What's the point of Nash Equilibrium in 3+-player games if it's unstable? For example, in a 3-player game, if 2 players deviate from the equilibrium strategy, there is no guarantees for the 3rd player that he maintains the equilibrium value playing a Nash Equilibrium strategy.

What exactly I'm asking is why it was defined as a set of mutually best responses for all the players, when in fact, each player plays against all the others. If we defined it as (Player1BestResponse; OthersBestResponse), which means player 1 tries to maximize his value and the others try to minimize it, we could maintain the properties of 2-player Nash Equilibrium. Player 1 could guarantee he would get a certain value playing this kind of strategy.

  • $\begingroup$ Your question falls within the partition function approach of R. M. Thrall and W. F. Lucas. N-person games in partition function form. Naval Research Logistics, pages 281–298, 1963. doi: 10.1002/nav.3800100126, and the formation of endogenous formation of coalitions studied by S. Hart and M. Kurz. Endogenous Formation of Coalitions. Econometrica, 51(4): 1047–1064, 1983. Applying this approach to the Nash equilibrium means that a post merger equilibrium emerges. This topic is studied in the field of industrial cooperation. $\endgroup$ May 21, 2017 at 15:21
  • $\begingroup$ Some of your concerns are addressed by a refinement known as "Coalition-proof Nash equilibrium". $\endgroup$
    – mlc
    May 22, 2017 at 20:18

1 Answer 1


The reason one defines Nash equilibrium as the non-existence of profitable unilateral deviations for each player is that anything more requires coalitional thinking on behalf of the other players, which, in a non-cooperative context, there is no reason to presuppose.

In your example, certainly two players could 'work together' to try to play against the third in some fashion, but one would have to ask also if there is any incentive for this teamwork to happen. If there is no incentive for either member of the two-player coalition to work together, then your example would seem normatively unreasonable as a solution concept for a game: it would require those two players to act against their self-interest. If, however, there is incentive for them to act as a coalition against the third (in the sense that both of these players lack a profitable deviation from doing so) then your solution concept reduces to Nash equilibrium anyway.

  • $\begingroup$ In most games the players are humans, they wouldn't play an exact Nash Equilibrium strategy (or anything near it). What I'm afraid of is their random deviations could hurt my strategy too much. They might not cooperate directly, but the existing meta could potentially make an equilibrium strategy be awfully losing. $\endgroup$ May 21, 2017 at 15:17
  • $\begingroup$ For example, 3 players bet 1 dollar each and have to choose heads or tails. If everyone chooses the same, the bets are returned. If not, the 2 players with the same outcome share the loser's bet. The Nash Equilibrium is (0.5;0.5) and the value is 0, but if two players always play tails, the one playing the equilibrium will be losing 0.5 dollars on average. $\endgroup$ May 21, 2017 at 15:17
  • $\begingroup$ @alain_morel Yes, but all that goes to show is that the tuple of strategies $(T,T, 50-50)$ is not a Nash equilibrium (and quite reasonably so as you point out). The issue is in your game there are a lot more Nash equilibria than just 50-50. All $T$ or all $H$ are also Nash equilibria. $\endgroup$ May 21, 2017 at 16:20
  • $\begingroup$ @alain_morel With regard to your first comment, if you're interested in extensions of Nash equilibrium there's a tremendous literature. I hope that the above convinces you that defining a solution concept as 'Nash equilibrium but robust to multilateral deviations' has properties that make it undesirable as a theory of non-cooperative behavior. If instead you are interested in variants of Nash equilibrium that are robust to certain beliefs or mistakes by players there are a number of other well-established refinements of the 'plain' Nash equilibrium. $\endgroup$ May 21, 2017 at 16:27

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