I wonder if there exists a game in which $\epsilon$ - Nash equilibrium leads to a much higher payoff than any other Nash equilibrium.

Show, for every $\epsilon$ > 0, a two player game where there is an $\epsilon$-Nash equilibrium in which both players have a much higher payoff then in any Nash equilibrium.

What is $\epsilon$ Nash - equilibrium.$\sigma$ is $\epsilon$-Nash equilibrium, if $ u_i(\sigma) \geq u_i(\sigma_i, \sigma_{-i}) - \epsilon, \forall \sigma_i \in \Delta(S_i), \forall i \in N$.

It sounds very counter-intuitive, in other word I have to show the game where $\epsilon$ - Nash equilibrium, which is according to the definition not the best strategy to play indeed leads to the best payoff.

Could you give such en example?


1 Answer 1


It is quite simple: if you enlarge the set of possibilities you can get better values.

Here, in place of requiring the possibilities to be equilibrium we require them to be almost equilibrium, a larger set of possibilities, therefore a higher possible value.

Here is a simple example:

$$\left[\begin{array}{cc} 5/5 & 0/6 \\ 6/0 & 1/1 \\ \end{array}\right]$$

Let $\epsilon = 2$. Now $5/5$ on the top left corner is an $\epsilon$-equilibrium where as the equilibrium is $1/1$.


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