Ace-King question

[kingstalker]

If you hold Ace King, what are the odds that somebody has pocket aces in a 9 handed ring game? (Texas Hold them)

[taxat]

I think its 33897:1.

[Ed Miller]

I think its 33897:1.

I must be the unluckiest person on the planet, because it seems to happen to me constantly. [img]/images/graemlins/tongue.gif[/img]

Since you have one of the aces, there are only three ways to make pocket aces out of 1225 total possible hands. Thus, it is about 407-1 that any one player has pocket aces, given that you have AK. That would make it somewhere around 50-1 that one of your opponents will show up with AA if you have AK UTG in a 9-handed game.

[bigpooch]

You're kidding, right? Anyone who knows anything should
know this isn't so!

If you have AK, there are now C(50,2) = 1225 combinations of
hands of which there are now only 3 combinations of AA. As
inclusion-exclusion does not apply here (at most one AA is
out against you), the chances are just 9x3/1225 = 27/1225
or about 2.204% or 44.37 to 1 against for 9 opponents.

For n opponents, it will be just nx3/1225; the game was only
9-handed so there would have only been 8 opponents and the
chances should be 24/1225 or about 1.959% or 50.04 to 1
against for the situation described.

[TaintedRogue]

What equation would you use for any pair?

Thanx,
Ken

[mojito]

something even much less probable happened yesterday in a tournament 9-handed table. 2 guys shared a pot , they had both K-K !!! now isnt that rare ? [img]/images/graemlins/smile.gif[/img]

[BruceZ]

What equation would you use for any pair?

Any pair, or any particular pair? For any particular pair, if the pair is AA or KK when you hold AK, it is n*3/1225, where n is the number of opponents. For any other particular pair, there are 6 possible ways to have the pair, but it is possible for up to 2 opponents to have that pair, so we have to subtract off that probability. There are C(n,2) ways to pick the 2 players that have it, and 1/C(50,4) ways for them to have all 4 cards of the same rank, so all together the probability that at least 1 particular pair is out is
n*6/1225 - C(n,2)*1/C(50,4).

For any pair, this becomes more complicated due the number of ways that multiple players can have a pair. See this post which computes the probability that at least 1 of 8 players being dealt a pair. This would require just a slight modification for the case where you take your hole cards into account.

[BruceZ]

Also, a good approximation to the probability that one of your n opponents has a pair, when you hold a non-pair, is 1 minus the probability that nobody has a pair, approximated as:

1 - [ 1 - (6*11 + 3*2)/1225 ]^n

= 1 - (1153/1225)^n

The stuff in brackets is 1 minus the probability that one particular player has a pair, which is the probability that he does not have a pair. 6*11 + 3*2 is the number of pairs remaining out of 1225 possible hands. Raising this to the nth power makes this approximate because the hands are not independent. That is, one player having a pair affects the probability of another player having a pair. The dependence is slight, so the approximation is quite good. For 8 opponents, this is 38.4%.

[balkii]

Ok the real question is, what are the chances of someone having AK (suited or unsuited) when I have AQ?

[trillig]

Just a little extra to think about...

If it's 50 to 1 that someone has aces, then it's also 50 to 1 that some has KK, so it's 25 to 1 that someone has KK or AA.
so it's also 2500 to 1 that you are up against KK and AA, right?

My last live tournament, I got knocked out with AKs, what are the odds that this scenario would happen?

AKs me, Op #1: JJ, Op #2: QQ

-t