Do "unexploitable" strategy exist in No Limit Holdem? By this I mean frequency-based mixed strategy that has non-negative expected payoff against any other strategy (let's assume the game is completely symmetric, seats are randomly assigned to all players).

From my very basic understanding of game theory, I would say that for a 2-player game the answer is yes because any equilibrium strategy (which should exist, right?) will do the job. But what if there are >2 players?

I'm interested in a proof for the (non-)existence of such a strategy.

EDIT #1: As pointed out by vadim123, we should assume collusion is impossible.

EDIT #2: After thinking about it for a bit, I'm actually not sure whether collusion matters at all. Maybe someone can show a concrete example (with some kind of formal proof)?

  • $\begingroup$ If there was such a thing Isildulr1 would be broke. Oh, wait a minute.. $\endgroup$ – Git Gud Jun 8 '13 at 15:09
  • $\begingroup$ If you can apply the strategy, then others cannot, which is impossible. Even in a company of 2 players, they cannot both apply the same 'unexploitable' strategy, because at least 1 should lose for the other to win, right? Therefore it cannot exist. $\endgroup$ – Mixxiphoid Jun 8 '13 at 15:16
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    $\begingroup$ Note I said "non-negative", not "positive". So if all players apply the same strategy, the expected payoff for everyone would be zero, which is non-negative. $\endgroup$ – zero-divisor Jun 8 '13 at 15:27
  • $\begingroup$ @Mixxiphoid No an unexploitable strategy means that your expected value playing against any other strategy is nonnegative. Playing against the unexploitable strategy would thus simply give you an expected value of 0. $\endgroup$ – Vincent Pfenninger Jun 8 '13 at 15:27
  • $\begingroup$ @VincentPfenninger I think unexploitable typically just means the strategy is minimax. If you always played in the big blind, your best strategy is still a losing strategy, but that doesn't mean people can exploit your play. $\endgroup$ – guy Jun 8 '13 at 15:38

The answer is no, without additional restrictions. Otherwise the other players simply collude, and only the best hand among them bets while the others fold. Hence you are betting your hand against the best of $(n-1)$ hands, and the expectation of that is negative.

Okay, here's another collusion strategy. The first hand when you fold, all other players go all-in. Consequently, instead of $n$ players with equal bankrolls, you have your same bankroll but now you have one opponent with a bankroll $(n-1)$ times larger than you.

There's a reason two-player games are well-studied, while $n$-player games are not. It's much harder, even with unintentional collusion that takes advantage of bad players.

  • $\begingroup$ Ok, your right, this is trivial. Let's assume collusion is not possible as it is against the rules anyway :) $\endgroup$ – zero-divisor Jun 8 '13 at 16:03
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    $\begingroup$ Even if you don't have explicit collusion, you can have implicit collusion, where some of the players play naively, betting exactly the value of their hand. If others exploit this knowledge, they will win more than their share. You, however, are playing indifferent to the betting patterns of others, so you will not. $\endgroup$ – vadim123 Jun 8 '13 at 16:05
  • $\begingroup$ But not winning more than my share does not mean I have negative EV, does it? $\endgroup$ – zero-divisor Jun 8 '13 at 16:10
  • $\begingroup$ If you allow one of the other players to exploit another, that is equivalent to playing with fewer players, where one of your opponents has a larger bankroll than you. This is a major disadvantage playing No Limit. $\endgroup$ – vadim123 Jun 8 '13 at 16:46
  • $\begingroup$ I'm not sure whether collusion matters here. In your example for instance, we're effectively playing against one other player with a very strong range, but who also pays $(n-1)$-times the blinds we do, so it's at least non-trivial that this constellation gives us a negative EV, isn't it? $\endgroup$ – zero-divisor Jun 8 '13 at 20:17

vadim123 is correct: with more than two players, you are vulnerable to collusion. This collusion may take place (illegally) during the game, or (probably still illegally) by agreements made before the game. It may even be unintended, as when you find yourself playing against a married couple who will obviously not want to step on each other's toes.

But, as you suspect, in a two-player game there is always a mixed strategy that gives you a non-negative expectation. (It may be -- I don't know -- that being first to deal gives you a very slight edge over your opponent, but we can ignore this subtlety if we start the game by drawing lots to see who deals.)

For a two-player game, the real question is: does there exist such a strategy that has a positive expectation if your opponent plays badly? For example, in rock-paper-scissors, the only good strategy from a game-theoretical point of view is always to choose your move at random. But this means that your expectation is 0, no matter how your opponent plays.

However, poker is more complex than rock-paper-scissors. It seems likely that if you play perfect poker (in the sense that your expectation is non-negative whatever your opponent does), then your opponent will have plenty of room for error, and you will win over the long run against anybody on the planet. But I can't back this up with anything like a convincing argument.


Unless I'm missing something, the answer is "no" via a very simple argument. (EDIT: I was thinking of "strictly positive" expected value, which the argument below argues against.)

Suppose the game is completely symmetric (i.e. the strategy set of each player is the same). Suppose the game is zero-sum (i.e. the total payoffs of all players add up to zero). Note this implies that, in expectation, the total payoffs added up over all players is zero (since it is zero in every case).

Then there is no strategy that guarantees positive expected value. If there were, then every player could play that strategy simultaneously, and they would all expect to win some strictly positive amount. So the expected total payoff (added up over all players) would be some positive amount, which contradicts the game being zero-sum.

The only question I have is if you think those assumptions are valid for this setting or not -- they seem to capture what you're asking.

  • $\begingroup$ Note that . I don't require the strategy to have a positive expectation, only a non-negative one. $\endgroup$ – zero-divisor Jun 12 '13 at 4:35

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