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/newforums/genpok.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.

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.


A's EV going down only assumes that B knows how to respond. If B calls at 0.5 or better all the time, A can change his strategy (ie no bluffing), to increase his EV. Of course if B knows that A has changed his strategy, and acts accordingly, then A's EV goes down, but that is only if B has worked it out. If B keeps calling at 0.5, and A stops bluffing and increases his high range, A's EV will go up I am quite sure.

... about two diferent problems.


Spikey's problem: (only one player can bet)


* 5/9 is not the correct betting strategy to A's 1/9 & 7/9. * No it's not ! 5/9 is a CALLING strategy.


George's problem: (both player's can bet)


* I suspect that it includes A sometimes checking hands he would otherwise bet. * Right ! Actually that is a key-point.


More later. I'm busy ;-)

Just a thought. Has it been considered that the best strategy involves more random play, such as "bet hand X with probibility X"?


Or maybe more complicated like "Bet hands X above 1/2 with probability X, and hands between 1/4 and..." etc.


Just a thought. Has it been considered that the best strategy involves more random play, such as "bet hand X with

probability X"?


Or maybe more complicated like "Bet hands X above 1/2 with probability X, and hands between 1/4 and..." etc.


I haven't read the original thread, so I don't know if this has been discussed, but it seems to me that maybe too much information is being given away by playing deterministically.

You're repeating yourself. ;-)


Yes it has been considered, but not yet calculated (by me). Jack seems to have a handle on it. Based on the solution of a problem from four years ago, the optimal solution will be just that--checking some hands (or all hands) you will sometimes bet, and at some optimal rate.


I'm having trouble grasping that concept mathematically. I lack the game theory knowledge others have so I'm trying to understand the problem from scratch. It's not easy.

I thought you might be interested in this. (This can be found at http://wolff.to/bruno/poker.html)


Analysis of a simple poker game

Last updated on January 4, 1999.


Two people play the following game:

Both players contribute to an ante of amount A (not necessarily equally).

Each player receives a uniformly distributed random number between 0 and 1.

Each player only knows the value of his number.

The first player may bet any nonnegative amount.

The second player may either match the amount bet or give the ante to the

first player (the second player may not raise or bet on his own).

If the second player matches the bet then the whole pot is given to the

the player with the higher number (which are revealed after the second

player decides what he is going to do).

The questions is: what fraction of the ante should be contributed by each

player to make the game "fair"?


The solution follows:


Optimal strategies for the players:

Let X and Y be player 1's and player 2's numbers, respectively. Let R be the

ratio of the bet to the ante.


2 ARCCOS(-SQRT(7X))

0 0 call if Y >= -------


1 + R


The first player should get back 4/7 of the ante, the second player 3/7.


How I derived the solution follows:


There are going to be many optimal solutions for each player for this game.

Many of these will not be analytic functions of the information available.

Finding one of these to start would be pretty difficult. The first step is

to put some extra conditions on the solutions to make the problem more

tractable. The extra conditions come from strategies that are good against

nonoptimal play by your opponent. Two conditions for player 2 are fairly

obvious. He should always call a bet of 0. For any particular bet he should

call on his best hands and fold on his worst hands.

For the first player things aren't as clear. However if the first player

only bets on his best hands he won't do very well since the second player

will only call when he has him beat. The first player can always guaranty

an expected value of his number times the ante by betting 0. Since when

bluffing you expect to lose if you are called you should bluff with your

worst hands and bet 0 with ones that are just poor.

At this point you want to find a pair of optimal strategies for playing

the game. These strategies have the property that neither person has any

incentive to change his strategy (though he may be indifferent to making

changes) to some other strategy if the other person is using an optimal

strategy. Searching for strategies that leave the other person indifferent

to his choices is often a very usefull way to search for optimal strategies.

Specificly, in this case the first player should bluff exactly enough to

keep the second player indifferent to calling or folding with the worst

hand he is calling with.

For the second player things aren't so clear. Two possibilities stick out,

either make the first player indifferent to where he bets on good hands or

to where he bets on bad hands (bluffs). The first option won't work since

the amount to bet for the maximum expected value for betting on good hands

depends on the hand, whereas for bluffing it doesn't. This means that the

betting will be spread out but that the bluffing won't. This will induce

the second player to change his strategy so that this won't result in an

optimal solution.

To see this, let E be the expected value divided by A and let F (a function

of R) be the smallest value the second player will call with.

Then for good hands (X > F), E = F + (1 + R)(X - F) - R(1 - X).

Simplifying gives E = X + 2RX - RF - R.


dE dF

-- = 2X - F - 1 - R-- which is a function of X.

dR dR


For bad hands (X -------- = (1 + R) = SQRT(--------)


2(1 - c) 2(1 - C)


1 - B

=> 1 - C = -----


2


To get another relation between B and C, I looked at the total amount of

hands needed for bluffing. The first player should bluff exactly enough so

that the second player is indifferent to calling or folding with his worst

calling hand. Let L be the ratio of bluffs to legitimate hands. The second

players expected return when calling is: E = (1 + R)L - R, and is 0 if he


R

folds. Setting these equal gives L = -----.


1 + R


/ 1 R(X) / 1 1

So that B = | -------- dX = | (1 - --------) dX


/ C R(X) + 1 / C R(X) + 1


/ 1 2(1 - X) 2 1 - B 3/2

= | (1 - SQRT(--------)) dX = 1 - C - - SQRT(-----)(1 - C)

/ C 1 - B 3 2


1 - B 2 2 1 - B 3/2 1 - B 2 1 - B 1 1

= ----- - -SQRT(-----)(-----) = ----- - - ----- = - - -B


2 3 1 - B 2 2 3 2 6 6


=> B = 1/7, C = 4/7


While against an optimal strategy it doesn't matter what hands the first

player bluffs with particular hands, in order to get an analytic function

for bluffing, we will use the following strategy: The first player will bluff

having his best hands with his worst hands. The rational for this is that the

second player is more likely to incorrectly call small bets with bad hands

(Y 0, so that the first solution is the one

we want to use when bluffing.


The expected value to the first player of the game (dividing by the amount

of the ante and not counting his contribution to it) is:


/ 1/7 / 4/7 / 1 1 - B

E = | 1/7 dX + | X dX + | ((2R + 1)X - 2R + 1 - B - -----) dX


/ 0 / 1/7 / 4/7 R + 1


/ 1 3 3 1

= 1/49 + 15/98 + | (2 SQRT(--------)X - X - 2 SQRT(--------) + 3 - - -


/ 4/7 7(1 - X) 7(1 - X) 7


6/7


______________) dX = 4/7


3


SQRT(--------)


7(1 - X)


So that the first player will win back 4/7 of the ante. Hence, the second

player will get back 3/7 of the ante. These are also the amounts they

should contribute to the ante to make the game fair.


Comments:


It is interesting that bluffing is naturally part of an optimal stratgey for

playing poker. It will be interesting to see the affects on the betting

stategies if the game is extended to allowing raising. I suspect that the

first player is giving away too much information about his hand, using the

current strategy, for it to be used if the second player is allowed to raise.

I am also curious to see if sandbagging (checking and raising) will be part

of an optimal strategy.

I believe that solving similar games with raising allowed is going to be

difficult (for me at least). Normally in multidecision games the trick is

to work backwards from terminal positions in the game or for cases with

loops, work backwards until the same position reappears. However, in this

case since betting reveals information about your hand distribution which

will affect future decisions, but the distribution is also based on the

expected value of the next stage in the game, it seems it will be difficult

to apply the above strategy in solving these kind of games.

Back in March or April of 1998 (see archives) David posted a problem he dubbed "The Ultimate Problem", or something like that. It was two players getting one last card, the first holding Aces, and the other holding kings. It was a real life situation, with check-raising, etc., except that both players knew for certain each others cards.


The problem was eventually solved by Paul Prudate (not sure of the spelling of his name). He posted his work and perhaps you will want to take a look at it.


Of course if differs from these problem in that it used discreet values. That simplified it in ways (and made it more complicated in others).

To get optimal strategies, in this game, you need to have Bayesian subgame perfection. Therefore you need to solve B's strategy as a function of A's strategy and keep. Assuming (and I think it's a fair assumption) that A's strategy will be to bet if he has a realization between [0,x) or between (1-2x,1] B's optimal strategy is to bet when he has a realization between [0,y) or (1-2y,1] where y=(15x+3)/24. Didn't bother to find out A's strategy analytically, but computationally I would say that x=.8876 approximately. Note that I am treating players as randomizing their calls when they have a realization between the two extremes and always calling when they are greater than 1-2x{1-2y} and always folding when less than x{y}.

And keep what??


My trying to solve this problem with my math background is akin to flying to the moon in 1950. I'm trying to invent a technique as I go along. Probably better that I get a book and learn game theory and then put my mental energy into going beyond that.