Please explain how this was derived. Please show your work and state your assumptions. Please don't think I'm being sarcastic or an ass, I am legitimately curious

I mean it's not necessarily true if there's a rake. If there is no rake, poker is zero-sum. Or are you asking about something else that I'm missing?

I mean we're talking about a hand starting on the river, so even in a 0 sum game both players should have been in +ev situations just to vpip in the hand (since the game would start out with money in the pot)

Even before any action is taken one player is in a more +ev spot due to position, so they were not in a 0ev situation since we're talking about a hand in isolation and not the overall gto strategies as a whole.

Or is this not the right way to look at it?

The problem here is that if P2 is playing his frequencies correctly his EV of bluffing is already $0 (ie. no different to if he folds), but he forces P1 to sometimes call his raises when he has the nuts.

If he only moves all in with the nuts and calls or folds with everything else, then P1 will have the same EV even if he doesn't adjust how he plays, but more importantly P1 can now exploit P2 by always folding to a raise. Eventually both players should reach the equilibrium if they keep adjusting to each other.

For a simple example of this let's look at a situation where P2 has a range of nuts + air only, and P1 has medium strength hands only. The pot is $100 and the effective stack is $100 as well.

As P2 we can move all in every time we have the nuts and check everything else, but if we do this then P1 will never call us and we make no extra money.

Instead, we can actually force P1 to call us 50% of the time (because if he calls us any less then he knows we could move in with anything and automatically make money off him). We want to force P1 into a situation where he calls us 50% of the time (because if he doesn't we will exploit him by overbluffing in the future), but we want to make sure our bluffs are still EV neutral.

Thus we move all in 50% of the time since we are getting 1-1 on a bluff, and we do that with a range of {66.67% value, 33.33% air).

P1 is forced to call us 50% of the time, so the EV of our value hands becomes:
P(nuts) * P(call) * (current pot + P1's call) + P(nuts) * P(fold) * (current pot) = 66.67% * 50% * $200 + 66.67% * 50% * $100 = $100 ie. the whole pot

and the value of our bluffs becomes P(air) * P(fold) * (current pot) + P(air) * P(fold) * (lost bet when called) = 33.33% * 50% * $100 + 33.33% * 50% * $-100 = $0

So we lose nothing when we bluff because the times our bluffs successfully win us the pot net out the lost bets when we are called and lose (but by forcing P1 to make these calls we eke out extra value when we have strong hands and there is nothing P1 can do about this)

If P2 only moves all in with the nuts and never bluffs, and P1 is playing GTO then P1 still won't care because his EV is the exact same (because he is never losing the pot to a bluff). However P1 can now adjust and exploit P2 by simply folding every time P2 bets, thus reducing P2's EV.

@papagavin, We can do this, determine the best GTO line to take, at the table in real time, against any one of the 9 Vs at are table..........how?

I think it's not. In what I bolded, the second statement does not imply the first statement, and also the first statement is wrong.

Let me give a simple example that shows why I think it's wrong. Imagine you're playing 4 handed and you are the CO. Do you have any part of your opening range that is folding to a 3bet? If the answer is yes, now imagine that you open and someone 3bets you. Obviously if the hand was in your opening range, you believed it was +EV to open (because folding is 0EV); but when you get 3bet, suddenly you fold and lose your open. That doesn't mean opening was -EV; it means that the EV of your open when you don't get 3bet outweighs what you lose when you do.

There is nothing about GTO play that stops anything similar from happening. A GTO player could easily wind up on some street in a spot where the best he can do is fold and lose his previous bets, because the EV of his line would have come from the opponent folding earlier (or making money on different runouts). In fact this could easily be true whether or not the opponent is playing GTO.

[QUOTE=Jay S;52675255]Thanks for writing this up, Vernon. Very interesting read, just as the COTM was.

I will point out one thing, which I also mentioned in that thread a while back: just as expanding a game from 2 players to 3 players changes the landscape in non-intuitive ways, expanding from a one-shot game to a repeated game does too. In a repeated game, strategies are allowed to consider what's happened in previous rounds, and all sorts of complicated equilibria become possible, even in a game as simple as the prisoners' dilemma.

To me, this just highlights how absurdly, ridiculously arcane the concept of a GTO strategy in poker is. When I see the term used in threads now, I just assume the person meant something about being balanced and let it go.[/QUOTE/]

My past conversations with CMV have led me to this exact place. He leads me to a place I don't fully understand and I realize that even if GTO in poker were knowable, it would not necessarily be profitable, and it WOULD necessarily require more work than is feasible. So, I just don't worry about it. I play as exploitably as possible. I mix up my play once in a while when there are regs that are paying attention and then snicker damn near every time someone mentions GTO on the forums.

I'm far from an expert, but I don't think the fact that poker is a repeated game is all that relevant. The thing about the Prisoner's Dilemma is that it isn't a zero-sum game, meaning that cooperation can ultimately benefit both players. Under those circumstances, whether the game repeats matters a ton. Cooperation is not a factor in a zero-sum game like poker. Multiway maybe it is, but it's going to be edge cases I think.

I think you're right that it wouldn't matter HU, since each player is always trying to "punish" the other player as much as possible; zero-sum game and all that. Multiway, actions are taken all the time that shift EV between opponents, as in Vernon's example. It's very possible there exist equilibrium strategies that involve punishing opponents in later hands for shifting EV away from you.

I don't actually want to go down that rabbit hole, because analyzing repeated game strategies gets messy really fast (nor am I advocating anyone trying to play this way, ldo). I'm just trying to underscore how insanely complex the notion of a GTO strategy is.

I'm confused, this seems to agree to me.

This is the statement we're looking at. In the example you just gave even though folding is the correct choice for the original raiser, his original bet was still +ev at the time.

So on your river decision example, assuming all of the actions up to that point were 0 sum doesn't make sense to me because when looking at any given hand in isolation there's money in the pot and some players have big advantages.


Sorry this is all excessively focused on details but I'd like to understand it.

Yeah. I used to play (10-man) SNGs professionally, years ago, they have very serious ICM considerations such that you really, really don't want to get allin on the bubble. So even heads-up hands on the bubble are not zero sum, they are negative sum, because equity shifts from the two of you to everyone not involved in the hand. One of the perks of being a big stack is that you get to shove preflop a lot without the risk of getting allin, because you cover everyone. The big stack tends to own the 4-way bubble.

So anyway, back when I played, some of the regs would use a punishing strategy when big stack (including me, sometimes, there were others who used it more). If one of the other regs dares shove and you can afford a call, you call with whatever random hand and type "my bubble" in chat by way of explanation. This is -EV over a single iteration, but it's usually even more -EV for the guy you snapped off, and because playing SNGs with other regs is such a heavily iterated game, it works well as a warning that you will not tolerate them trying to muscle onto your turf, leaving you free to cheerfully bully everyone on future bubbles. I'd be surprised if a GTO strategy for SNGs didn't include some amount of this spite calling.

It strikes me that the the Prisoner's Dilemma game is actually a 3 player game. The police automatically can only win by making the offer. The prisoners can only lose.

No, the officer always offers

Sounds like a classic GTO move. No matter what happens, the officer isn't going to be worse off by making the move. A reason for making this point is that denouncing isn't the best solution for the two prisoners combined. They are better off together not talking in relation to the officer.

These two quotes make me wonder how many people actually understand what zero-sum means. I skated over it in my COTM because I assumed it would be an easy concept that anyone could look up. Maybe I was wrong.

A game is zero-sum if and only if the sum of all players' utilities from playing is zero. That's the entire definition.

The examples above do not in any way contradict that definition despite them being used to claim that they do.

BGP's example certainly does not contradict the definition. His statement would make more sense if you replaced the word "sum" with "EV". But zero sum and zero EV are NOT AT ALL the same thing.

Every game you can play in a casino is zero-sum. When you lose money, the house wins exactly what you lost, and vice versa. You don't need to be 0EV to be zero-sum. Zero-sum is a much less restrictive condition.

ChrisV's example actually disproves itself. If two people are playing their hands in such a way that they both lose ICM, but the other players are gaining ICM in such a way that the ICM lost by the two players equals the total ICM gained by the players who folded, the game is still zero-sum. Just because it's not zero-sum in a way that feels fair doesn't mean it's not zero-sum.

An actual non-zero-sum game would be poker played with a rake. There, the winner of a pot doesn't win the sum of what the losers put in--he wins that amount minus the rake. So that is a real negative-sum game.

The officer is part of the construct of the game. It has zero bearing on it. To bring it to poker (in a loose analogy), the officer is essentially the dealer. The officer will always produce the offer just like the dealer will always deal the cards.

I guess I fail to see what point your making with bringing up what the best solution is in relation to if the cop is a 3rd party

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Fwiw, there is a "N Person Prisoners Dilemma", which does exist and is used to model various different real life situations.

I feel like you are being a nit here, obviously the SNG as a whole (discounting vig) is zero-sum. My point is that say it's 4-handed and the CO and button fold, if you consider the remaining heads-up encounter in isolation, as its own game, then it is (frequently) negative sum. It's actually similar to rake in that the more they put in the pot, the more tournament equity they collectively lose. Yes, that equity goes to other players in the SNG, but I'm not saying the SNG is negative sum; I'm saying this heads up hand, considered in isolation, is negative sum.

While we are nitting it up over terminology, "ICM lost" and "ICM gained" are not things.

But that's a moot point. Sub-games of zero-sum games are not necessarily zero-sum. This is obviously true in your example and in BGP's example as well (where the sub-game of just playing the river is positive-sum because the pot should be considered dead money).

If we were playing 4-handed without a rake and I said that any CO vs. button hand heads-up was positive sum, I'm guessing you would contradict me, saying that it is still zero-sum because the blinds lost the dead money that makes up the initial pot. You would be right; and your example is similar.

I wouldn't argue that you're wrong that a hand with dead money is positive-sum in isolation, I'd just say that's not a very helpful way of looking at it. It doesn't have any impact on how the game should be played iteratively.

In the situations I'm talking about, it's the fact that heads-up confrontations are negative sum that creates the strategy where the big stack can threaten to make a -EV play in order to inflict even worse pain on opponents. That's just not a possible strategy in a zero-sum game, if you accept a -EV play there then it perforce means your opponent is winning. So I'm not looking at the heads-up game in isolation because I think that's the one true correct way to look at it, I'm looking at it that way because it makes the strategy easier to analyze.

Thanks, I was definitely confusing ev and zero sum, but now I'm confused about other stuff.

Taking roulette for example, if I win on a 38 to 1 bet and am only paid 35 to 1 is that 0 sum?

Let's say that I'm playing another game that is identical to roulette but it's against another player and he pays 38 to 1 and I pay the same losses to him, but the house takes a 3/38 fee for letting us play in their casino out of my winnings. Is that 0 sum?

Zero-sum has no connection to EV or payouts whatsoever. All zero-sum means is that if you win, it's because someone else lost, and vice versa. When you play the house at casino games, whatever money they get comes directly from you, and whatever money you get comes directly from the house. Whether your roulette payout is 38-1, 35-1 or 15-1 isn't relevant.

An example of something that (frequently) doesn't operate this way is the stock market. If I own a stock and the company finds a way to cut costs (for instance, via more efficient production of some good) then the stock goes up in value and/or I will get higher dividends. Me getting that win doesn't require that someone else lost; instead, wealth was created.

Who we consider to be a participant in a given game is a matter of how we define the game. For instance, poker with a rake is a zero-sum game if we consider the house to be a participant in the game, but that would be ridiculous since they don't make any decisions or have impact on any outcomes. So we consider poker with a rake to be a negative-sum game.

Edit:

Yeah that's an example of the paragraph above. If you consider yourself and the other player to be the only participants in the game, which seems reasonable, then it's negative-sum.

OK thanks, that makes sense. I was under the impression that someone's loss would need be the exact same as someone else's win and vice versa.

x-sum game checklist:

Can I win in this game without other player(s) supplying that money (or points, or whatever victory is denominated in)? Then it's a positive-sum game.

Can I lose in this game without other player(s) winning that money? Then it's a negative sum game.

Is neither true? Zero-sum.

Not really sure what you'd call it when it's both. The stock market is an example of both, since everybody can win if a factory gets more efficient, but everybody can lose if an asteroid hits the factory. That might be outside the wheelhouse of game theory, not sure.

I mean, that does have to be true. I know what you mean, but a 35-1 payout can't have both outcomes happen simultaneously. Either you lose $1 and the house wins $1, or you win $35 and the house loses $35. It will never be the case on a single iteration of the game that you win $35 but the house only loses $1, or something.

I meant that I was under the impression that my payout would need to be equal to the probability of my winning. I'd need to be paid an amount that makes the game 0 ev, which 35 to 1 doesn't accomplish.