0
$\begingroup$

A Nash equilibrium is defined to be a (mixed or pure) strategy for which no player have incentive to unilaterally deviate from his or her current strategy.

In this definition, there is nothing said about the payoff to each players having being equal at the Nash equilibrium. But in most games, such as matching pennies and prisoner's dilemma. The Nash equilibrium gives equal payoff to each player.

Can anyone resolve this question for me?

Put it in mathematical terms. Consider a game $G$ with players $P1,P2$, with utility fuctions $U_1, U_2$. The players attempt to maximize his or her own utility. Then the mixed strategy $p^* = (p_1^*, p_2^*)$ is a Nash equilibrium if for any mixed strategies $p_1, p_2$,

$U_1(p_1^*, p_2) \geq U_1(p_1, p_2)$ and $U_2(p_2^*, p_1) \geq U_2(p_2, p_1)$

How does this then imply $U_1(p_1^*, p_2) = U_2(p_2^*, p_1)$?

$\endgroup$
  • 3
    $\begingroup$ It doesn't. Games can be massively unfair. $\endgroup$ – spaceisdarkgreen Jul 1 '17 at 22:13
1
$\begingroup$

Your two examples are symmetric games, a special case where the strategies and the payoffs of the players are the same up to appropriate permutations. More formally, assuming two players for simplicity, they have the same set of pure strategies $(S_1=S_2)$ and their payoffs are such that $u_1(s_1,s_2) = u_2(s_2,s_1)$. As a consequence, at a symmetric equilibrium where both players play the same strategies, they have the same payoffs. It is only the symmetry assumption that drives the equality of payoffs.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

I think it is false in general.

Consider the trivial game $G$ with players $P_1,P_2$, each of them having only One strategy $S$. Suppose the payoff are $0$ for $P_1$ and $1$ for $P_2$. Then you have $(S,S)$ as the only Nash Equilibrium, with different payoffs. Am I misunderstanding the question posed? Otherwise I think you will get other nontrivial examples by adding an arbitrary $c\neq 0$ to the utility function of one of the two players in a game with equal payoff for the Nash Equilibrium.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

In general it is not true. Generically,

  1. Payoffs to players in equilibrium will differ.
  2. If there are many equilibria, then some equilibria will generically payoff dominate other equilibria.

Here is an example. Consider the following game where player $1$ can play Top or Bottom, and player $2$ can play Left or Right.

$\begin{matrix} & L & R \\ T & 3,2 & 0,0 \\ B & 0,0 & 2,1\end{matrix} $

The game has two pure strategy Nash equilibria {T,L} and {B,R}. We see that payoffs in each equilibrium to both players are different. Moreover, both players earn higher payoff in equilibrium {T,L} than in {B,R}; in other words equilibrium {T,L} Pareto-dominates equilibrium {B,R}.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.