Simplified Sklansky Problem Solved

This is the solution to the simplified Sklansky Problem in which there is only one betting round. However, it should be noted that I have yet to explore whether slow playing will increase expectation. In the problem, Player A and player B were each dealt a real number between 0 and 1. Both had to ante $1. A had the option of betting the pot ($2) or checking. If A bet, B could either call or fold. If A checked, B could bet or check behind him. If A checked and B bet, A could only call or fold. It should also be noted that I figured A’s betting strategy independent of B’s betting strategy. That is, I figured A’s best betting strategy, ignoring what B might do when A checked, then figured B’s best strategy based on that. It may be possible that they both should be combined into one equation with two variables. If so, I doubt if I have the fire power to figure it. Anyway, here is what I have so far.


B will be getting 2:1 on his call so in order for A to maximize expectation, he must give B exactly 2:1 based on hand value, as much as possible (The optimal strategy will put B in a position where it doesn’t matter what he does, as much as possible). So A will be betting his best hands along with a percentage of his worst hands. In order to give him that 2:1 odds as much as possible, he must bet his best hands twice as much as his worst hands. In other words, if he bet his best 1/3 hands, he must also bet his worst 1/6. If X is the lower range of his best hands, and Y is the upper half of his worst range, then Y=(1-X)/2.


B must call all hands X and higher, as he will be getting 2:1 pot odds, and his chances of winning will be between 1/3 and 1 (because of A’s bluffs). Actually his eq in that upper range will average 1 bet (half the antes) against A’s upper range. If A didn’t bluff, then B could tweak this by folding the lower part of this range. With hands below Y, B must fold, as he loses against all in the upper range, as well as some bluffs, giving him less that 1/3 chance of winning. With hands between Y and X, it doesn’t matter whether B calls or folds, as he has 1/3 chance of winning and is getting 2:1 pot odds. B could lower his swings by folding all of the time, but in the real world, he should call, to prevent A from changing strategy.


To calculate A’s equity, we must add all the combinations together. That is, A’s bluff hands against B’s low range, middle range and high range. And also A’s high range against B’s same ranges. To do that we add up the likelihood of each combination multiplied by the amount A will win or lose. It should be noted that total equity will add up to $2, because of the dead antes. Subtract $1 from the total to see how much A or B will win on average for this hand.


That can be represented by the following formula:


2Y**2 - 2Y(1-Y) + 2Y(1-X) + 4(1-X)(X-Y) + (1-X)**2, which reduces to


4Y**2 - 4Y +2XY -3X**2 +2X +1, then substituting Y=(1-X)/2 = 1/2 - X/2


4(1/2-X/2)**2 -4(1/2-X/2) +2X(1/2-X/2) - 3X**2 + 2X +1, which reduces to


-3X**2 + 3X = A’s equity for his betting hands.


To find the highest equity for value X we need to find the point on it’s graph where it’s tangent is zero.


The derivative gives us the formula for the tangents.


The derivative of -3X**2 + 3X is -6X+3 (I forget the exact way to represent this, it’s been a long time :-( )


Solving for zero:


0=-6X +3


X=3/6=1/2


So A’s proper betting ranges are 1/2...1, and 0...1/4.


A’s equity for the betting hands is $3/4 (Just plug 1/2 into -3X**2 +3X).


A’s equity for the checking hands is figured the same way. This time, however, B will be trying to put A in the position where it doesn’t matter whether A calls or folds, as much as possible.


B knows that A only has hands 1/4...1/2 left. He will obviously bet all hands higher than this and bluff too. B’s goal is to give A a 1/3 chance of winning. B will bet hands 0...Y and X...1. (Note: we are using the variables X and Y again, but they are not the same X and Y in the above formula.) A will fold all hands less that Y for the same reasons B did. We know that Y=(1-X)/2 and that both Y and X will be between 1/4 and 1/2, possibly those exact two numbers. The math will tell us exactly where. Again, we will be figuring for A’s equity (checking hands).


4Y(1/2-Y) + (X-Y)**2 + 2(1/2-X)(X-Y) + (1/2-X)**2 - 2(1/2-X)(1/2) -2(X-Y)(1-X), which reduces to:


-3Y**2 -2XY + 3Y +2X**2 -X -1/4, substituting Y=1/2-X/2 and reducing we get,


9/4X**2 - 2X +1/2 = A’s equity on his checking hands


Solving the derivative for zero we get


9/2X -2 = 0,


X= 4/9


B will bet 4/9...1, and 0...5/18. A will fold 1/4...5/18, call 4/9...1/2 and shows the same equity 5/18...4/9 whether he calls or folds, so he will call.


A’s equity (checking hands) = $1/18

A’s equity (betting hands) = $3/4

A’s total equity = $29/36


B’s equity = $2 - $29/36 = $43/36.


A will lose 19 16/36 cents per hand.


B will win 19 16/36 cents per hand.


To re-cap:


A should bet 0...1/4, and 1/2...1

B should call 1/4...1 and fold 0...1/4

If A checks B should bet 0...5/18, and 4/9...1

If A checks and B bets then A should call 5/18..1/2 and fold 1/4...5/18


I guess David can consider this my undergraduate study in preparation for my Masters Degree at Sklansky University.

* I figured A’s betting strategy independent of B’s betting strategy. * Actually it is the other way round.


* In order to give him that 2:1 odds as much as possible, he must bet his best hands twice as much as his worst hands. * Right ! The point is to make the calling hand indifferent to call ANYWERE in the middle range.


* B could lower his swings by folding all of the time, but in the real world, he should call, to prevent A from changing strategy. * Right ! You have to 'keep him honest' - not only in the real world.


Nice, that you actually is producing an answer one can relate to:


* A should bet 0...1/4, and 1/2...1

B should call 1/4...1 and fold 0...1/4

If A checks B should bet 0...5/18, and 4/9...1

If A checks and B bets then A should call 5/18..1/2 and fold 1/4...5/18 *


... but your solution is wrong. It's not stabilezed. Just one example: Why would B bluff ?


*A’s total equity = $29/36

B’s equity = $2 - $29/36 = $43/36. *


Could be your method of calculation is right. EV for the optimal solution is:


EV(A)= $11/12

EV(B)= $13/12


Have you solved the problem - in the case where only B is allowed to bet ?


Let me know if you want more - and how much - I don't wanna spoil your fun ;-)

can someone repost this "sklansky problem"? i'd like to take a shot at it.

A and B each ante $1 and are each dealt a random number.


A can check or bet the pot. If A bets, B can call or fold. If A checks, B can check or bet the pot, then A can call or fold.


If both players remain, they are each dealt a second random number.


A second round of betting follows, same rules as the first.


If both players remain, there is a showdown. Player whose two random numbers add to the higher total wins the pot.


Optimal strategies? Value of the game to each player?


--JMike

p.s. I forget exactly how to say it, but let 'random number' = a real number drawn from a uniform distribution on the interval [0 1]

Players A and B ante 1$. Both players are "dealt" a real number from 0 to 1. Player A can check or bet. Player B can check, bet, or call, but not raise. A second number is dealt in round 2. The same rules apply (A can check or bet. B can check, bet or call). All bets are for the size of the pot. Best hand is the sum of the two numbers.


What is the optimal strategy for the players, and what is their EV?

http://www.twoplustwo.com/cgi-bin/ne....pl?read=31855

By all means, spoil my fun.


B needs to bluff for the same reasons A needed to bluff. B will not get paid off if he bets his best hands only. A will gain more than enough equity laying down his 1/4...1/2 against B's low hands (1/4 * 1/4 * 2 = 1/8 compared with 1/18 otherwise) and folding otherwise. So B must do some betting with hands lower than 1/4.


I already stated that I hadn't considered A slow playing his hands. I've yet to do that. It may very well be that A can do better by checking some, if not all, of his best hands. Also, I stated that I figured A's best betting stratery first, then B's response to that. Combining the betting and checking formulas into one (with two variables) may yield a different result.


However, you posted the solution previously as A betting 0...1/9 and 7/9...1, with B responding by betting with 5/9...1. You failed to mentinon their calling hands. But assuming the best B is allowing A too much equity by not betting some of his lower hands too.


I suspect that combining the two formulas into one will yield a better result. I have no opinion on the slow playing but will look into it.

"B could lower his swings by folding all of the time, but in the real world, he should call, to prevent A from changing strategy."


If B calls too much, then A will change strategy, and not bluff as much, but play more hands in the high range. That way B will will lose out by calling too much. Similary in the real game, against a calling station, wait for cards, they are real value.


It works both ways, if B calls too little, A will bluff more, and B's EV will go down.


I can't comment yet on the rest, I'm the one having fun playing with my algebra. I'll hopefully post my answer on A's best strategy in the simplified game in a week or two. Every day I discover new angles.

At the moment though I think The Poker Player Formerly Known As Jack is correct with his values of 1/9, 7/9, and 5/9 (sorry).

No! No! No !


That refered specifically to the range where the equity would be the same whether B called or not. If A changed his strategy when B called in this range his expectation would go down! B's calling forces A to stay honest.


5/9 is not the correct betting strategy to A's 1/9&7/9. B's correct strategy would be to bet 0...15/81, and 51/81...1. That assumes that A always bets in that range.


By the way, 1/9:7/9 (68/81) does have a higher expectation than 1/4:1/2 (29/36). And always checking yields a better expectation than both (8/9). "Jack" is claiming an expectation of 11/12 for A. He has not stated the strategy that would yield that. I suspect that it includes A sometimes checking hands he would otherwise bet.