Pocket kings vs. the ace on flop

Kweejibo · Apr 25,2005, 11:56 PM

If you have pocket kings in a limit game, raise pre-flop and get 3-4 callers in a loose game and the ace falls, are you generally better off to fold at this point to someones bet or stay with it? Does position make much of a difference? what type of situations would a call or a raise be recquired?

Dave Scharf · Apr 26,2005, 01:19 PM

Generally, fold. You're beat, fold.

Roy Cooke, in answering this same question said (paraphrasing): "If you fold K-K in a limit hold'em game anytime there is an ace on board for the rest of your life, you will save a lot of money."

I agree. Players LOVE aces. If there is any card that your opponents are most likely to have, it's an ace. Certainly, you are in classic spot of "winning small pots" when you have the best hand because they can't call, versus losing big pots if you doggedly hang onto K-K.

Fold'em.

Lamb_of_john · Apr 27,2005, 01:48 AM

of course it's an all togather different story when that ace hits the flop accompanied by a big fat smelly KING.

WHOOOOOOWEEE!!!Â* What a great feeling it is to see your cowboys make a set.Â* No one sees it coming!!

thanks
johnny

Flint Bones · Apr 27,2005, 08:34 AM

In a regular table of 9 the odds are that someone has at least one ace (I believe that Math states there is a 70% chance and only 30% when playing less than 5).

So folding isn't a bad play.

I like John's comment, however, I am still hurting from having pocket kings and hitting my king on the flop and being beaten by a straight. The guy called a 3XBB in EP with 53o. Who does that?

Back on topic, it may be hard to muck the KK but the odds usually are that someone has the Ace. It sucks but it happens a lot.

Dave Scharf · Apr 27,2005, 10:05 AM

Quote:

In a regular table of 9 the odds are that someone has at least one ace (I believe that Math states there is a 70% chance and only 30% when playing less than 5).

My instinct is that it's higher than that. I am having supper with a couple of PhD math guys tonight, I'll ask.

Also note that it's not how frequently an ace WILL be present, it's how frequently a player will PLAY an ace. Aces get played. So there is a basic chance of "will an ace be present" followed by "will it be played."

Flint Bones · Apr 27,2005, 10:17 AM

Quote:

My instinct is that it's higher than that. I am having supper with a couple of PhD math guys tonight, I'll ask.

Please do. I think I have it right but I am not positive. I am certainly not aÂ* Mathmagician or even a Mathlete. Perhaps it seems higher due to your other point that they are played far more than any other card.

Dave Scharf · Apr 27,2005, 10:21 AM

http://www.math.sfu.ca/~alspach/comp47.pdf

Brian's work on the distribution of aces pre-flop at a 9 handed or 10 handed hold'em game.

Go slow...

ScottyZ · Apr 27,2005, 10:27 AM

As Dave already pointed out, the probability that someone was dealt an Ace is easy to compute, but figuring the probability that a player with an Ace sees the flop, turn, river and showdown is very complicated.

In low-limit, it may be reasonable to assume that a typical player will see a flop 100% of the time with any Ace, and see the showdown every time he flops an Ace. With those assumptions, the two probabilities1 above are the same.

The probability of no-one being dealt an Ace at a 9 player table (assuming you have KK) is

(34 choose 4) / (50 choose 4) = ~20%

since there are 34 cards remaining in the stub of undealt cards.

So, the unconditional probability of at least one Ace being out in the hole cards when you have KK is ~80%.

This hasn't quite answered the question, since we are looking for the (conditional) probability of an Ace being out in the hole cards given that at least one Ace has appeared on the flop.

By Bayes' rule2:

P(Ace in hole cards given Ace on flop)

= P(Ace in hole cards and Ace on flop) / P(Ace on flop)

The numerator is a little tricky to calculate since the events are not independent.

ScottyZ

1Conditional on flopping an Ace that is.

2Technically, by definition.

CurlingZone · Apr 27,2005, 03:24 PM

What about a re-raise in this position? If you take the pot, great.

If you get some callers again (with low kicker) you get to see one more card, but you know he's got an ace. One more chance to hit the K, and you'll pay off nicely if you do.

If you're re-raised, you know its time to get out.

MickeyHoldem · Apr 27,2005, 06:58 PM

Mmmmmmm.... math!!

Given you have KK and the flop has one ace... the odds that atleast one player at a full table (10 player) has been dealt an ace is equal to 1 minus the odds no one was dealt an ace ... calculated the following way:

atleast one ace = 1 - (44 choose 18)/(47 choose 18) = 1 - 0.225 = .775 or 77.5%

(47 choose 18) = the ways to choose 18 cards out of the the remaining 47 unseen cards to give to the 9 opponents
(44 choose 18) = the way to choose 18 cards out of the 44 unseen non-ace cards to give to the 9 opponents

whether or not said opponent will play any ace... I leave to more knowledgable minds!!

Now the more tricky implied question...

Given you have KK, what are the odds an ace will flop (or atleast one ace will flop), when atleast one opponent holds an ace?1

There are (50c18) ways to choose cards for nine opponents = 18053528883775 (1)
There are (46c14) ways to choose cards for 9 opponents that contain all 4 aces = 239877544005 (2)
There are 4*(46c15) ways to choose cards for 9 opponents that contain 3 aces = 2046955042176 (3)
There are 6*(46c16) ways to choose cards for 9 opponents that contain 2 aces = 5948963091324 (4)
There are 4*(46c17) ways to choose cards for 9 opponents that contain 1 ace = 6998780107440 (5)
Finally there are (46c18) ways to choose cards that contain no aces = 2818953098830 (6)

as a check we can see that (2) + (3) + (4) + (5) + (6) = (1)

Now in all situations there are (32c3) possible flops = 4960
We can ignore (2) & (6) since the opponents hold all the aces or none

In (3)... (31c3) flops will contain no ace = 4495
In (4)... (30c3) flops will contain no ace = 4060
In (5)... (29c3) flops will contain no ace = 3654

Now we con compute an answer...

= (3)/(1) * (1 - 4495/4960) + (4)/(1) * (1 - 4060/4960) + (5)/(1) * (1 - 3654/4960)
= 0.172496743 or 17.25%

Whewww... that was fun!!

1 assuming of course that an opponent will play any ace they are dealt... to simplify the question!

boxcard · Apr 27,2005, 10:44 PM

Great post Mike. It seems to me your calculations are bang on given the assumptions you made.
To respond to CurlingZone's post, I don't believe that you will find many low limit players who will lay down their pair of aces to your re-raise. Fishing for a 2-outer with only two cards to come would require a hefty sized pot to make this a +EV play. Also, you are going to pay the price even worse if you hit your set, only to find that your opponent has a set of aces(just as probable as your set). I think that in the long run, you will make a lot more money by folding.

Yorkpoker · Apr 28,2005, 02:57 AM

If I see someone with a slide rule at the table, does that mean he's holding kings?

FrogKing · May 17,2005, 11:10 AM

We can ignore (2) & (6) since the opponents hold all the aces or none

(6) the opponents hold all the aces.

But you do have to be carefull, cuz some of these time one opponent will have pocket aces.