Found an input error in the matrix. Got the same solution now:

x1 = 8/245
x2 = 494/735
x3 = 24/35
x4 = 223/245
x5 = 227/245
x6 = 47/49
x7 = 4/35
x8 = 29/35
x9 = 31/35
x10 = 2/3
x11 = 24/35
x12 = 33/35

Im getting this book used because i am cheap. lol

The work in this thread is mostly from a paper published which was cited as a source for the book. The paper is free, and located here:

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

OK, so let's put these equations for use in Pot-Limit Omaha. The answers for a full-pot bet and a full-pot raise are already in the solution of the paper, as that was basically the point of the paper. However, sometimes expert players choose to bet 1/4 pot on the river and raise full-pot because of complicated strategy. Now we will model this as well.

P=4 B=1 R=6

This means Player1 places a bet of 1 into a pot of 4 and Player2 puts in a raise of (1+6) and creates a pot of 12 and offers the Player1 a call of 6 more or a fold.

OR..

Player1 checks, Player2 bets 1 into 4, Player1 checkraises (1+6) and then Player2 can call 6 or fold.


This sequence of action creates the following ranges:

Player1:

0-bluff-(4.039)---checkfold---(22.525)raisefold(25.128)------------checkcall-----------(74.596)--betfold--(84.694)--betcall--(94.793)--checkraise--1


Player2 when Player1 checks:

0-----bluff-----(8.205)-----------------------check----------------------(58.974)-----------betfold---------(79.487)----------------betcall-----------------1


Player2 when Player1 bets:

0-----------fold-----------(20.000)-----bluff-----(25.128)----------------------------------call--------------------------(89.743)-----------raise------------1





Some observations:

These ranges assume equality, neither player has 'range advantage'.

The initial bet of 1/4 pot immediately folds out the exact same portion of range here as it did in the 1/4 pot bet and 1/2 pot raise calculations. An application of this is that whenever a player places a bet, an exact portion of the opponent's range is immediately folded. If that opponent chooses instead to bluff with those hands, he must do so with the very worst, and do so at a frequency that balances with his value-raises.

When facing a small 1/4 size bet, Player1 checkraises slightly more often (94.793 vs 96.000) while P2 raises the 1/4 bet from P1 much more often (89.743 vs 93.333) than if the initial bet was full pot.

So, we see that the 1/4 size bet leads to a cheap showdown or a large raise more often than a full pot size initial bet.

This is known as 'polarized'.

It's amazing to me that nobody had worked out the asymmetric bet ranges since the Ferguson paper, or found the mistake in it!

Only a typo (+/-), I should not have called it a mistake.

I am sure that some of these equations are in the ‘guts’ of solvers, so they were solved by others after Ferguson published the paper.

Does the value of B have to be same in all the equations? Like can you use a different bet size for the probe bet equations?

How would you modify these equations to account for rake?

If your calculation involves the pot some where just subtract rake from it (i.e some constant like P-.05P) or use a percentage of it instead) (i.e .95P).

Does the same logic apply to the equations from MOP?

alpha = s/(1 + s)
cbet = (1 + s)/(1 + 2s)

If I'm understanding the equations correctly, the 1s refer to the pot size. So would you simply subtract the rake %?

alpha = s/(1 - rake + s)
cbet = (1 - rake + s)/(1 - rake + 2s)

If you're going to add a rake term you want to avoid adding an additional unknown variable since it will make solving harder.

Since we can express rake in terms of pot, we can avoid this issue.

If the convention in the equation is that P = 1 and it stands for the amount you can win just subtract some 1/x where x varies between some number between 0 and 1, which represents no rake and 100% rake.

So everywhere there is a 1, put in (1-1/x) instead. 1/20 would represent 5% rake for example.