Happy new year! Any luck?

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Asymmetric range game theory is still in the early solver phase of evolution. It's currently not possible to calculate gto check raising ranges without solvers, and even then the solutions will depend on the preflop ranges, which are still debatable.

The first decision point on the flop looks like this:

.do I have a donking range on this flop?
.....yes*....................................no**
....../.....\....................................\
check ..bet..................................check
............................................../.........\
...................................call/fold......check raise

notes:

*checking range consists of (slightly profitable checking hands), (draws that earn exactly the fraction of the pot by betting as these same hands earn by checking) and (very profitable checking hands). The check raise range should then be selected from the top of the (very profitable checking hands) range segment, as well as some of the stronger (draws that earn the same fraction by betting and checking) at frequency <100%. Perhaps a few of the strongest draws to check the flop should be bet 100% of the time.

**checking range consists of (very profitable check raising hands get check raised 100%), (marginally profitable check raising hands get check raised <100%), (unprofitable check raising hands that call or fold), (very profitable draws that get check raised 100%), (slightly profitable draws that get check raised <100%), (slightly profitable draws that check call 100%), and (slightly profitable draws that check call <100%).

I think that the above is true for any holdem game no matter the structure. However, different structures will influence the size and makeup of these range segments and the profitability of these range segments.

True but I'm looking for the naive solution as a baseline

I know it has been a while, so if you will, please restate the terms exactly of the toy game you are studying, and then if you want the frequencies for 1/2 pot bets, I will provide you a solution. The only calculus involved is with calculating optimal bet size, so if you have already decided on the bet size then it is simply plug-n-play once you have the equations.

You're a life saver:

Probably cant put it all into one post, so let's break it up and double check each step and give references along the way. Some of this may be a review for you, if you have studied the referenced paper before. It has come up is some lengthy threads here in the Poker Theory section before. I am not teaching you, but wording this in a step by step manner for others that may need it broken down into steps.

***Attention for this examination of (0,1) toy poker games, zero is nut-low and 1 is the "nuts"***

For starters, you want one complete street of poker, where each player can check or bet, and raise or checkraise, and the action ends with either a checkback, a fold, or a call.

This is the game described and solved in section 4 of the paper:

UNIFORM(0,1) TWO-PERSON POKER MODELS
Chris Ferguson, TiltWare, Los Angeles
Tom Ferguson, Mathematics, UCLA
C´ephas Gawargy, Mathematics, Universit´e Paris 1

which can be located at link:

https://www.math.ucla.edu/~tom/papers/poker2.pdf

However, the authors have left some computations up to the reader, for brevity, as the method is thoroughly documented earlier in section 2 of the paper.

You want the solution for 1/2 pot bets and raises, instead of the full pot size used in the paper. The equations in the paper have already had the full pot bets inserted and simplified the equations so we will actually have to do some algebra on our own. But hey, it's just algebra!

We get to recreate the equations from scratch. How do we do this? We use the principle of "indifference".

The authors have already mapped out all the different actions that can take place, in an extensive form game tree diagram in figure 4.1

Player 1 can take these 5 actions, which will create different sections within his range of hands. They are:

bet-fold
bet-call
check-fold
check-call
check-raise

But, how many regions are there and how do we figure out where they go?

First, you have to remember some bets are bluff and some bets are value. Also, some checkraises are bluff and some are value. We do not allow Player 1 to bet-reraise as a re-bluff, since we do not allow this for value either. It is part of the strategy in the paper, that balanced players checkraise with the very top of their range for more profit than just "peddling the nuts" by bet-raise.

So lets rename our actions to denote the different behaviors:

Bluff-Fold
ValueBet-Fold
ValueBet-Call
Check-Fold
Check-Call
BluffCheck-raise
ValueCheck-Raise

Now we have 7 strategies for player 1.

Next step is to order these strategies into regions and figure out which ones have higher hands than the other and put them in order. No hands are in more than one region, and each region is "bounded" by one or more "indifference points".

Which section has the lowest hands? The Bluff-Fold region. We scrape the bottom of the barrel for these otherwise worthless hands, and use them for profit! However, we will never ever win with these if called.

Next lowest is the check-fold region. These hands are bad and not worth a call, but not bad enough to use for the bluff-fold.

Note: These regions are ordered logically so that P1 can expect extra profit from "mistakes" by P2. If P2 checks-back with a hand that should have been bet (for bluff or value), then maybe our lowest check-fold region will get a chance at a free showdown for instance.

If our last region was check-fold, then lets make (at least temporarily) the next region the check-call region. These are not worth a bet, but are worth a call. Pretty straight-forward logic.

Next higher region is the ValueBet-Fold region. These are worth leading with a bet, but not calling a raise.

Next is the ValueBet-call region. We bet these and then call a potential raise from P2.

So far we have 5 regions put in logical order, but we still have to locate our two checkraise regions. Remember, these are Bluffcheckraise and Valuecheckraise.

The ValueCheckraise region should be the very best hands we have. We use these hands to get more profit from P2, and also to "protect" our entire check-call region from exploitation by P2. When P1 checks as first to act, he does this with the widest range of hands and the "nuts" is always part of these hands in this toy game (and balanced poker players will have some combinations of the "nuts" in the check region to checkraise with at least some of the time).

So far so good, six out of seven regions are in order, and it was all pretty logical. But what about the Bluffcheckraise region? Hmm...

Where do we put it? Recall the note from earlier about increased profit from possible mistakes. If P2 checks-back with an above marginal hand that should have been bet for value, or checks-back with a total trash hand that should have been a bluff, then P1 could get a free showdown with a hand from the check-fold region of P1.

What mistake can P2 make after P1 BluffCheckraises that might give extra profit to P1?

The only possibility is an overcall. If P2 calls with the correct hands, and folds the correct hands, then P1 gets nothing extra. However, the farther down from the "nuts" a hand is (and the farther up from "nut low"), the harder it is in practical application for a poker player to judge its worth. Remember, these (0,1) real distribution toy games are eventually intended to represent actual hands dealt with actual cards. The farther into the middle of our range we look, the more our mind's eye trembles.

But we will not try to capitalize on this by altering P1 range in error. P1 will still only bluff with a hand that serves no other purpose. P1 will Bluffcheckraise with the very top of the check-fold region. P2 can only call, and these hands are right in the middle of range, so if P2 is going to make a mistake, this is the most likely candidate.

This is a long winded way to say "we pick the better check-fold hands and Bluffcheckraise them since they might possibly extract free profit from a mistake".

So Bluffcheckraise goes back in between check-fold and check-call regions... Now we are set with our seven regions of P1 range, and we know what order in which they go.

Now lets name the separation points between the regions variables: a, b, c, d, e, and f. The range of player 1 looks like this:

0<-------a<-----------b<---------c<-------d<-------e<-------f<---------1
Bet-F Chk-F Chk-R Chk-C Bet-F Bet-C Chk-R

Next post: The ranges of player 2 and the full set of indifference equations....

Fantastic post. I'm giddy with anticipation

Just a quick correction this morning, before diving into the indifference equations tonight.

The BluffCheckraise region is drawn from the bottom hands of the check-call region. The location does not change, it is between check-fold and check-call, but the final answers indicate it is comprised of hands that would otherwise be check-called. This corrects my previous statement, and restores a correct understanding of bluff-checkraising strategy. We bluff-checkraise with the worst hands with which we might check-call for value. The location of these hands is still near the middle of range, which still makes them harder to see within range.

***Attention for this examination of (0,1) toy poker games, zero is nut-low and 1 is the "nuts"***



Picking up where we left off in the last post, we are investigating section 4 of the paper:



UNIFORM(0,1) TWO-PERSON POKER MODELS

Chris Ferguson, TiltWare, Los Angeles

Tom Ferguson, Mathematics, UCLA

C´ephas Gawargy, Mathematics, Universit´e Paris 1



which can be located at link:



https://www.math.ucla.edu/~tom/papers/poker2.pdf



Next task is to define Player 2's ranges and then use the principle of "indifference" to generate the full set of indifference equations.



The ultimate goal is to put 1/2 pot size bets (or any bet size) into the equations and look at the resulting frequency of checks, bets, raises, calls, and folds. After all the equations are defined, then we can look at results and theorize about why players make bets less than pot size and what an optimal response should be.



From the game tree diagram in figure 4.1, P2 can take the following actions. Note that P2 acts second, so each action is relative to a first action from P1.



If P1 checks, then P2 can:

check
Bet-Fold
Bet call



If P1 bets, then P2 can:

Call
Raise
Fold



It is possible to put all these into one (0,1) range diagram, but it is easier to keep track of what is going on to examine them separately, which is the method in the paper.



When P1 checks, P2 has 4 strategies, since she can Bet-fold as bluff and Bet-fold as value.



So we have:

Check
Bluff-fold
VBet-fold
Vbet-call

These define 4 regions in the range, let's rank them by strength of hands:

Lowest hands are the Bluff-fold, as with P1 earlier.
Next is the Check region, these are not worth a bet and not bad enough to bluff.
Next is the ValueBet-fold, these are good for a bet but not calling a checkraise.
Last is the ValueBet-call, these are good for a bet and calling a checkraise.

Now we give the "indifference points" between them variables, called g,h,and j.

P2's range, after a check from P1, looks like this:

0<----------g<-------h<-----------j<--------1
Bluff-f, chk, bet-f, bet-call

When P1 leads with a bet, then P2 can take these actions:

Call
Raise
Fold

And has 4 regions which are:

Call
Bluff-Raise
Value-Raise
Fold

We rank each region by hand strength:

Which region is the lowest? At first glance we might think the Bluff-raise would have the lowest hands, but here P2 is acting second in response to a bet from P1. Remember the previous discussion of logically capitalizing on a potential mistake from P1, and that P1 might also make an overcall.

When checkraising for bluff, we do this with the very bottom of our check-call range, to earn more profit with hands that are likely losers.

So, the Fold region is the bottom of P2's range when facing a bet from P1.
Next is the Bluffcheckraise region.
Next comes the Call region, these are worth a call but not a raise.
Last is the Value-Raise region. These are the top hands she will raise with when facing a bet from P1.

Give the boundary indifference points variables k, m, and n.

P2's range when P1 bets first looks like this:

0<---------k<----------m<------------n<---------1
Fold, Bluff-Raise, Call, Value-Raise

Alright, now we have three diagrams of ranges for all of the strategies involved in the hand. Each region is "bounded" by indifference points or 0, or 1. Further, each indifference point now has a variable name, and we can generate our indifference equations.

Lets pause for a second and talk a little bit about the principle of indifference.



Indifference is NOT required to solve these types of poker games. Further, these diagrams are also not required, and the solving can be done in other ways. Indifference, and these diagrams has the following advantages: Half as many equations, half as many variables, and a better depiction of the relationship of the ranges of each player. Lastly, to do this requires the pot to belong to neither player and some important assumptions about the order of the variables are also required, to solve it this way.


When the original half-street of (0,1) poker was presented by John von Neumann and Oskar Morgenstern in the seminal groundbreaking work 'Theory of Games and Economic Behavior' (1944), which is listed as a reference in the paper, there was no application of indifference and no range diagrams, per se'.

Indifference and these diagrams help us to "see" what is going on between the players more clearly!

When 2 regions of a player's range are separated by an indifference point, this means that, when the opponent is optimal, then the player is indifferent to taking either action in either range. We take the potential outcome of the action in each region and set them equal to each other.

The pot is 2 units and belongs to neither players.

Further it is assumed that:

a<g<b<c<h<d<j<f and a<k<m<d<e<n<f


While you can start anywhere and get to the same end, for simplicity we start with the first variable, "a" and work our way down the list, as is the method in section 2 and 3 of the paper. These first two can fit in the end of this post. Then maybe two or three per post. There are a couple of mega equations and those will take up a whole post. When we have all twelve equations calculated and double-checked, I will use the aid of computer to solve the twelve equations in twelve unknowns.


For the sake of further discussion of strategy, I will keep the bets and raises separated by player so that we can alter the results asymmetrically to show the value of the game change if one of the players decides to bet more or less optimally than the other.

So, a bet from Player 1 is called B1 and a raise from player 1 is R1. Similarly, a bet from player 2 is B2 and a raise is R2. For now though, we will just say that B1=B2 and R1=R2



For Player 1:

Player 2 makes Player 1 indifferent to bluff-fold and check-fold at indifference point "a" by:

If P1 bets amount B1 at "a" then he wins the pot of 2 with probability k and loses B1 with probability (1-k). If P1 check-folds he gains zero, since a<g. This means P1 bluffs at hand "a" and gains if P2 folds, and always loses if P2 calls. "k" is where P2 decides to fold or do anything other than fold. Now we start to see the relationship of the ranges.



By indifference:

Player 2 makes Player 1 indifferent at "a" by:

EV of Bluff-fold = EV of Check-fold

2k - B1(1-k) = 0

2k - B1 + B1k = 0

2k + B1k = B1

This is the first indifference equation, and a quick double-check with pot-limit of B1=2 yields the equation in the paper: 2k=1

For the next equation, Player 2 makes Player 1 indifferent between check-folding and bluffcheckraising at b:

When player 1 check folds, he wins 2 with probability (b-g) and zero all else.
When player 1 checkraises he also wins the 2(b-g) above plus 2+B2 with both probability g and also (j-h) and loses 2+R1 with probability (1-j)

EV of Check-fold = EV of BluffCheckraise

2(b-g) = 2(b-g) + (2+B2)(g+j-h) - (2+R1)(1-j)

0 = (2+B2)(g+j-h) - (2+R1)(1-j)

(2+R1)(1-j) = 2g + 2j - 2h + (B2)g + (B2)j - (B2)h

2 - 2j + R1 - (R1)j = 2g + 2j - 2h + (B2)g + (B2)j - (B2)h

2 + R1 = (R1)j + 2g + 4j - 2h + (B2)g + (B2)j - (B2)h

(R1)j + 2g + 4j - 2h + (B2)g + (B2)j - (B2)h = 2 + R1

This is the second indifference equation, and when we substitute pot limit where B2=2 and R1=6, we get the same answer as in the paper.

So far so good, more equations tomorrow!

Ok, Houston we have a problem. It would seem that the chef has given us the wrong recipe to make our pie! Let us just say that the authors want to make it hard on us, lol!

If so, we can only proceed with the assumption that the final answers given are correct, but a few of the indifference equations are changed (to throw us off).

example:

In section 3.2 the indifference equations for section 3 are derived.

The important rules given are:

The pot is 2 and belongs to neither player,
where 0 <a<e<b<c<f< 1 and 0 < d < e.
d<a is allowed.

For P1 to be indifferent at a:

If P1 checks, P1 wins 2 with probability a ....ok
If P1 Bets amount B, P1 wins 2 with probability d and loses B with probability (1-d) ...ok
What is left out is that d<a is allowed, so P1 can also win 2+B with probability (a-d) if P2 ever calls with d<a.

However, we know from the answers that d>a for all d that we will be examining, so we must amend our rules accordingly.

So also we use: a<d<e<b<c<f for all values which we are examining, which makes things a whole lot easier, and would appear to be the method in the paper.

Back to the equation:

Setting these equal gives 2a = 2d-B(1-d) which is exactly the expression in the paragraph. By simple algebra this resolves to:

2a = 2d-B(1-d)
2a = 2d-B+Bd
B+2a = 2d+Bd
B = -2a+2d+Bd
B = -2a+(2+B)d which is the proposed corrected indifference equation at point a.

however the paper lists this as −2a − (2 + B)d = B

which is correct? Lets insert the answers from the solution and see...

For pot-limit where B=2, if a=(2/21) and d = (23/42) then...

Our equation B = -2a + (2+B)d

2 = -2(2/21) + (2+2)(23/42)
2 = -4/21 + 92/42
2 = -8/42 + 92/42
2 = (92-8)/42
2 = 84/42
2 = 2

The equation given in the paper −2a − (2 + B)d = B

B = -2a - (2+B)d
2 = -2(2/21) - (2+2)(23/42)
2 = -4/21 - 92/42
2 = -8/42 - 92/42
2 is not equal to -100/42

So, this means that, if we are to proceed, we have to trust the final answers (likely from a computer) and will use the final answers to check our indifference equations, instead of comparing our indifference equations to the equations in the paper.

Now that I am assured that I have not forgotten high school algebra, things will proceed at a faster pace!

Sometimes I must admit that poker theory is much more difficult than I imagine it to be.

I am having trouble with the indifference equations in section 4. Most likely, due to the added strategy of check-rasie “sand bagging”.

However, stubbornness I have in abundance, so I will pause with section 4, and go to section 2 and walk through each equation there and verify that I do have all the regular betting/bluffing EV understood. This *should* be brief, since they are all there in full form in section 2.

I feel better now that even you're struggling. Frankly thanks to your efforts, this has the makings of thread of the year

Ok I am going to walk through the easier indifference equations from section 2.

Some of these are just word-for-word the same as the paper, some of them are just easier to compute my way. I don't know if the last three are calculated in the manner found within the paper for any particular reason, or not. Below works just as well, since we have the shortcut of the order of the variables.

Paragraph 2.1 has the description of the actions taken by both players and the range diagrams. Paragraph 2.2 is the indifference equations.

***Note*** The pot is 2 and belongs to neither player. If the action goes check-check, bet-fold, or check-Bet-fold, the winning player always wins 2. **

We assume a<e<b<f<c and a<d<c





Player 1 indifferent between bet and check-fold at a:

If P1 bets B at x=a he wins 2 with probability d and loses B with probability (1-d)

If P1 check-folds he wins 0. This requires a<d and a<e.

Set these two expressions equal to each other:

2d-B(1-d) = 0

2d = B(1-d)

2d/(1-d) = B

the paper solves for d:

2d –B +Bd = 0

(2+B)d = B

d = B/(2+B)



Player 1 indifferent between check-fold and check-call at b:

If P1 check-folds at x=b, P1 wins 2 with probability (b-e) and 0 otherwise.

If P1 check-calls at x=b, P1 wins 2 with probability (b-e), wins 2+B with probability e, and loses B with probability (1-f)

This requires a<d and a<e.

So, we see a single decision by P1 interacts with the entire range of P2...

Set these expectations equal:

2(b-e) = 2(b-e) +(2+B)e - (1-f)B

*note* if the action goes check-check, there is no difference between check-fold and check-call. The 2(b-e) cancels out immediately.

0 = (2+B)e - (1-f)B

0 = (2+B)e –B +Bf

B = (2+B)e + Bf



For Player 1 to be indifferent between check-call and bet at point c:

If P1 check-calls at x=c, he wins 2+B with probability e, wins 2+B with probability (c-f), wins 2 with probability (f-e), and loses B with probability (1-c).

**Note** this section of range is check-call but we must include the EV to P1 if P2 checks back 2(f-e) ***

If P1 bets at x=c, he wins 2 with probability d, wins 2+B with probability (c-d) and loses B with probability (1-c)

Set these expressions equal:

**note** we keep the expressions "as is" because several terms will cancel immediately! ***

2(f-e)+(2+B)(e+c-f)-B(1-c) = 2d+(2+B)(c-d) -B(1-c)

Quick cancellation of redundant values gives: (this saves time but is not required)

2(f-e)+(2+B)(e-f) = 2d+(2+B)(-d)

2f-2e+2e-2f+Be-Bf = 2d-2d-Bd

More cancellation...

Be-Bf = -Bd

---> e – f = -d

---> d + e – f = 0 (This requires f<c and d<c)

*** Interesting, this is not in terms of B. No matter what the bet size, any B>0, these three indifference points are ALWAYS related in this ratio! ***



Now for Player 2...



For Player 2 to be indifferent between fold and call at d:

P2 wins 2+B with probability a and loses B with probability (1-c)

She wins zero by folding...

Set these equal:

(2+B)a-(1-c)B = 0

(2+B)a –B +Bc = 0

(2+B)a + Bc = B This requires a<d<c.



For Player 2 to be indifferent between bet and check at e:



If P2 bets amount B at e, she wins 2 with probability (b-a) and loses B with probability (c-b).

If P2 checks at e, she wins 2 with probability (e-a), zero otherwise.

Set these equal:

2(b-a)-B(c-b) = 2(e-a)

2b-2a-Bc+Bb = 2e-2a

2b-Bc+Bb –2e = 0

(2+B)b -Bc –2e = 0 (This requires a<e<b.)



Lastly, For Player 2 to be indifferent between check and bet at f:



If P2 checks at f, she wins 2 with probability (f-a), and zero otherwise.

If P2 bets amount B at f, she wins 2 with probability (b-a), wins 2+B with probability (f-b), and loses B with probability (c-f).



Set these two equal:



2(f-a) = 2(b-a) + (2+B)(f-b) - B(c-f)

2f-2a = 2b-2a + 2f-2b+Bf-Bb – Bc+Bf

Cancellations...

0 = Bf-Bb-Bc+Bf

0 = 2f-b-c

Or

0 = b+c-2f (This requires b<f<c.)



So, we arrived at each of the six equations as provided in section 2.

d = B/(2+B)

B = (2+B)e + Bf

0 = d + e – f

B = (2+B)a + Bc

0 = (2+B)b -Bc –2e

0 = b+c-2f

All right! I was stuck for a while on the third indifference equation from section 4.2 of the paper.

What was tripping me up is I was trying to include every previous branch decision in the game tree in the EV calculation at every point in the tree. Actually, quite a silly mistake.



So, since others might have the same trouble, let's look at both the incorrect calculation as well as the correct, to avoid the repetition of the error in the rest of the equations.



***NOTE***The following is an INCORRECT way to attempt to calculate indifference at c*****



For P1 to be indifferent between check-raise and check-call at point c:

If P1 check-raises, he wins:

2 with probability (c-g)… "check-back from P2"

(2+B) with probability g... "fold out P2 bluffs"

**(2+B) with probability (j-h)**… "fold out some P2 Bets"

And loses R with probability (1-j) "lose to top P2 range"



If P1 check-calls, he wins:

2 with probability (c-g)… "check-back from P2"

(2+B) with probability g... "Calls P2 bluffs"

And loses B with probability (1-h) "loses to all P2 value bets"



Set these two equal:

2(c-g) + (2+B)g + (2+B)(j-h) - R(1-j) = 2(c-g) + (2+B)g – B(1-h)

(2+B)(j-h) - R + Rj = Bh-B

2j+Bj-2h-Bh –Bh +Rj = R-B

Logical, right?.... *****No this is wrong****

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





***NOTE the following is the correct calculation of the correct equation as shown in the paper***



The problem with the above was the "(2+B) with probability (j-h) "fold out some P2 Bets"

This should correctly be said: "wins an 'extra' B with probability (j-h)" [NOT wins (2+B)(j-h)]

The reason is that the EV of previous decisions in the game tree that are not located at this decision point are NOT to be included. In the game tree diagram (Fig 4.1), we are evaluating the decision of Player 1 after P1 checks, then P2 bets. Further, we are only comparing raising versus calling. In the diagram, this decision will lead to either the +/-3, +3, or +/-9 outcomes. We are not considering the +/- outcomes in terms of the antes in the diagram. We are only considering the PLUS OR MINUS EV OF THESE THREE OUTCOMES.

So, the correct indifference equation at point c:

For P1 to be indifferent between check-raise and check-call at point c in the game tree diagram:

If P1 raises, he wins:

(B) with probability g

(B) with probability (j-h)

And loses R with probability (1-j)


If P1 calls he wins:

(B) with probability g

And loses (B) with probability (1-h)


Set these two equal:

Bg+B(j-h)-R(1-j) = Bg-B(1-h)

Bj-Bh-R+Rj = Bh-B

Bj-2Bh+Rj = R-B ….which is the correct third indifference equation



Substituting B=2 and R=6 gives the equation shown in the paper, and substitution of the answers checks also.

Continuing on with the indifference equations in section 4.2 ….



For Player 1 to be indifferent at between check-call and bet-fold at point d:

***NOTE*** it is a given that d<j and d<n. These </> relationships will be important for calculations of small ranges near the top of both players range. Also, now we are evaluating decisions that happen in different branches of the game tree diagram in fig 4.1, since one range is after a P1 check and the other is after a P1 bet. Unlike the previous equation, we must include all the EV of every previous action.



If Player 1 check-calls at d, he wins:

2 with probability (h-g)

2+B with probability g

2+B with probability (d-h)

And loses B with probability (1-d)



If Player 1 bet-folds at d:

He wins 2 with probability k

He wins 2+B with probability (d-m)

He loses B with probability (1-d)

And loses B with probability (m-k)



Set these equal...

2(h-g)+(2+B)g+(2+B)(d-h)-B(1-d) = 2k+(2+B)(d-m)-B(1-d)-B(m-k)

2h-2g+2g+Bg-2h-Bh = 2k-2m-Bm-Bm+Bk

Bg-Bh = 2k-2m-2Bm+Bk

Bg-Bh-2k+2m+2Bm-Bk = 0 which is the fourth indifference equation



Substitution of B=2 yields the equation in the paper, and substitution of the final answers checks also.

For Player 1 to be indifferent between bet-fold and bet-call at e:



***NOTE*** once again we are back on just one side of the tree, and at the last possible decision. This makes it somewhat easier.



If P1 bet-folds he loses B with probability (1-n+m-k)

If P1 bet-calls he wins (2+B+R) with probability (m-k)

And loses B+R with probability (1-n)



Set these equal:

-B(1-n+m-k) = (2+B+R)(m-k) - (B+R)(1-n)

-B+Bn-Bm+Bk = 2m-2k+Bm-Bk+Rm-Rk-B+Bn-R+Rn

R = 2m-2k+Bm-Bk+Rm-Rk+Bn+Rn-Bn+Bm-Bk

R = 2m+2Bm+Rm-2k-2Bk-Rk+Rn

R = (m-k)(2+2B+R)+Rn ….which is the indifference equation at point e.

Substitution B=2 and R=6 yields the equation in the paper and final answers check also.

Can't wait for the 6th equation!

For Player 1 to be indifferent between bet-call and check-raise at point f:



***NOTE*** Once again, dealing with separate sides of the game tree diagram (fig 4.1), so all of the EV has to be accounted for.



If P1 Bet-calls at f, he wins:

2 with probability k

2+B with probability (n-m)

2+B+R with probability (m-k)

2+B+R with probability (f-n)

And loses B+R with probability (1-f)



If P1 check-raises at f, he wins:

2 with probability (h-g)

2+B with probability g

2+B with probability (j-h)

2+B+R with probability (f-j)

And loses B+R with probability (1-f)



Set these equal:



2k+(2+B)(n-m)+(2+B+R)(m-k+f-n) - (B+R)(1-f) = 2(h-g)+(2+B)(g+j-h)+(2+B+R)(f-j) - (B+R)(1-f)



Cancellations....



2k+(2+B)(n-m)+(2+B+R)(m-k-n) = 2(h-g)+(2+B)(g+j-h)+(2+B+R)(-j)



2k+2n-2m+Bn-Bm+2m-2k-2n+Bm-Bk-Bn+Rm-Rk-Rn-Bk=2h-2g+2g+2j-2h+Bg+Bj-Bh-2j-Bj-Rj



Cancellations...

Rm-Rk-Rn-Bk-Bg+Bh+Rj = 0 ....which is the indifference equation at point f



Substitute R=6 and B=2

6m-6k-6n-2k-2g+2h+6j = 0

6m-8k-6n-2g+2h+6j = 0

3m-4k-3n-g+h+3j = 0 = g-h-3j+4k-3m+3n .....which is the equation in the paper.

Substitution of the final answers checks also:

0 = 20 – 110 – 3(130) + 4(75) -3(80) + 3(140) = 20-110-390+300-240+420 = 0

Picking up where we left off in the last post, we are investigating section 4 of the paper:

UNIFORM(0,1) TWO-PERSON POKER MODELS

Chris Ferguson, TiltWare, Los Angeles

Tom Ferguson, Mathematics, UCLA

C´ephas Gawargy, Mathematics, Universit´e Paris 1

which can be located at link:

https://www.math.ucla.edu/~tom/papers/poker2.pdf



So far, we have 6 indifference equations for Player 1, and when we finish with 6 more equations for Player 2 we can test our results to make sure we are correct.



For Player 2 to be indifferent between bet-fold and check at point g after P1 checks:



If P2 bet-folds, she wins:

2 with probability (b-a)

And loses B with probability (1-f+d-b)



If P2 checks, she wins:

2 with probability (g-a) and zero otherwise.



Set these equal:

2(b-a)-B(1-f+d-b) = 2(g-a)

2b-B+Bf-Bd+Bb = 2g

(2+B)b –2g +Bf -Bd = B ...which is the indifference equation at g

Substitution of B=2 gives the equation in the paper and final values check also.



For Player 2 to be indifferent between check and bet-fold at h after a check from P1:



If P2 checks she wins:

2 with probability (h-a) and zero otherwise



If P2 bet-folds she wins:

2 with probability (b-a)

2+B with probability (h-c)

And loses B with probability (1-f+d-h+c-b)



Set these equal:

2(h-a) = 2(b-a) + (2+B)(h-c) - B(1-f+d-h+c-b)

2h = 2b+2h-2c+Bh-Bc-B+Bf-Bd+Bh-Bc+Bb

B = 2b-2c+2Bh-2Bc+Bf-Bd+Bb …..which is the indifference equation at h

Substitution B=2 yields the equation in the paper and final values check also.

Hmmm... I suspect some of these equations may, in fact, *require* a method used earlier in the paper, in section 2, which seemed unnecessary to me back in section 2.

With regard to the last three equations in section 2, the authors are demonstrating how to account for allowing variables to move around and within the >/< relationships which are given.

example:

in section 2.2, equation 4...

The authors write:

"For Player II to be indifferent at d: If II folds a bet at d, she wins nothing. If II calls a bet at d, she wins 2+ B with probability a/(a+1 − c) and loses B with probability (1 − c)/(a + 1 − c). Equating the expectation to zero gives (2 + B)a + Bc = B. (This requires a<d<c.)"

by comparison, the simplified version of indifference that I arrived at:

For Player 2 to be indifferent between fold and call at d:

P2 wins 2+B with probability a and loses B with probability (1-c)
She wins zero by folding...

Set these equal:

(2+B)a-(1-c)B = 0
(2+B)a –B +Bc = 0
(2+B)a + Bc = B This requires a<d<c

While these two equations arrived at the same answer, since the added denominator cancelled out, the method in the paper is taking into account that Player 1 may be bluffing MORE OR LESS often that betting for value. We know by experience that pot size bets for value are twice as frequent as pot size bets for bluff, to make the caller indifferent. We also know that any optimal strategy will hold this to be true, and the final answer for bluff frequency is 1/12 and the value bet frequency is 1/6=2/12. However, I suspect that when it comes to the formal proof of a conjectured strategy such relationships can not be assumed.

So, the method in the paper is definitely more thorough, and it seems required when it comes to evaluating the optimality of the strategy in piecewise linear graphical form, and in calculating the final value of the game.

Now for OUR purpose of simply wanting a working set of indifference equations, this may not be necessary at all since we have accepted the conjectured strategy as optimal and are not proofing the value of the game either.

But, only one way to find out!

My naive solution thus far has been:

1/9 * call
call = 1 - alpha
alpha = s / (1 + s)
s = bet size

This takes folding frequency, which is correlated with bet size, into account. So for a 1/2 pot bet, you would check-raise ~ 7.4% of your range. Curious to see how close my answer is to the actual solution.

Well, I will put the last equations together tonight and post, as is.

I wager one potato that they don’t work on the first try.

For Player 2 to be indifferent between bet-fold and bet-call at j after a check from Player 1:



If Player 1 folds at j after a check-raise from Player 2:

She loses B



If Player 2 calls at j after a check-raise from Player 2:

She wins +R with probability (c-b)/(c-b+1-f)

She loses R with probability (1-f)/(1-f+c-b)



Set these equal:



-B = R(c-b)/(c-b+1-f) - R(1-f)/(1-f+c-b)

-B(1-f+c-b) = R(c-b) - R(1-f)

-B+Bf-Bc+Bb = Rc-Rb-R+Rf

R-B = Rc-Rb+Rf-Bf+Bc-Bb which is the indifference equation at j



Substitute R=6 and B=2

4 = 6c-6b+6f-2f+2c-2b

4 = 8c – 8b +4f

1 = 2c – 2b +f which is the equation in the paper and final values check also.









For player 2 to be indifferent between fold and raise at k after a bet from player 1:



If Player 2 folds, she wins zero....

If Player 2 raises, she wins:

2+B with probability (a+e-d)

And loses R+B with probability (f-e)



Set these equal:



0 = (2+B)(a+e-d) - (R+B)(f-e)

0 = 2a+2e-2d+Ba+Be-Bd-Rf+Re-Bf+Be

0 = 2a+Ba-2d-Bd+2e+2Be+Re-Rf-Bf ...which is the indifference equation at k



Set B=2 and R=6

0 = 4a-4d+12e-8f

0 = a-d+3e-2f …..which is the equation in the paper and final values check also







For Player 2 to be indifferent between raise and call at m after a bet from Player 1:



If Player 2 raises, she wins

2+B with probability (a+e-d)

And loses R+B with probability (f-e)



If Player 2 calls, she wins:

2+B with probability a

And loses B with probability (f-d)



Set these equal:

(2+B)(a+e-d)-(R+B)(f-e) = (2+B)a – B(f-d)

(2+B)(e-d) - (R+B)(f-e) = Bd-Bf

2e-2d+Be-Bd-Rf+Re-Bf+Be –Bd+Bf = 0

2e+2Be+Re-2d-2Bd-Rf = 0

0 = 2d+2Bd-2e-2Be-Re+Rf ...which is the indifference equation at m



Substitute B=2 and R=6

0 = 2d+4d-2e-4e-6e+6f

0 = 6d-12e+6f

0 = d-2e+f …..which is the equation in the paper and final values check also







For Player 2 to be indifferent between call and raise at n after a bet from player 1:



If Player 2 calls, she wins:

(2+B) with probability (a+n-d)

And loses B with probability (f-n)



If Player 2 raises, she wins:

(2+B) with probability (a+e-d)

(2+B+R) with probability (n-e)

And loses (B+R) with probability (f-n)



Set these equal:

(2+B)(a+n-d) - B(f-n) = (2+B)(a+e-d) +(2+B+R)(n-e) -(B+R)(f-n)

(2+B)(n) - Bf +Bn = (2+B)(e) + 2n-2e+Bn-Be+Rn-Re-Bf+Bn-Rf+Rn

2n +Bn –Bf +Bn = 2e +Be +2n-2e+Bn-Be+Rn-Re-Bf+Bn-Rf+Rn

0 = Rn-Re-Rf+Rn

0 = 2n-e-f = e +f –2n ….which is the indifference equation and is in the paper.

Final answer checks also.




So that gives us 12 indifference equations that, if correct, can produce optimal betting frequencies for one complete street of poker including check-raises.

It is too late to put these into a solver tonight. For now, here is the set as best as I can transcribe them. If even one "B" is mixed with a "2" then the set would be broken. We will have to try them out with different bet and raise sizes to see if they duplicate the table produced in the paper.

2k+Bk=B

Rj+2g+4j-2h+Bg+Bj-Bh=2+R

Bj-2Bh+Rj=R-B

Bg-Bh-2k+2m+2Bm-Bk=0

R=(m-k)(2+2B+R)+Rn

Rm-Rk-Rn-Bk-Bg+Bh+Rj=0

(2+B)b-2g+Bf-Bd=B

B=2b-2c+2Bh-2Bc+Bf-Bd+Bb

R-B=Rc-Rb+Rf-Bf+Bc-Bb

0=2a+Ba-2d-Bd+2e+2Be+Re-Rf-Bf

0=2d+Bd-2e-2Be-Re+Rf

0=e+f-2n