OT: Texas Hold Em Puzzle

[primelord]

As usual this puzzle was not of my own creation. As a matter of fact I am not even certain I completely understand the answer yet. :) But atleast I have the answer. It is just a straight math problem, but with all the poker interest on the board I thought some may find it interesting.

It's possible some of you have run across this before so if you have seen it before please give the people who haven't a chance. This problem was originally created by David Sklansky.

One and two dollar blinds. You are in the small blind. Everyone folds. (Infer nothing about card distributions from that). Your hand is A8 offsuit. Your opponent in the two dollar big blind has more money than you. He is an expert.

You have only two options. Fold, or turn your cards face up and move in. He sees your A8 before he acts. Whether you should fold or move in depends on how much more money, besides the dollar blind, you have in front of you. Hopefully it is obvious that you should not move in if your bankroll is above x dollars. What's x?

[MJ4H]

I believe I just saw this the other day on 2+2. I will not further comment :)

[primelord]

Originally posted by MattJones4Heisman
I believe I just saw this the other day on 2+2. I will not further comment :)

You were the one I was thinking about when I said some of you may have run across this before. :)

No takers? Quik?

[QuikSand]

I'm not sure I understand the puzzle correctly.

[QuikSand]

I hve trouble seeing how there is a perfect answer here, but in saying so, I possibly am revealing weaknesses in my analytical game.

If there is a magic number (as the puzzle heavily implies), then I'm guessing it resides in some ratio that involves some copmbination of your chip count, his chip count, the amount of the remaining blind, and perhaps the likelihood of A8 beating a randomly selected two-card hand (which I suspect is pretty good, but Ii don't know what it is). My guess is that it's two of those factors in a fraction one over the other... but intuitively, I'm at a loss to see what factors make up that deciding ratio.

While this may be useful as an exercise, I can't help of my college days when I would play pool at a student union hall, and I'd frequently see a guy who practiced hour after hour of certain trick shots - especially shots involving shooting balls as they were rolling across the table. I once commented to my friend "Yeah, that's going to come up in a game, so you have to be ready!"

My only choices are to reveal my hand and go all in, or fold it entirely... really? Uh, okay... for science, right.

[primelord]

Sklansky does come up with an exact number that you shouldn't move all in with. I am still having trouble myself determining exactly how he came up with the numbers he did. Perhaps I shouldn't post a puzzle that I can;t even explain the answer for. :)

I'll try to give a hint, but again being that I don't even fully understand the problem it's entirely possible I'll either give the problem away by giving the hint or give a hint that is of no benefit what so ever.

It appears that you don't need to know offhand what the likelihood of A8 beating any two random cards is, but rather what percentage of hands would make it correct for the big blind to call with.

As for whether or not being able to solve this will benefit you at the tables I have no idea. :) Sklansky labeled the problem as "Important No Limit Math Problem" however it's possible he has just finally covered every relevant topic in poker and is now just moving on to obscure situations for his own enjoyment. :)

[QuikSand]

I also assume that we are at a no-limit cash table, rather than in a no-limit tournament (where chip position and bust sequence are more important considerations). We're counting the chips as nominal value, right?

The problem is lacking many details otherwise, so I feel secure in this assumption.

[primelord]

Originally posted by QuikSand
I also assume that we are at a no-limit cash table, rather than in a no-limit tournament (where chip position and bust sequence are more important considerations). We're counting the chips as nominal value, right?

The problem is lacking many details otherwise, so I feel secure in this assumption.

I made the same assumption although he doesn't specify in either the problem or the answer.

[QuikSand]

Here's my worry with this problem (and I may be missing something entirely). You put your stack in, and reveal your A8. Then, it's your expert opponent's decision whether to call the bet against the A8.

Shouldn't you then know that you will either be (a) getting his $2 blind, in the event that the expert player sees that he has the losing end of this bet, or (b) called in a situation where you are the underdog, and your expected losses are possibly rather significant?

So, I guess that's the puzzle... figure out how many hands the big blind wouuld call with on what is essentially a 50-50 bet. Then, you could back out the expected losses by figuring your likelihood of winning the bet against each of them (weighted for their likelihood) and impute the likelihood of winning that bet overall. From there, you could get the implied ratio of how much it's worth risking to try and steal this guys measly two bucks.


So... I suppose you need to first know exactly what hands are worth him calling here. All pairs, I reckon, and all aces with a better kicker, and suited A8 as well. That is a total of 90 hands, if my quick count is correct.

I'm not sure whether it's correct to call here heads up with anything less than an ace, suited or otherwise. (Shabby math guy that I am) Does he call with KQs? As an expert player, he probably shoudl factor in his expected winnings from just sticking to basic plays, rather than pushing in on a big near-even split... but I reckon Sklansky wants this down to the penny.

I'm done. ::white flag::

[primelord]

Before revealing the answer. Which I assume I can do soon since you are the only one who has taken a crack at it. You probably shouldn't call with something like KQs. The odds are in the favor of neither hand improving so calling with anything that doesn't win if neither hand is improved is probably a mistake.

[Maple Leafs]

Originally posted by QuikSand
My only choices are to reveal my hand and go all in, or fold it entirely... really? Uh, okay... for science, right. And yet you don't mind when Bono can walk across a bridge five times faster than The Edge?

[primelord]

Ok here is the answer. I will try and make any future puzzle I post better than this one. :)

The opponent should not call big bets with worse than A9 or a pair. Disregard A8 for simplicity. That's about one tenth of all hands. When he calls he will win about about 70%. The bettor, with x more dollars to bet, thus has an expectation of .9 times 2, plus .03 times (x+1), minus .07 times (x+1). Folding gives him an expectation of minus one.

Seting them equal and multiplying by 100 gives us:

180 + 3x + 3 -7x -7 = -100

4x = 276

x = 69

[QuikSand]

That (exactly 69, as opposed to "about 70") is an awfully precise answer for supporting data like "abot one tenth" and "about 70%." Don't they teach significant digits in poker shool?

[primelord]

Originally posted by QuikSand
That (exactly 69, as opposed to "about 70") is an awfully precise answer for supporting data like "abot one tenth" and "about 70%." Don't they teach significant digits in poker shool?

So you're saying my puzzle sucked. Fair enough. :)

[cuervo72]

It would have been better had the answer been x-3.

[Celeval]

Originally posted by QuikSand
That (exactly 69, as opposed to "about 70") is an awfully precise answer for supporting data like "abot one tenth" and "about 70%." Don't they teach significant digits in poker shool?

But 69 is within one significant digit of 2/3, so the solution holds.

[sabotai]

"x = 69"

I think Sklanksy has a dirty mind...

[Vegas Vic]

Sklansky and his sidekick, Mason Malmuth, are fond of coming up with these esoteric Holdem problems, but they don't have much practical use. I don't know too many players who have access to a computer or calculator (or the time) during the course of a game to do these complex calculations.

[Butter_of_69]

Churchill!

[QuikSand]

Originally posted by Vegas Vic
Sklansky and his sidekick, Mason Malmuth, are fond of coming up with these esoteric Holdem problems, but they don't have much practical use. I don't know too many players who have access to a computer or calculator (or the time) during the course of a game to do these complex calculations.

I can see the point. If solving this problem requires you to properly understand the notion of "when would he call" and "how likely am I to be beaten if he calls" then it does encourage elements of the correct poker mindset, I suppose.

However, the actual math of the thing surprises me... from the puzzle setup, I expected there to be a rather elegant solution, and am disappionted that in this case Sklansky was willing to round off all the edges of the math (and still come to a precise answer, sich still irks me).

I think I was on the right track with my musings above, but basically abandoned them because I figured I was getting too complicated. Turns out Sklansky reached the same conclusion, used a couple of thumbnails instead, and "completed" the math that way.

[Raven Hawk]

x = your mortgage payment

[Ufer]

Originally posted by QuikSand
I can see the point. If solving this problem requires you to properly understand the notion of "when would he call" and "how likely am I to be beaten if he calls" then it does encourage elements of the correct poker mindset, I suppose.


I agree. The top players often think in terms of "what does my opponent want to do and how can I minimize his options." So in this case, even if Howard Lederer can read your mind, it doesn't help him. Sure, only the hard-core pros have ready access to the probabilities and the math (because they think about these problems all the time).