Quiz: Bob Ciaffone: AK in SB

Surfing the Net I found an interesting Quiz By Bob Ciafone. Here is one hand.


Bob Ciaffone's Limit Hold-em Quiz. April 17, 2001


You are playing in a typical $10-20 limit hold-em game, fairly tight and aggressive, blinds $5-10. The game is ten-handed and the rake is 5%, max. $3. You're playing in seat # 10. There's ... a professional player in seat # 2 ... All other players can be considered average or a little below average.


6) You're in the small blind, holding Ah Ks. Seat # 2 has raised, everybody has folded. What do you do?


fold (7 points)

call (5 points)

raise (4 points)


The raiser figures to have a high pair or AK as well (remember, he's a professional player raising under the gun). Why get involved when your hand is an underdog to begin with? You don't know where you're at in the hand: if you flop a king, you might lose a lot of money; if you flop an ace when the raiser has in fact a wired pair, you might not get any action; if you flop nothing but he doesn't have anything either he might outplay you and make you lay down the best hand.


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Any comments ?

I'm going to let you guess what my answer is. But I've found this problem to be so easy and instructive to analyze that I wrote my next Card Player column on it. But I will give one hint. I don't know any professional players that won't raise UTG with AQ as well.

I'd raise.

I'll re-raise. Even if the pro has a wired pair, the likelihood of him having AA or KK is diminished by my holding. As Mason pointed out, a pro may raise with AQ, or IMO, QQ or other pocket pair.


Incidentally, can somebody show the calculation of the probability that the pro is holding AA or KK? I'm no math whiz.

List of hands pro might raise with in a fairly tight game: AA, KK, QQ, JJ, TT, AKs, AQs, AK.

In a fairly tight game, the pro better be raising with a lot more hands i.e. playing well dictates that you loosen your preflop open-raising requirements in a tight game. So, the pro should be raising with all kinds of hands that he might not raise with in a loose game such as 88, AJ off, KQs, perhaps even JTs etc. These hands are too good to throw away. But if you limp with it in a tight game, you are likelky to get raised from someone behind you and find youreslf in a headsup tilt out of position. Better to raise yourself if you are going to play at all. If the game is such that you feel that you must muck these hands, the game is way too tough to play. Find another game, come back another day...do anything other than stay in the game.

In a tight game, a pro should also be mixing in a few limps with AA/KK.

Ciaffone's rankings are in reverse order.

Ironwood,


It's not usually necessary to be a "math whiz" as most of the significant numbers have already been done, verified, and reverified.


However it might help your game to know HOW this particular figure was obtained.


On any given hand there is a 1/221 chance of being dealt "AA"; obviously the chance is the same for "KK".


This figure is deduced by figuring out how many ways there are to make "AA" in two cards -


There are six ways for this to occur (c/d, c/h, c/s, d/h, d/s, h/s).


First though, we have to know how many possible two card hands a player can start with in holdem.


You could figure it out the same way butwith a 52 card deck it would take quite a while.


(Yes, there is a shortcut and it is 100% accurate).


Anytime you want to know how many combinations of "x" there are in a set of "y" you do the following:


The first card can be any of 52; the second card can be any of 51. Multiply these two numbers and you get 2,652.


However, we don't care about what order they arrived so we must divide by the number of possible orders there are. In this case there are two; I can get black pocket aces either by getting the As then the Ac or vice versa.


2,652 divided by 2 = 1326


1,326 divided by 6 = 221


Remember there are 6 ways to be dealt "AA".


By the way, you can use the shortcut to find out how many ways there are to be dealt "AA"; we know there are four aces, the # of different ways to be dealt AA is:


4 times 3 (or 12) which we must then divide by 2 since there are 2 orders in which each combination can occur.


Using either method we arrive at the answer of 6.


(Relax, we're almost done :>).


The question you asked is slightly more complicated since you might say, "we're not playing with a full deck".


You want to know how often the pro will be dealt "AA" or "KK" when YOU are holding AK.


This changes two things;


#1 - there are now only 50 cards from which he can receive his hand (since you hold 2 cards), and


#2 - it is harder for him to have "AA" or "KK" since you have one of each.


50 times 49 = 2,450


Divide this by the # of orders he can receive his cards (2) and we get 1,225.


Also, there are only 3 ways for him to get "AA" and 3 ways for him to get "KK".


WE'RE HERE ! :>)


The odds against him having either "AA" or "KK" when YOU hold "AK" are 1,219 to 6.


It's usually more convenient to use percentages so 6/1,225 = (roughly) .49 %


Using odds, it's 203 to 1 against him having the two hands that really cripple yours.


There is also a .74 % chance that he ALSO has "AK".


In odds that's 135 to 1 against.


I hope that was worth wading through.


It really is useful to be able to do calculations of THIS type fo the following reason.


If you have even a rough idea of what "his" raising hands are you're able to make more informed choices regarding how to proceed in situations like this - both before and after the flop.


If you have seen that a particular player will raise UTG with AA, KK, QQ, JJ, plus AQ, AJ(suited) or KQ(suited) I think calling - or possibly re-raising - seems to be the right choice.


- BTW, these are "reasonable" raising requirements; I would say the majority of players use a list close to this one. (Some add TT and even 99; others won't raise with KQs, but you get the general idea).


If you know that another player will only raise with AA, KK, or AK (and I know THIS player personally) your "AK" is close to worthless.

is C(Y,X)=Y!/X!(Y-X)!

*****************

To find out how many combinations can be made with 5 cards taken 3 at a time,we would obtain:

C(5,3)=5!/3!(5-3)!

5!=5x4x3x2x1

3!=3x2x1

(5-3)!=2!=2x1

*********************************************

We're assuming that there is no distinction between Ax and xA.

*********************************************

If we want to differentiate between Ax and xA and assume that they are distinct,then we would use the permutation formula:P(Y,X)= Y(Y-1)(Y-2)...(Y-X+1) .

Happy pokering,

Sitting Bull

Yeah, but how many times do you get dealt AA and KK to be able to "mix in a few limps"?

I feel very strongly about never limping with those...

Well done, Larry - nice job.


For someone who describes themself as "not a math whiz", I thought keeping it simple might be the better approach.


As for me, a million thanks for confirming what I thought to be true.


P.S. I didn't take my math as seriously as I should have when I was in school. May I contact you with a few "problems" that I am not sure I figured out correctly ?


If so, I can be reached at:


lundellchris@hotmail.com


Chris L. - U.N.C. '86 (go "Heels)


P.P.S. Your posts have been and I expect will continue to be welcome additions to this forum.


Thanks for all you've added and figure to add in the future.


I especially like your upbeat "I'm going to have fun with this no matter what" while still treating the topic at hand with the appropriate degree of seriousness.


That plus the fact that I am yet to read one word of malice in any post you've authored.


Been meaning to tell you that for a while; glad I finally got around to it.

hehe...good point....but the point still is that if I am in a rock garden and I get AA UTG, you can bet that I will limp. Nothing worse than seeing 7 guys fold and then seeing both blinds shrug their shoulers and also muck. You almost want to toss the big blind one red chip so that he will call.

probably you know more mathematics than I do at this point in my life.

I flunked graduate school about 13 years ago when I made a "C" in "The Theory of Probability".

This course was indeed "fuzzy mathematics'.--(LOL)!

I'm learning from U and the other forum members about Texas Hold'em. I certainly appreciate your sharing your knowledge.

Writers such as Malmuth,Sklansky,Jones,Dr Alanschoonmaker,"Ace"Slotbloom,and others have given me food for thought.

In general,there are no strictly Hold'em games in my area;there are only multiple hold'em and Omaha games.

Since I have NO experience in omaha,I do not play the multiple games.

My "specialty" is no ante 1-5 7-card stud.

There is no higher limit stud game in my area.

I will record your E-mail address. Keep me informed on how you are doing. Thanks for the E-mail.


Now go merrily poker along(LOL)!


Sitting Bull

Ps Take that last cheeseburger with you when you go home and turn the lights off on your way out!


E-mail me some french fries at duple1@hotmail.com!

Thanks to you guys. I have taken some Statistics too but it was eons ago and I'm quite lazy to go back to my books to refresh my brains with permutations/combinations, et. al.

Given the game setting, we can say that the player UTG will raise with AA, KK, QQ, JJ, AKs, AQs, AJs, AKo, KQs, and (in rare cases) with a lower hand (just to "mix things up" and keep us guessing).


Given that we have AKo, his most likely hands are AKo (7); QQ and JJ (6 each); followed by AQs, AJs, KQs, AA, and KK (3 each); AKs (2); and finally rags. Or, put another way, he is most likely to have Queens or Jacks (12), followed by AK (9), followed by an A or K (9), and finally a high pair (6). A third way: he is twice as likely to have A* or K* (possibly a pair) as a smaller pair (24 vs 12). From the get-go, he has a better hand than you half the time. Will he win with that hand? See below.


Throw in AQo (9) as a hand that your opponent could raise UTG with, and life gets more interesting. He's more likely to have an ace (27 possible hands) versus other possible hands (18: QQ, JJ, KQ, KK). He's very likely to have an ace or a high pair.


If an ace does show up on the flop, odds change a bit. He could have AA (1), AKs (1 or 2), AKo (5 or 4), AQs or AJs (2) -- all of which are better than your hand. However, his chance of KK (3), KQs (3), QQ (6), and JJ (6) go up relative to A*. In other words, if an ace shows up, it is more likely to help you than to help him. (Aces show up more often when your opponent doesn't have one, than when he does.)


As for how AKo does against each of these hands: well, jeez, I would think someone somewhere has simulated this, but I have no idea where. I'll do some research and get back to y'all on it. I'm currently working on my own poker simulator (much harder than craps, btw), so maybe in a month or two when I get finished, I'll be able to contribute more here... =)


-Rapid