Nash Equilibrium in 3+-player games with more than 1 players deviating from it

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.

• 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. May 21, 2017 at 15:21
• Some of your concerns are addressed by a refinement known as "Coalition-proof Nash equilibrium".
– mlc
May 22, 2017 at 20:18

• @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. May 21, 2017 at 16:20