so there is a %3.6 chance someone could have them. Now a good question would be, what are the chances someone will get KK and someone else getts AA. 2 hands, so 4 total cards. 4 cards possible for each person. 4 * 3 + 4 * 3 (add the two combos of cards for each player, these are the amount of combos that will succeed / 52 * 51 * 50 * 49 (all four card combos possible) 24 / 6497400 or 1 / 270725 Buts thats only before we know what people have, if you have KK already, it don't mean its 1 / 270725 that the other guy has AA.
Someone let me know if I did this wrong

eOXevious
05-05-2005, 01:24 PM
Wait I think I did something wrong, but I know the best and easiest way to figure out probability is to find the total number of possible ways to succeed and divide by the total number of combos (includeing the succeeds)

Gavagai
05-05-2005, 02:10 PM
It seems I've been given 3 different answers, which is right? Was the difference between the 1st and 3rd answers accounted for by the fact that they referred to 10 person and 9 person tables respectively?

Thanks,

Gavagai

etgryphon
05-05-2005, 02:15 PM
Closer is correct...

The others are wrong. Read the link to convince yourself.

-Gryph

BruceZ
05-05-2005, 02:42 PM
[ QUOTE ]
Closer is correct...

The others are wrong. Read the link to convince yourself.

-Gryph

[/ QUOTE ]

Actually no one gave the right answer since he asked about a 9 person table, though closer gave the correct answer (mine) for a 10 person table. For a 9 person table, the exact answer is 8*6/C(50,2) - C(8,2)/C(50,4) =~ 3.9% or 1 in 25.6 = 24.6:1.

etgryphon
05-05-2005, 02:46 PM
[ QUOTE ]
[ QUOTE ]
Closer is correct...

The others are wrong. Read the link to convince yourself.

-Gryph

[/ QUOTE ]

Actually no one gave the right answer since he asked about a 9 person table, though closer gave the correct answer (mine) for a 10 person table. For a 9 person table, the exact answer is 8*6/C(50,2) - C(8,2)/C(50,4) =~ 3.9%.

[/ QUOTE ]

Oops...At least closer's (yours) math is the correct math for figuring the answer out.

Not to hijack the thread but right now I'm trying to derive a chart that will list the pocket pairs v. # oppoents with hands += to your PP and AK. Not a trivial problem. It gets really hard above 2 opponents.

-Gryph

LINES
05-05-2005, 03:04 PM
The chances that someone holds KK or AA when you have either or, is 44:1. No matter what

BruceZ
05-05-2005, 03:22 PM
[ QUOTE ]
The chances that someone holds KK or AA when you have either or, is 44:1. No matter what

[/ QUOTE ]

Sounds like you're ready to make some side bets! I'll let you give me 40:1. /images/graemlins/cool.gif

binions
05-06-2005, 12:37 AM
[ QUOTE ]
The chances that someone holds KK or AA when you have either or, is 44:1. No matter what

[/ QUOTE ]

Um, no. The chance of AA when you have KK is highly dependent on how many hands are dealt. The others have pegged it. It's ~24:1 at a 9 person table.

At a 2 person table, it would be rarer - 203:1 to be exact. After all, it's 216:1 to get AA before any cards are dealt.

At a 26 person table, AA over KK would be more common than 24:1.

Gavagai
05-07-2005, 10:48 AM
[ QUOTE ]
Actually no one gave the right answer since he asked about a 9 person table, though closer gave the correct answer (mine) for a 10 person table. For a 9 person table, the exact answer is 8*6/C(50,2) - C(8,2)/C(50,4) =~ 3.9% or 1 in 25.6 = 24.6:1.

[/ QUOTE ]

Could you explain this calculation? What is 'C'? I'm afraid I never took maths to a very advanced level.

Thanks,

Gavagai

BruceZ
05-07-2005, 03:46 PM
[ QUOTE ]
[ QUOTE ]
Actually no one gave the right answer since he asked about a 9 person table, though closer gave the correct answer (mine) for a 10 person table. For a 9 person table, the exact answer is 8*6/C(50,2) - C(8,2)/C(50,4) =~ 3.9% or 1 in 25.6 = 24.6:1.

[/ QUOTE ]

Could you explain this calculation? What is 'C'? I'm afraid I never took maths to a very advanced level.

Thanks,

Gavagai

[/ QUOTE ]

Try this post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=2139261&amp;page=&amp;view=&amp;s b=5&amp;o=&amp;vc=1) or one of a million others that discuss the same thing.

J.Copperthite
05-09-2005, 11:14 AM
Of course dealing a flop in a 26 person game would be impossible... /images/graemlins/smile.gif /images/graemlins/smile.gif

eOXevious
05-09-2005, 12:49 PM
[ QUOTE ]
[ QUOTE ]
Actually no one gave the right answer since he asked about a 9 person table, though closer gave the correct answer (mine) for a 10 person table. For a 9 person table, the exact answer is 8*6/C(50,2) - C(8,2)/C(50,4) =~ 3.9% or 1 in 25.6 = 24.6:1.

[/ QUOTE ]

Could you explain this calculation? What is 'C'? I'm afraid I never took maths to a very advanced level.

Thanks,

Gavagai

[/ QUOTE ]

C stands for combination, google combinations and permutations... its very interesting and easy stuff

Cobra
05-09-2005, 07:51 PM
I did something similar. I don't know if this is what you are looking for.

http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=1412067&amp;page=&amp;view=&amp;s b=5&amp;o=&amp;vc=1

Cobra

etgryphon
05-10-2005, 10:48 AM
Haha

Yeah that was the orginial attempt at it. I kinda left it alone for a bit. Can you either PM or just post it here and explain all the terms?

Thanks,

-Gryph

SrGuapo
05-11-2005, 05:33 PM
I follow the math on the AA vs. KK example that closer posted, and was thinking about the probabilty someone will have AA or KK if I have QQ on a ten person table. This math becomes another level more complicated, right?

BruceZ
05-11-2005, 06:19 PM
[ QUOTE ]
I follow the math on the AA vs. KK example that closer posted, and was thinking about the probabilty someone will have AA or KK if I have QQ on a ten person table. This math becomes another level more complicated, right?

[/ QUOTE ]

If you want an exact answer, then you need 4 terms since up to 4 people can have AA or KK, but as a practical matter, 2 terms will still get you to within 0.001%. Here is the exact calculation:

P(AA or KK vs. QQ) =

9*12/C(50,2) -
C(9,2)*(12*7)/C(50,2)/C(48,2) +
C(9,3)*12*(6*2 + 1*6)/C(50,2)/C(48,2)/C(46,2) -
C(9,4)*12*(6*2*1 + 1*6*1)/C(50,2)/C(48,2)/C(46,2)/C(44,2)

=~8.6% or 1 in 11.6 or 10.6-1.

As an explanation of the third term, notice that the first player has a choice of 12 hands, the second player has a choice of either 6 of one hand or 1 of the other, and if the second player chooses one of the 6, then the remaining player has a choice of 2 hands, otherwise he has a choice of 6 hands. As these problems get more complicated, you can draw a tree-like structure to describe the various possibilities.