The easiest way to check these equations in a manner readily verifiable by a reader of the thread is to use a Gauss-Jordan elimination calculator such as this one:

https://www.emathhelp.net/calculator...on-calculator/

This readily accepts matrices of up to 9 equations and 9 unknowns.

Using this tool gives the following results:

The indifference equations in section 2 when B=2 give the values solved in section 2.

Recall we did find an error in the indifference equations in section 3.
Fortunately, our correction of the +/- typo in section 3.2 of the paper is proven to be correct!

The first equation should have been published as:

−2a + (2 + B)d = B

Substitution of B=2 and R=6 yields the values given in table.

Those two sections were sets of six equations in six unknowns.

However, section 4 has a set of 12 equations in 12 unknowns, so we must reduce this to 9 equations if we want to use the readily available matrix solver. Substitution of values for B and R immediately gives a solution to the first equation, so the value for k is first substituted into the other equations. Examining the remaining equations and selecting the variables g and h to be substituted into the remaining equations requires the least effort. That leaves just 9 equations.

We can test the indifference equations in section 4.2 rather quickly this way. Using the above method yields these equations:

a-d+3e-2f=0

-2b+2c+f=1

2b-d+f+j=2

2b-3c-d+f+4j=3

d-2e+f=0

e+f-2n=0

3j-3m=1

6j+3m-3n=4

2m+n=2

...and entry of these equations into the matrix gives the answers for pot-limit found in section 4.2.

Unfortunately, using this method to test our full set of indifference equations that we have arrived at shows that we are incorrect, and it does not require the matrix calculator to show this.

My equation for indifference at point c:

Bj-2Bh+Rj = R-B

To test the equation for limit poker where pot=B=R=2

would give j=h. This is just incorrect.

So, I am pretty sure where I went astray, and remain determined to get the full set of working indifference equations. Just not there yet.

Curious as to where you think you erred for my own edification

The check-raise equations. This was the first incorrect one, I think. I have double checked and the method is not nearly as hard as I was making it to be.

If I get this one equation right tonight, then its easy peasy and double checking the entirety of the set will produce a working set. I reallllllly hope I get these done tonight!

Just checking in with your progress. Looking forward to putting these solutions to work

Well, it just so happened I was typing this up just now!

Finally, the complete set of twelve Ferguson indifference equations, non-simplified and usable for any size bets, raises, and check-raises.

These are verified and double-checked by the solutions given for limit poker at the end of the paper, as well as via Gauss-Jordan elimination method for solving the corresponding 12 equations in matrix form.

First, a few notes on potential pit-falls when working with equations of this type, then the equations themselves, and finally the values for half-pot bets and raises.

1) Absolutely use the variable P instead of the numerical value for the size of the pot when deriving the equations. If you don't you will mix up some pots with a number ratio that just happens to equal pot.

2) The wording used in section 2 for player 2 uses the complete expression for probability, in the form (P+B) with probability (x-y)/(1-y) etc. This is done only to make the sentence grammatically correct according to game theory, since every result player 2 achieves is in reaction to the dual purpose bets/raises of player 1. So, technically, every outcome of player 2 is a ratio of the differing purposes of player 1. None of this is required to solve the indifference equations, provided the relationships of the variables stay in the order given (which is required by GTO).

3) Absolutely remember that every raise also has a corresponding "call" simultaneous to the raise. If P1 bets 2, and P2 Raises 6, Player 2 is actually putting B+R = 2+6 into the pot.

example:

Player 1 bets, player 2 raises, player 1 calls the raise...
If Player 1 wins, he wins a net increase of P+B+R which is the pot plus the money Player 2 added to the pot (B+R).
However, if Player 1 loses, then he has a net decrease of -B-R which is the money he put into the pot and lost (The pot itself belongs to neither player and is never a loss)

4)When error checking with the limit-poker values at the end of the paper, the pot size is always 2 for all values computed in the table. The Bet/Raise sizes are what changes. Leftmost column is P=2,B=2,R=2. Next left-most column is P=2, B=.6666, R=.6666, next column is P=2, B=.3333, R=.3333.

5)As with (4), be careful when choosing a number for R, to take into account the correct method of computing a raise. A raise is a call, plus a raise. For half-pot bets and raises, both the bets and the raises should lay 3:1 odds for a player to call.

Example:

If Pot =2, then a half-pot bet of B=1 will give 3:1 odds on a call.
If Pot=2 and B=1, then a raise of R=2 gives the correct 3:1 odds to a call.

For the expression Pot+Bet+Call+Raise to give 3:1 odds, then it is required that P=2, B=1 (the call=B), and R=2

To call the raise at the end of this action is to risk 2 to win 6, 3:1 odds.


The equations below are the complete indifference equations from 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

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



Section 4. Two Rounds of Betting With Sandbagging.

One complete street of (0,1) poker with bets, raises, and check-raises, the indifference equations, for any pot size, any bet size, and any raise size (raise must be equal or greater than bet):



Player 1 indifferent at a:

k = B/(P+B)



Player 1 indifferent at b:

(P+B)g - (P+B)h + (P+2B+R)j = B+R



Player 1 indifferent at c:

-(P+2B)h + (P+2B+R)j = R



Player 1 indifferent at d:

Bg – Bh - (P+B)k + (P+2B)m = 0



Player 1 indifferent at e:

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



Player 1 indifferent at f:

Bg – Bh – Rj + (B+R)k – Rm +Rn = 0



Player 2 indifferent at g:

(P+B)b – Bd + Bf – Pg = B



Player 2 indifferent at h:

(P+B)b - (P+2B)c – Bd + Bf +2Bh = B



Player 2 indifferent at j:

-(P+2B+R)b + (P+2B+R)c +Rf = R



Player 2 indifferent at k:

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



Player 2 indifferent at m:

(P+2B)d - (P+2B+R)e + Rf = 0



Player 2 indifferent at n:

e + f – 2n = 0



Solving the above 12 equations in 12 unknowns for half-pot limit, P=2, B=1, and R=2 yields these results:



a = 3/48 = .0625

b = 17/48 = .354

c = 18/48 = .375

d = 36/48 = .750

e = 39/48 = .8125

f = 45/48 = .9375

g = 6/48 = .125

h = 30/48 = .625

j = 36/48 = .750

k = 16/48 = .3333

m = 18/48 = .375

n = 42/48 = .875

POTY. This has been the final elusive piece of the puzzle for me. Going to work on the practical implementation side of things

Before I reinvent the wheel, has anyone worked out the equations for other bet sizes? Particularly looking for 1/4x and 2x pot. Thanks!

What are the correct set of 9 equations? And how are you reducing from 12 to 9?

the reduction from 12 equations to 9, in order to use an online matrix solver, is done by algebraic substitution.

Before you take that step, you have to decide what bet size you want to solve for.

Pick a bet size, and I will reduce from 12 to 9 equations, and show the steps. Then you can try to solve the rest, I will only tell you if you are correct or not.

I was first trying to replicate your process for 1/2 pot. I can solve for k, but then I'm left with this:

(1/3) = (1)/((2)+(1))
((2)+(1))g - ((2)+(1))h + ((2)+2(1)+(2))j = (1)+(2)
-((2)+2(1))h + ((2)+2(1)+(2))j = (2)
(1)g – (1)h - ((2)+(1))(1/3) + ((2)+2(1))m = 0
((2)+2(1)+(2))m - ((2)+2(1)+(2))(1/3) + (2)n = (2)
(1)g – (1)h – (2)j + ((1)+(2))(1/3) – (2)m +(2)n = 0
((2)+(1))b – (1)d + (1)f – (2)g = (1)
((2)+(1))b - ((2)+2(1))c – (1)d + (1)f +2(1)h = (1)
-((2)+2(1)+(2))b + ((2)+2(1)+(2))c +(2)f = (2)
((2)+(1))a - ((2)+(1))d + ((2)+2(1)+(2))e - ((1)+(2))f = 0
((2)+2(1))d - ((2)+2(1)+(2))e + (2)f = 0
e + f – 2n = 0

Then once I checked my answers against yours, I wanted to solve for 1/4 pot next. Thanks again

Firstly, I found an online matrix solver that will accept up to 100 indifference equations in 100 variables! So, algebraic reduction is unnecessary (but good practice). The hard part, as evidenced by this thread, is the derivation of the indifference equations by hand with pen and paper.



Secondly, before I spoon feed you the method, I must caution you. I don’t know why you are seeking these solutions, only that it was very interesting to solve. If you are wanting to model real poker between humans, it is unlikely that a human would ever respond to a ¼ size bet with a ¼ size raise. 2X pot size is more likely to simulate two players “getting it in” when deep stacked. Remember, this is just a model. Actual poker ranges are asymmetric and have point masses with blockers. This could, in theory, be why a player would choose a bet size that maximizes the utility of a certain portion of range.



Anyway, here is the method that I used for half-pot bets and raises:



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

Here is again the 12 Ferguson indifference equations, neatly organized and ready for any size bet and raise (although I have not tried asymmetrical sizes those should work also).



k = B/(P+B)

(P+B)g - (P+B)h + (P+2B+R)j = B+R


-(P+2B)h + (P+2B+R)j = R


Bg – Bh - (P+B)k + (P+2B)m = 0


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


Bg – Bh – Rj + (B+R)k – Rm +Rn = 0


(P+B)b – Bd + Bf – Pg = B


(P+B)b - (P+2B)c – Bd + Bf +2Bh = B



-(P+2B+R)b + (P+2B+R)c +Rf = R


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


(P+2B)d - (P+2B+R)e + Rf = 0



e + f – 2n = 0



The first step, is to pick a bet/raise size. Remember, a raise is a call and a raise, so the value for raise R must offer the same odds at the end of the action as the initial bet B. This makes the game a form of limit, either half pot, 2x pot, etc. The initial pot size is always 2.



For half pot bets and raises, the action is P1 bets 1 into a pot of 2, and P2 then calls the 1 and raises another 2. This creates a pot of 6 and P1 then can call the 2 or fold.



So, for half-pot limit, B=1 R=2 and P=2



Now, take these values and put them into all 12 equations, but don’t simplify. Keep every variable with a coefficient and keep them in alphabetical order, left to right. Iirc, this is how I have them ordered for the purpose of creating a matrix easily.



k = B/(P+B)

k = 1/3



(P+B)g - (P+B)h + (P+2B+R)j = B+R

3g – 3h + 6j = 3



-(P+2B)h + (P+2B+R)j = R
-4h + 6j = 2



Bg – Bh - (P+B)k + (P+2B)m = 0

g – h – 3k + 4m = 0



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

6m – 6k + 2n = 2



Bg – Bh – Rj + (B+R)k – Rm +Rn = 0

g – h – 2j + 3k – 2m + 2n = 0



(P+B)b – Bd + Bf – Pg = B

3b – d + f – 2g = 1



(P+B)b - (P+2B)c – Bd + Bf +2Bh = B

3b – 4c – d + f +2h = 1



-(P+2B+R)b + (P+2B+R)c +Rf = R

-6b + 6c + 2f = 2



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

3a – 3d + 6e – 3f = 0



(P+2B)d - (P+2B+R)e + Rf = 0

4d – 6e + 2f = 0



e + f – 2n = 0 <-------- This last one is a balance ratio equation, independent of bet size. Every set of indifference equations that I have seen has at least one of these.



Now, if we have a very large chalkboard or a 12X13 matrix solver, these equations will plug right in immediately. There are 13 columns across, which stand for each variable a to n (no i or l) plus the whole number to the right of the equals sign makes the last column. Equations with no value for a certain variable get a zero in that cell of the matrix.



This one here

https://matrix.reshish.com/gauss-jordanElimination.php

Accepts up to 100 linear equations, and it will even show you the steps, in case you have that giant chalkboard and want to practice. Plus, it accepts fractions as coefficients which will give the actual fractions as answers. Neat!



First, go through your equations and draw the matrix like this:







Go to the link above, and select 12X13 matrix:







Then, input them into the solver like this:







Click “fill empty cells with zero” and click “very detailed solution” because awesome:






click solve and boom:



these are the values from a to n, top to bottom.

Now, it is your turn. Do this for 2X pot. This is your homework.

Meanwhile, I will try some asymmetric bets.

ALSO, BIG THANK YOU! It would seem that dusting off this in my brain has detected a solution waiting in my subconscious to a much more complicated problem that eluded me a year ago.

There's no need to thank me as you did all the work. I've been building a simple human model to mimic AI solutions, so I just knew the right questions to ask. I'm off to do my homework, and looking forward to the asymmetric bets which was the next logical step.

You appear to swapped the values for Player 2 indifferent at g & h:

3b – d + f – 2g = 1
3b – 4c – d + f +2h = 1



I get this solution set:

x1 = 11/96
x2 = 35/288
x3 = 7/48
x4 = 7/12
x5 = 67/96
x6 = 89/96
x7 = -7/48
x8 = 7/16
x9 = 5/8
x10 = 1/3
x11 = 19/48
x12 = 13/16

Nah, I double checked everything before I poasted (it was correct the first try).

Try again amigo.

You list the equations as
7: 3b – d + f – 2g = 1
8: 3b – 4c – d + f +2h = 1

Yet input the values
7: 3, -4, -1, 1, 2, 1
8: 3, -1, 1, -2, 1

So either you swapped the values or I'm missing something.

The order in which the equations are input in the matrix does not matter, so long as they are correct. When drawing the matrix, I skipped 7, and drew 8 first, then 7, then 9-12.

Ahh I screwed up the second equation with a 2 at the end instead of 3. Solving for 1/4 pot next as it has the most immediate application for me.

So I jumped straight ahead to asymmetric bets, and here's my solution for 1/4 pot bet vs 1/2 pot raise (P=4, B=1, R=3):

k = 1/5
5g - 5h + 9j = 4
-6h + 9j = 3
g - h - 5k + 6m = 0
9m - 9k + 3n = 3
g - h - 3j + 4k - 3m + 3n = 0
5b - d + f - 4g = 1
5b - 6c - d + f + 2h = 1
-9b + 9c + 3f = 3
5a - 5d + 9e - 4f = 0
6d - 9e + 3f = 0
e + f - 2n = 0

x1 = 54/1225
x2 = 11/49
x3 = 26/105
x4 = 174/245
x5 = 192/245
x6 = 228/245
x7 = 3/35
x8 = 4/7
x9 = 5/7
x10 = 1/5
x11 = 26/105
x12 = 6/7

A verify would be appreciated

Will check this evening. Very exciting!

Confirmed. Well done!

Also, I am pleased that the ordering of the variables set forth in the paper holds true, even in asymmetry. I was slightly concerned that some of the ranges would disappear entirely if this was broken somehow.

Here is a rough look at what these ranges look like.





We can see, as expected, there is not a lot of folding going on. P1 check-calls a huge chunk of middle range, and P2 calls an even larger chunk of middle range.

Not a lot of bluffing happening either. Mostly the first player to place a bet is seeking either a cheap showdown, or to entice a large raise which will be called. Or, the first player to bet is trying to squeeze some value out of that huge call range of the other player.

I don't recall ever seeing such ranges published. Bravo!

The "great game" of Pot-Limit Omaha features some rivers where a 1/4 pot bet is placed, and gets either a call or a full-pot raise. This has some of the motivations mentioned above.

Care to do those next?

Tried 2x pot bet vs 1/2 pot raise (P=2,B=4,R=5), but some of the frequencies don't appear to make sense:

k = 2/3
6g - 6h + 15j = 9
-10h + 15j = 5
4g - 4h - 6k + 10m = 0
15m - 15k + 5n = 5
4g - 4h - 5j + 9k – 5m +5n = 0
6b - 4d + 4f - 2g = 4
6b - 10c - 4d + 4f + 8h = 4
-15b + 15c +5f = 5
8a - 8d + 15e - 9f = 0
10d - 15e + 5f = 0
e + f – 2n = 0

x1 = 487/980
x2 = 134/735
x3 = 4/35
x4 = 103/245
x5 = 167/245
x6 = 59/49
x7 = 4/35
x8 = 29/35
x9 = 31/35
x10 = 2/3
x11 = 24/35
x12 = 33/35

Thoughts?

This is incorrect.

Fixed but some values are still problematic:

k = 2/3
6g - 6h + 15j = 9
-10h + 15j = 5
4g - 4h - 6k + 10m = 0
15m - 15k + 5n = 5
4g - 4h - 5j + 9k – 5m +5n = 0
6b - 4d + 4f - 2g = 4
6b - 10c - 4d + 4f + 8h = 4
-15b + 15c +5f = 5
6a - 6d + 15e - 9f = 0
10d - 15e + 5f = 0
e + f – 2n = 0

x1 = 128/245
x2 = 134/735
x3 = 4/35
x4 = 103/245
x5 = 167/245
x6 = 59/49
x7 = 4/35
x8 = 29/35
x9 = 31/35
x10 = 2/3
x11 = 24/35
x12 = 33/35

here is what I get:



The problem might be that I originally wrote equation 5 with m before k which is not convenient when inputting into the matrix from left to right. When paying attention to the column variables, all is fine, but when just looking at the coefficients and plugging in, transposition is easy. Switch equation 5 to:

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

for easier plugging into the matrix.

Try again and see if you get usable values.