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. /images/graemlins/tongue.gif

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 ? /images/graemlins/smile.gif

[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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=396659&page=&view=&sb =5&o=&vc=1) 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

[BruceZ]

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

With 9 opponents, using inclusion-exclusion (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=417 383&Forum=probability&Words=inclusion-exclusion&Match=Entire%20Phrase&Searchpage=0&Limit =25&Old=3months&Main=416981&Search=true#Post417383 ):

9*(12/1225) -
C(9,2)*3*4*2*3/C(50,2)/C(48,2) +
C(9,3)*3*4*2*3*1*2/C(50,2)/C(48,2)/C(46,2)
= 8.6% or 1 in 11.6

There are 3 terms since up to 3 opponents can have AK. Note the standard inclusion-exclusion form. The first line is for 1 of 9 opponents having it, so 9 is multiplied by the probability of a particular player having it or 12/1225. The second line subtracts off 2 players having it, so the leading term is C(9,2), and the final line is for 3 players having it, so the leading term is C(9,3). The denominator in each term is the total number of ways to deal hands to the players who have it without regard to order. The numerator in each each term is the number of ways to deal AK, without regard to order, starting with 3 aces and 4 kings in the deck.

Or you can use this quick and accurate approximation:

1 - (1213/1225)^9 = 8.5% or 1 in 11.8.

Almost the same result, for a lot less work.

BTW, I stated in the post I referenced above that the text by Feller doesn't cover the inclusion-exclusion principle. I started to think that it must cover the probability of a union, and in fact it does in chapter IV.1, and it even mentions the inclusion-exclusion principle by name. It just didn't have it in the index. Also, DeGroot has the general formula in 1.10, though it doesn't refer to it as the inclusion-exclusion principle. However, neither book gives detailed examples of the type we have covered many times on this forum.

[BruceZ]

Of course if you have AQ, you are also concerned about AA and KK being out, as well as AK. This post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=213933&page=&view=&sb =5&o=&vc=1) has the probabilities for this and several other similar match-ups, for 9 handed, and for 4 handed. I computed these for use on the WPT TV show on the travel channel awhile back. Though the figures are correct, I have since found much simpler ways to express some of these probabilities, and you can find those in more recent threads about inclusion-exclusion.

[trillig]

I had some AK sweet revenge the other night.

I got AKo, got All In called by AA and KK6 flopped and then rags, oh that felt good. 8)

-t