The chance someone has AA if you have KK at a 9 person table

[Gavagai]

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]

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/...;o=&fpart=

[LINES]

44:1

[eOXevious]

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]

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]

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]

Closer is correct...

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

-Gryph

[BruceZ]

[ 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]

[ 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]

The chances that someone holds KK or AA when you have either or, is 44:1. No matter what