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Game theory post-grad reporting.

In my current research I have stumbled upon a problem that seems simple, but managed to block me for weeks already.

Let's say we have $n$-player game with $2$ pure strategies available to each player, and we know for sure that this game has at least one Nash equilibrium in totally mixed strategies. What would be good sufficient conditions for such game to have exactly one equilibrium in totally mixed strategies? Please, take note, I'm talking about totally mixed strategies for all players, and do not care of situations where one or more players use pure strategies.

On a hunch it feels like a class of $2 \times 2 \times ... \times 2$ games with exactly one Nash equilibrium in totally mixed strategies should be very large. But to my surprise I can't outline even single meaningful subclass of it.

EDIT 1: My current best guess is that at least following should be provable:

If in any two situations that are distinct by action of only one player no player have same payoff, then game can't have more then one equilibrium in totally mixed strategies.

But unfortunately, even for such weak assumption I don't see how.

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  • $\begingroup$ It does sound very complicated. The payoff table has $2^n$ numbers. What about $n=2$ and $n=3$? $\endgroup$
    – usul
    Feb 1 '17 at 2:59
  • $\begingroup$ @usul, in $2 \times 2$ everything is simple - we do have explicit formula to calculate probabilities for unique totally mixed equilibrium (if it exists) that works in all non-degenerate cases. So, the only possibility for other totally mixed equilibrium is zero payment matrix for one of players, which gives us whole line segment of them. For $2 \times 2 \times 2$ things are becoming more complex, but I haven't managed to come with any non-degenerate cases of multiple totally mixed equilibrium. And I have suspicion that for higher dimensions interesting games can't have more then one too. $\endgroup$
    – Doktor Diagoras
    Feb 1 '17 at 12:34
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I hope the following will provide a nontrivial game with continuum of completely mixed equilibria.

Let us consider the payoff function of a single player, and later we will extend the analysis to a multiplayer game.

Consider the 2x2 game in which the payoff is 1 in the two diagonal entries, and 0 in the two off-diagonal entries. Once the other player plays each pure strategy with probability 1/2, you are indifferent among your pure strategies.

We can have a similar property in a 2x2x2 game: denoting the two pure strategies of each player by $\{A,B\}$, the payoff to the player is 1 for the following strategy profiles: $(A,B,A), (A,A,B), (B,A,A), (B,B,B)$; and 0 otherwise. In this game, once at least one of the other players plays each pure strategy with probability 1/2, our player is indifferent among his two pure strategies.

We can extend this definition to any number of players: the payoff of our player is 1 if the number of players who choose A is even, and 0 otherwise. Suppose that this is the payoff function of ALL players. Then any strategy profile in which at least two players play $(1/2,1/2)$ is an equilibrium. This hints of a plethora of other nontrivial games that have multiple strictly mixed equilibria.

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  • $\begingroup$ That's clever! I have sidestepped around this problem in my research already, but your answer is very insightful. I need to think if this can change anything in my current course. $\endgroup$
    – Doktor Diagoras
    Apr 10 '17 at 14:55

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