I did a little searching on whether or not GTO play can even be solved for poker with 3 or more players and thought I got a definitive answer but maybe not.

IMO, a Nash equilibrium exists if there's a fixed number of players and finite actions. So since 3 handed poker cash game poker (lets ignore rake for now) meets these requirements there will be a GTO solution.

What this logically means (at least in my mind) is that assuming symmetrical conditions, two players can't collude against a GTO playing player in order to gain more money for themselves. By collude I do not mean sharing hole card information in any way, I mean only using betting strategies (squeezing for example). So the GTO playing player is guaranteed not to lose money (luck aside).

So:
A. the above is true
B. its not necessarily true, this is an open game theory problem
C. its not true at all, GTO play has been proven impossible multiway
D. GTO play and successful collusion can coexist somehow (what?)

My understanding is that a Nash Equilibrim exists but all that means is that each player has no incentive to unilaterally change his strategy. Two players on the other hand might (and probably will) gain and advantage by simultaneously changing their strategies at the expense of the 3rd player.

Yeah the question which was spurred by this thread is most answered there.

So basically strategies at "Nash equilibrium" can be exploited by two or more players. But I think poker players mean something different when the term "GTO" is used.

+1 to bobf

Lots of poker players dont understand game theory at all, so they could mean whatever they want by saying "GTO". Nash is exploitable, but only with "implicit" colluding where one player plays a -ev strat.
Morton's theorem also gives examples of this kind of exploitation.

lol just search don colluder in google
your answer

True that 2 players can play in a way to have an edge

This must be true if gto exists for 3+ way poker. However, the only way that I see the possibility of gto being impervious to collusion is that gto is psychic and will know if and when opponents are colluding. Thus the alliance will be dissolved before it ever comes to fruition.

This is basically it.

Poker players who understand some game theory will often pay lip service to the idea that GTO can be beaten by collusion, but then act like as long as there is no active collusion, GTO can't lose. This is also false, as players might end up playing the same strategies that colluders would play totally by coincidence.

In fact, it goes even further than that. The definition of a Nash equilibrium, as was said above, is that a player can't unilaterally gain by deviating. However, there is nothing in the definition that says the utility of the Nash equilibrium is locked in once another player deviates.

In more concrete terms, let's say you have a 3-player game with a Nash equilibrium whose payout is (0,0,0). It is possible that the third player could deviate from the NE by playing a strategy such that the combination of the other two GTO strategies and the deviated strategy pays out something like (2,-1,-1) or even (1,-1,0). In other words, the second player could take a loss from only the third player deviating from GTO. (Note that the deviating player isn't gaining--the utility is shifted to a different player!) There is nothing in the definitions that stops that kind of scenario.

It seems to me that rather than (or perhaps in addition to) focusing on 'collusion', it's more helpful to say that in multiplayer games there must always exist 'suicide strategies', and that a Nash strategy will remain unbeatable provided there are no masochists in the game.

Not sure what you mean by this but it seems like whatever you mean by it has got to be false.

There was an old thread in this forum where someone was looking at a toy game where 3 players each hold a coin and set it to heads or tails (secretly), and if anyone's coin was different than the other 2, that player wins $2 and the other two players each lose $1. In that game, despite it being completely symmetrical, there exists a Nash equilibrium where one player wins and the other two players lose 100% of the time. So there exists a zero-sum game that is symmetrical (i.e. all players have the same strategies available to them and they pay out the same), but where everyone can be playing GTO and one or more players is still going to lose.

^Can you find the link? The way you've described it doesn't seem to support any kind of argument for anything as far as I can tell.

The point I was making was that in a game like poker a player can make it impossible for you to win in a spot, but they have to sacrifice their own ability to win in the process. The only way to take EV directly away from a Nash-playing opponent and give it to yourself is to flat-out cheat by sharing hole-card info etc. Or I suppose in some rare circumstance two players' strategies could unwittingly align such as to have the effect of deliberate collusion vs a third. Not sure about that though really, without there being an ex-game-rules info-sharing component.

Obv you have things like implicitly colluding to face a player with calling two bets cold, but that's an example of something everyone gets a 'turn' to do, and if when it's your 'turn' the guy who's supposed to collude with you screws it up, that's an example of them hurting you by hurting themselves.

http://forumserver.twoplustwo.com/15...24/?highlight=

The argument is hashed out a little better over the course of several posts in there. The point is that H/T/anything is actually a Nash equilibrium in that game, including H/T/T. So every strategy is GTO and GTO can lose even when other players are sticking to Nash equilibrium strategies.

OK, I see. But is there anything analogous to this in poker? I would argue that the quoted example is a special kind of game in which collusion is kind-of built in to the way it is scored. Whereas in poker it's just the (lone) best hand that wins. Sure it's often possible for player C to do something wacky that causes a pot that 'should have' gone to player A to get shipped to player B instead, but C has to pay heavily out of his own stack to effect that outcome. Right?

Interesting that you phrase it this way. It's FAR from clear that this is right. I actually worked out an example of this on my own a few months ago because I was curious, and I think it applies here.

This is a toy example that's analogous to poker. Let's say there are 3 players who I'll refer to as P, A, and B. P is holding a number in the range (-0.4,0) U (1,1.6). A and B are each holding a number in the range (0,1). The pot is 1 and each player has a stack of 1. A and B have checked, and it is on P to act. P can check or shove. If he checks, showdown. If he bets, then A can call or fold, and after that B can call or fold. Highest number wins at showdown.

It turns out that the Nash equilibrium is for P to shove the top 90% of his range--meaning he value bets 60% of the time, bluffs 30%, and checks back 10%. Then, if P bets, A never calls--that's right, literally never calls--and B calls 50% of the time, but if A happens to call, he folds 100% of the time. (That last part is a necessary contingency to make it an equilibrium.)

The EVs for each player are 0.9 for P, 0.05 for A, and 0.05 for B. (This is easy to see because when P bets, A and B both have 0EV, so the 90% of the time that P bets, his EV is 1. The 10% of the time that he checks, A and B win half the time each because they have identical ranges.)

Now suppose P decides to deviate from this equilibrium by shoving his entire range instead of checking back 10% of the time. Now he is exploitable--but suppose that A and B continue playing their GTO strategies. Since P now never checks, A never sees a showdown--and since all his EV came from showdown equity, his EV goes down to 0. Meanwhile, however, since B's GTO calling frequency is designed to make P indifferent to bluffing, P's EV actually does not change! So now instead of .9/.05/.05, it's .9/0/.1.

So here P's deviation from the NE is shifting EV from A to B, without P losing any EV, even though P has deviated from the NE and A and B have not. In light of this example I think it is extremely hard to make the case that this kind of situation couldn't happen in real poker.

Excellent posts guys.

Interesting. So P did have to open himself up to exploitation/loss of EV with respect to B, but you've shown that even if B were not to sense P's imbalance and actively exploit it, he's stll gaining EV from A without any inherent cost to P. So perhaps in addition to suicide strategies there are also playing-god-for-the-lulz strategies. Certainly such things can arise when bubble pressure comes heavily into play in DoNs or satellites, where a very big stack can choose to favour or pick on certain short stacks on a whim.

Still, while your example shows that a set of ranges can align in such a way as to allow for god-playing on a particular street in normal poker, it does I think remain an open question whether player P can arrive at that street with the (presumably) highly unusual range construction required, without having had to sacrifice EV on earlier streets.

That is a really great example. Complete and intuitive - you don't have to do any calculations to see that it's right.

That's actually irrelevant. You can give P any polarized range you choose and repeat the example. It just means that P will be checking back in a Nash equilibrium a different percentage of the time, and therefore the EV for A and B will go up. It will still be the case that when you have P bluff more often than his GTO strategy, he shifts EV from A to B without losing EV himself.

Right, yeah I see, all that matters is that P is the one who's completely polarised. So we're seeing that there's an inherent disadvantage to being in the middle in the betting order, because you're at the mercy of others as to whether you get your fair share of checkdowns. Although I suppose everything is still going to even out over time as we experience our fair share of seating arrangements.

I was curious to see what happens in your toy game if A knows P is overbluffing and responds by always calling... Doesn't go well for A.

Not really sure what B's Nash overcalling freq should be given that A isn't ever supposed to call, but if B operates on the assumption that P is playing Nash and A is calling 1/2 then we have B overcalling 1/4 and EVs are now:
P 0.95
A -0.05
B 0.1

If B knows what's up then he should overcall 3/8 and now EVs are:
P 1.025
A -0.1375
B 0.1125

I said this in the example. B overcalls 0%. That is included in his Nash equilibrium strategy. The reason is because if B ever overcalls, then not only is A's calling range not at equilibrium anymore, but P has an incentive to change his bluffing strategy (since now his bets have a chance of getting called twice).

So in fact, if A knows P is bluffing too much, he can start calling 100% of the time, and if B doesn't deviate from GTO, it is HIS EV that now drops to 0, and the EVs are distributed .8/.2/0. So this then becomes an example of one player exploiting another player and freezing out a third player who is playing GTO.