I'm not sure exactly what "as balanced" means... but I am saying that in that example, the GTO's betting range is going to be different than his calling range. Likewise with villain's "expected" ranges for those two lines. So the strategies for future betting rounds are going to be different.

To me it's intuitive, not counter-intuitive. Perfect strategy depends on the two ranges. Different actions (betting lines) imply different ranges even if they lead to the same pot/stack sizes.

Sorry, it's hard to visualize in a GTO v GTO scenario. I understand what you're saying.

Im just having trouble, because a Check should be protected in GTO.

Keep riffing. I'm learning a lot and appreciating very much. =D

Sorry for the double-post, i can't edit anymore.

I have a really stupid question...

Does Chubukov require that the hands referenced are shoved, or is that a proof that it is unexploitable to be AI at the designated stack depths in a HU scenario with the given holdings?

I'm not sure if, for example, it is a proof that raise, 4bet calling TT HU for 100bb (more, I just don't have the chart in front of me) is unexploitable, or if it's just a proof that open-shoving TT HU for 100bb is unexploitable.

I have no clue. 0.o

It's just a proof open-shoving is unexploitable. An open-shove is necessarily up against a range of ATC, whereas a 4bet-shove or a 4bet-call is not.

IMO you should just try to understand Nash Equilibrium in a mathematical sense. It will answer all of your questions and it's not that complicated.

BTW 3+ player games usually don't have Nash Equilibria, so it doesn't make sense to talk about GTO outside HU situations.

GTO is just a fancy word for one of the strategies of the Nash Equilibrium.

This is very helpful! - "Just try and understand! Then you will understand!"

Perhaps you'll enlighten us all about the "uncomplicated" relationship between an abstract mathematical proof about games in general, and the conditions under which Q5o forms part of a GTO UTG open-shoving range.

"Our survery says these things are unusual. It is therefore meaningless to talk about them".

Ok. This isn't going to be as thorough of a response as I hoped for. I'm just going to spew out some stuff before I have to go back to studying/working and grading 80 papers over the weekend.

A pure strategy is a special case (degenerate) of a mixed strategy. You can tell the difference based on the probabilities assigned to the actions. It's not really a slight of hand, its just that you are moving to a more general thing which includes the specific case of a pure strategy.

I don't think so, but yes typically when you say mixed strategy you are referring to non degenerate cases but not excluding them.

The proof I provided is for the existense of a mixed strategy NE. To me it seems pretty clear there won't be a pure strategy NE to poker (you can probably show it if you had to, but its not worth the time). The proof posted also doesn't say anything about the number of equilibrium, it just guarantees that there is at least one.


All strategies are decided in advance. A strategy is a complete plan of what to do in any and all situations. If you gave someone else your strategy and went and did someething instead of playing they game they could play it identical to you based on your instructions. They are deciding between the actions available at a given node, but the probabilities assigned to each node have weights that were assigned in advance.


How good are you at math? What are the highest courses/subjects you know?

It's easier to prove it's not pure by contradiction. I think it could be done, but that's just my initial impression.

You see what your opponent does. If he bets an amount x you are at set of nodes contained within the same information set where all you know is that the opponent bet x. If he bets y=/=x then you are at a different information set.

They are different strategies? I mean it seems like you want to be somewhere in the middle of the game tree rather than at the beginning, then the answer is it depends.

No, the opponent should be indifferent between his choice of strategies against our strategy (note this is before nature decides who goes first, deals the cards, determines the order of following cards on the flop, turn, river, etc). The strategy we play, since it isn't optimal, doesn't adjust to information revealed about the type of strategy our opponent is playing.


I see no reason to believe this. Your opponent doesn't know what you hold, that's why you can do this. He's at an information set, but doesn't know which node within the set he is at. He has no idea if you are 3 betting to X with great cards or with marginal holdings.

Uhh, can you prove that there is a NE here? Infinite stacks=infinite actions. Not to mention potential problems of infinite pay offs, and having to do dumb stuff that won't work out nicely with math. Pretty bad example, and wrong stuff in this post.


I concur. Once you understand what NEs are then it should be clearer the role each of the things you ask about should be to you. What do you mean about 3+player games not have NE? They can if they meet certain conditions which a whole lot of games do...

GTO is a misnomer for one of the strategies in a symmetric NE(not just any NE) from my take on it.

Edit:
Go bulls. Ronin, do you play at hammond? I'm from/my family still lives in NW IL.

A Nash equilibrium is a set of strategies such that no player can do better by unilaterally changing strategies. So this concept does exist for multi-player games, but choosing a NE strategy only guarantees you'll be unexploitable in a two player context. With 3+ players you could be playing a NE strategy and still get exploited if multiple opponents deviate from NE, but not if only one of them does (hence the "unilaterally" caveat).

I was answering the specific question in OP. If you actually understand what Nash Equilibrium is it's clear that it will depend only the starting SPR and the players' relative position. Of course it doesn't tell what the specific strategy is but that wasn't the question.

"Our survery says these things are unusual. It is therefore meaningless to talk about them".[/QUOTE]

Alright, let me repraphrase. In 3 or more player poker games with reasonable stack sizes there is no strategy that guarantees a 0 payout. There's always a chance that the other players are colluding against you. Possibly unintentionally. I don't have a mathematical proof for this and it's possible this isn't true for very short stacks (like 5BBs or less). But if you've played any poker at all, I'm sure you agree it's true.

I suspect there aren't Nash Equilibria either, but even if there is it's not a very useful concept. If a player realized he was stuck in a NE that gave him a negative payout he would either leave the game or played a strategy that lowered his payout in the short term hoping that it would lead to others changing their strategy.

I'm pretty sure there are even 3 player zero-sum games that are symmetric and have multiple NE, each with non-zero payouts.

I agree that it's implausible that a pure strategy NE for poker exists.

As I understand it though, at present it simply hasn't been proven that a (non-degenerate) mixed stratgegy NE exists for poker. This isn't to say it does not exist, but just that there is no *proof*.

I think it's important to recognise this if only because there is a widely held assumption that the proof now exists - and it doesn't (as far as I can see).

It's been proven that 2 player fixed-sum games have mixed strategy NE. Since HU poker is a 2 player fixed-sum game (ignoring rake) it has a NE.

Sure, but what I was responding to...
"3+ player games usually don't have Nash Equilibria"
They definitely do have NE. Possibly multiple NE as you later stated.

I think there are two ways of looking at this.

(1) A "GTO-bot" playing multiple opponents who were all colluding (in some optimal fashion) would have a negative winrate over any random sample of hands. A negative win-rate in the blinds is important for this to be the case.

(2) A "GTO-bot" can guarantee a payout of 0 for itself by folding. Folding is 0EV, even from the blinds. A game theoretic definition of NLHE(Cash Game) would be that it is *not* a zero-sum game.

Yeah, I guess that's wrong. I guess I was thinking along the lines of 3+ player poker games have no unexploitable strategies. I'm still 99.99% sure that's true (for realistic stack sizes).

Could probably construct trivial examples where there are unexploitable strategies but I think that's what you're trying to eliminate with the "realistic stack sizes" clause. But generally collusion (implicit or explicit) is going to be a problem with multiplayer equilibrium strategies, so to that extent I agree.

Wasn't this covered by Sklansky in one of his books? In this game you fold all your hands except AA where you shove.

I wasn't disputing this.

The point was that it's not been proven that 2 player fixed-sum games have non-degenerate mixed strategy NE.

What does non-degenerate mean here? Not pure strategy?

Yes - a mixed strategy in which no pure strategy is assigned a probability of 1.

Also in retrospect it was incorrect to include the qualification "2 player fixed-sum games". I should just have said "games", but just copy and paisted from your post.

Ok, yes, infinite stacks was a bad example. But that got me thinking about what happens to the NE's as the stack size tends to infinity: Will there be some kind of convergence of the NE's? Doubtful, since as stack size grows we keep adding new possible actions.

My intention was to counter the notion that GTO actions are based only on (stack size, pot, position, board, hole cards) but not on the past betting for the current hand.

So my second example was better: For finite stack sizes heads up if we consider these two betting lines:

a) Villain checks | GTO bets pot | Villain calls
b) Villain bets pot | GTO calls

The stacks, pot, position, etc. are the same for (a) and (b). Yet it seems likely and intuitive, at least to me, that the GTO strategy for future betting after (a) will not be the same as after (b) even for the same exact hand, because:

After (a) Villain supposedly has calling range, GTO has a betting range
After (b) Villain supposedly has a betting range, GTO has a calling range

Since each player has a different range of hands in (a) vs (b) one would not expect the GTO strategy to be the same following (a) vs following (b).

The bolded is wrong. In a 3+ player game you could get less than your equilibrium EV by playing the NE, even if only one player deviates, and even if the game is zero sum. All the NE guarantees is that if only one player deviates it won't benefit them, but their deviation might hurt both you and them while benefiting a third player. This can happen for example in a 3 player poker tournament where the top two positions pay. From the small blind you go all-in, the big blind calls wider than he should, hurting you both while benefiting the third player.

Well, consider this: You are facing a bet on the river. Your calling frequency is determined by size of the bet in relation to the size of the pot. Unless the number of hand combinations line up perfectly there's going to be a hand that you're calling sometimes and folding sometimes.

Given that there are millions of different boards and thousands to millions of ways to reach the river on each one it's practically impossible that it always just works out.

This is convincing.

Clearly, the existence of a pure strategy NE for poker is unlikely. I just thought that it was important to note that the fixed point theorems' proofs of the existence of mixed NE in general might not in themsleves be sufficient to prove the existence of a non-degenerate variety for poker. Something else would be required - such as an argument along the lines that you've just offered here.