Yes, i think its clear that the person with highest equity will go all in. Some hands might have odds to call. Those hands could also preemptively go all in first if they are earlier to act, without gaining or losing EV.

I think I should re-emphasize the distinction between heuristic correctness, and computational correctness.

I agree, as a matter of intuition, that the person with the most equity goes all-in. But this isn't an intuitive effort that I'm asking about, it's a computational one. The only heuristic shortcuts I'm willing to take are shortcuts that are, as a mathematical matter, provably true.

What I was hoping to start was a discussion about known algorithms for solving these situations, known optimizations to those algorithms, and if I got really lucky, a pointer to a code library that already has what I need implemented.

I think I have found a theoretical situation where it is optimal to make a non all-in bet. Here's the setup:

3 players, call them A, B, and C. The turn has just been dealt. The pot is 100 chips and the stacks are each 100 chips. 3 handed hot and cold, A has a 46% of winning, B has a 24% chance of winning, and C has a 30% chance of winning. If it's heads up between A and C, then A has a 46% chance of winning and C has a 54% chance of winning hot and cold (so C gets all of B's equity if B folds). The other heads up matchups don't matter.

If A bets all-in, then B will fold and C will call. The EV to A will be 38.

If A checks, then what will B, and then C do? If B moves all-in, C calls and A calls, for an EV to B of -4. If B checks, C will bet all-in, and A will call and B will then have to fold, for an EV to B of 0. If B makes an intermediately sized bet, then he will get moved in on by C or A, depending. If he is not pot committed, then C will move in and he'll have to fold, and the bet will have been -EV. If he is pot committed, then C will call and A will move in, and he'll have to call, and the bet will have an EV of -4 (same as if he moved in). Therefore, B must check following an A check. After check-check, C moves all in, A calls, and B folds - for an EV to A of 38. Therefore the EV of checking to A is 38.

A would ideally like to get some action but keep B in the hand. He can do this with an intermediately sized bet, say, of 40 chips (just as an example - the optimal bet size is larger but for simplicity let's use 40). If B calls, then B will be pot committed, and C will just call - he won't move in because he has less than the 33% pot equity required to make a raise that will be called in 2 places profitable. Therefore B will only have to call the 40, which he can do profitably. C will call. Therefore if A bets 40, B calls and C calls, and A's EV is 61.2 .

This is all complicated enough that there could be an error somewhere, but everything looks OK to me right now. Even though it's not a concrete holdem example, it has me confident that the "all-in is always the optimal bet size" conjecture is false.

I disagree. B must have made an error by calling. "A would ideally like to get some action but keep B in the hand." yes because he gets more of his chips that way. Therefore B, know what is gonna happen should be folding.

Since it is a 3 player hand, A's increase in EV doesn't imply a decrease in EV for B. In my situation, both A and B's EVs increase when A makes a 40 chip bet versus going all-in or folding. The EV comes at the expense of C.

aahhhh i see.... very smart.

but i still dont see something. If A & B are each gaining EV by making a small pot vs C then they gain even more by making a big pot from C. So they want to be all in.

I don't know if you are also interested in this in the case if of incomplete information. If so, here is a nice review paper of the state of the start in poker AI:
http://www.cs.auckland.ac.nz/~jrub00...reprintAIJ.pdf. IIRC it also has some examples of how to calculate EV, which would be applicable in this case.

A betting small instead of checking does not increase the betting - it reduces the betting by disincentivising C from going all-in when it's his turn. B benefits from reduced betting, and as such he benefits from A betting small over checking, while he would not benefit from all-in instead of small betting.

I just realized that carlop posted a counterexample to the "someone should always push" conjecture back in post 20 (somehow missed it when first reading the thread, sorry carlop). Anyway, that already serves almost the same purpose as my example. I was aiming at the slightly different "all-in is always the optimal bet size" conjecture, but disproving either one establishes that you probably need to do some kind of brute force search (and that the non brute force search ideas that have been suggested in this thread are not correct). His example is also better than mine because it's simpler and is a real holdem example (rather than a theoretically constructed situation).

Ok I looked at the details more carefully. Hopefully I'm not missing something and saying something stupid here.

Assuming high hand wins, the following table seems compatible with your description.

prob (a,b,c)
46% (1,0,0)
24% (0,2,1)
30% (0,1,2)

This table describes 3 possible outcomes, assuming nobody folds. For example the last entry means their is a 30% chance that A will have the worst hand, B the middle hand, C the winning hand.

You said "If A bets all-in, then B will fold and C will call. But, if A bets all-in, then B will call with 54% equity knowing that C will fold with 30% equity.

Also, if A raises to 40 then B will go all in with 54% equity, again knowing C will fold and A will call.

I think the optimal outcome is:
- if B bets first he will go all-in. A will call. C will fold.
- if C bets first he will go all-in A will call. B will fold.
- if A bets first, it doesn't matter what he does because he will end up all in vs either B or C, whoever acts next.

I don't follow your table exactly, but if A bets all-in and B calls, then C is getting 3 to 1 on his money and he calls (he needs only 25% pot equity to make the call, and he has 30%). Same deal if A bets 40 (after which he would definitely call an all-in) and B goes all-in.

Edit: Figured out how to read your table, it's fine. My other comments still stand.

Yes, my mistake, was thinking 2 to 1 and 33%. I will revisit when I get some time.

Yes, interesting, I think you are right. 3-way A is the only one who wants to get more money in, but 2-way he does not. So he wants to bet as much as he can without causing anyone to fold.