Hello, 2+2

I had a debate with a friend of mine, and what he said blew my mind.

I cannot quote exactly what he said because we talked in a different language, but here is a example.

If I use solver to solver some game trees, using BN's and BB's GTO preflop range, like this image(maybe its not perfect GTO preflop range, what you get the idea)



If I memorize all the trees/decisions from flop to river on all the board, in real life, when I open on the button with GTO preflop range, and no matter what range BB uses, I can do the exact same thing as what I memorized, and in the long term, I will still win money. (of course I need to count the money I earn from preflop steal, 3bet pot, 4bet pot, etc.)

The reason why I thought this is ridiculous is that, I thought to be at Nash equilibrium, you need to adjust to villain's range, your won range and change the decisions, you cant just solve own tree and call it the perfect solution.

I dont even know anymore because what he said makes sense. the NE is for the whole game not only post flop, which means If I do the exact same thing, no matter what range villains uses, he should lose because of the deviation from NE

If you can play a perfect GTO strategy all the way from preflop to river, then yes you will make money passively exploiting their mistakes. This is true no matter what range they use.

Let's say BB calls too tight. The GTO strategy will overextend a bit against this tight range, but it will more than make up for it by stealing blinds.

Let's say BB calls too wide. The GTO strategy won't steal as many blinds, but it will make up for it in postflop EV.

What your friend said is true

I think any 1v1 game has an optimal strategy that you can use and write down and even if your opponent can see it, he cannot beat you. Poker is no different

Villain's range doesn't matter.

Your friend is right. You would make money playing GTO. You MIGHT be able to make more money, though, by deviating from GTO to exploit your opponent’s mistakes. GTO is unfortunate terminology (the correct term for what’s called GTO is Nash Equilibrium). It implies that the strategy is optimal in all situations, but that simply isn’t true. What is true is that of you play GTO perfectly, there’s nothing an opponent can do to reduce your win rate. If you deviate from GTO to exploit an opponent, that opponent can deviate right back and exploit you. That can’t happen if you play GTO.

To help understand what GTO really is, suppose you’re playing HU NLHE against a smart opponent who has no experience with poker. He tells you up front “I don’t see why you’d play anything but AA and why you’d do anything but go all in when you get AA.” What strategy should you use? Clearly GTO is not optimal. When in BB you get a walk unless he shoves, in which case fold (unless you have AA). In the SB you should minraise ATC and steal his blind almost all the time, folding to his shoves.

Your opponent is smart though, so he soon realizes that his strategy is colossally bad and gradually adjusts. You see his adjustment and come up with a new counter strategy. He adjusts to improve his strategy and you likewise do the same. Assuming both of you adjust properly, this doesn’t go on forever as it might seem it could. You will both be homing in on The GTO strategy as an equilibrium. Once you reach that equilibrium, no further adjustments will improve your win rate and adjustments will cease.

The bolded is where you have misunderstood Nash Equilibrium.

In poker it is correct to say that a NE solution does not need to adjust to any change in the opponents strategy. So long as you stay firm, the opponent only loses more (or stays the same) by changing. Thus, neither player has any incentive to change strategy when both are playing the NE solution.

Now, in poker you CAN adjust to observed flaws in your opponents strategy. This is exploitation. This is more profitable than staying with NE, IF your assumption of the flaws are correct, AND you play a correctly modified exploitive counterstrategy.

I think OP is missing the part where it's impossible to actually memorize the full poker tree.

These trees are not some simple thing you can just take a look at and then try to copy while playing.

They're immensely complicated, taking mixed actions with nearly every single hand in your range.
Even with a tight range, that's a lot of numbers to remember for just one very specific situation.

They change on every flop, they change with effective stack sizes, they change depending on your position, on your opponents position, on the amount of players in the hand, even on the amount of rake in the game you're playing.


So yes, trying to win at poker by memorizing GTO trees would be ridiculous.

OP is definitely not missing anything. It's just a theoretical question

If you "adjust to villain's range" you aren't playing the Nash Equilibrium, you are playing exploitively. This can be more profitable, but if you deviate from Nash and play to exploit, then you also become exploitable and then you are in a contest of adjustments.

To understand this concept better, think of rock-paper-scissors. Nash GTO is 1/3 random of each for both players, because neither player can improve their EV by unilaterally changing this. But let's say your opponent isn't so good with the random numbers and they play rock too much. If you play GTO, you still randomize uniformly and your EV is still zero. You can exploit their play maximally by playing paper 100% of the time. If you opponent realizes this though, they can change their strategy and play scissors 100% of the time and own you back. You might try instead to exploit him sub-maximally by doing paper a little more often but not 100%. This becomes a game of adjustment and counter-adjustment. This type of play is NOT the GTO/Nash equilibrium.

Excuse my ignorance, if memorizing the trees is impossible, but in order to play it correctly you cannot deviate from the algorithm, then how can anyone play GTO?

No one can

The skill is to play MORE CLOSELY to a theoretical GTO solution, than your human opponent. Let's not assume this is understood when the question is asked.

You can try your best to play closer to it. If you're playing closer than villain, you'll play better than him.

I just want to highlight that this GTO thing is not about memorizing stuff, because if we change the solver's premisses (as sizings and ranges), the solutions will be different. So, the idea is to study why things are what they are. The idea is not about memorizing solutions.

Well said. The "why" is fundamentally so much more valuable than the "what"

Is there such proof? Why it can not be so, that GTO solution is infinite, and thus you can not write it down?

I think it depends if you can use continuous betting size, or if it is limited.

Another question. Is there any theoretical need there to be one GTO solution? Why not many, or infinite many optimal solutions?

Yes, every non-cooperative two player game has a Nash equilibrium. This is actually based on Nash's original work. A somewhat technical tutorial and proof can be found here: https://www.cs.ubc.ca/~jiang/papers/NashReport.pdf

I am not a 100% sure but I think even with infinitely many betting sizes, all betting actions are still a bounded, compact space. The space of all betting actions is always bound by the stacks. Thus, even with continuous betting sizing, there exists a Nash equilibrium for the two player game (https://ocw.mit.edu/courses/electric...4S10_lec06.pdf).

However, to date no AFAIK no proof exists whether there exists a Nash equilibrium in multiplayer games.

With respect to uniqueness, a game may have several Nash equilibria. Whether Poker has more than one, I do not know.

Does actual computation take infinite time, or can we somehow speed up computation? And if we could get result is result size infinite?

If you actually wanted to model a version of Poker with an infinite amount of continuous betting sizes, then you could never compute that exactly as you can't store that game tree in memory.

i think its easily disproven that no nash exists in multiplayer games

If the actions available to the players are finite, it has at least one NE.

A deck of cards is rigidly mathematical, 52 cards with two discrete degrees of freedom each.

Even versions of poker that would be considered 'not fun' or not socially acceptable would still have at least one NE with n players.

Nash Equilibrium exists in multiplayer games, but it's possible to exploit the solution through collusion (intended or unintended).

Finite; but it's an approximation by limiting the game tree. Even a few well-chosen sizes per node is often enough to get a really strong solution.

Many spots have multiple equilibria. But some spots only have one solution. It's like any other strategy game in that regard.

But is there a proof that a NE always exists in a multiplayer game analog to the proof in the two player game?

See Nash's Existence Theorem. It works so long as the game is finite and mixed strategies are allowed, regardless of how many players.

Poker is technically a finite & discrete game since betting increments can be no smaller than the smallest blind.