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I'm looking at the following game:

2 symmetric players $\{A,B\}$, each choose a number $x_a,x_b\in [0,1]$.

The utility of player $A$ is: $$U_A = \ \begin{cases} p\cdot x_a &\mbox{if } x_a> x_b \\ \frac{p\cdot x_a+ (1-p)\cdot (1-x_a)}{2} &\mbox{if } x_a= x_b\\ (1-p)(1-x_a)\ \ \ \ \ \ \ \ \ \ & \mbox{else}\end{cases} $$

For some constant $p\in[\frac{1}{2},1]$.

Is there (always) a symmetric (mixed-strategies) equilibrium for the game? Can we find one?

If no equilibrium exist, what is the equilibrium of the discrete version of the game (i.e. where $x_a,x_b$ are chosen, say, from $\{0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1\}$).


Observations about the game:

  • If $p\geq \frac{2}{3}$ then $x_a=x_b=1$ is a symmetric equilibrium.
  • If $p\in [0.5,\frac{2}{3}]$, $x_a = 1$, $x_b = 0$ is a non-symmetric equilibrium.

What can we say about the symmetric equilibrium in the $p\in (0.5,\frac{2}{3})$ case?


The motivation for the game comes from a larger family of games, in which I'm trying to show that the price of anarchy is at most $1.5$ which is what we get for $p=\frac{2}{3}$.

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    $\begingroup$ I don't understand what you say about $p = 1/2$. It seems to me that $(1/2, 1/2)$ is never an equilibrium: it's better for either player to choose either $0$ or $1$ (to get payoff $1/2$) instead of payoff $1/4$ at $(1/2, 1/2)$. $\endgroup$ Commented Nov 13, 2014 at 18:44

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Since you say it's symmetric, I'm assuming the utility of player $B$ is $$U_B = \ \begin{cases} p \cdot x_b &\mbox{if } x_b> x_a \\ \frac{p\cdot x_b+ (1-p)\cdot (1-x_b)}{2} &\mbox{if } x_a= x_b\\ (1-p)(1-x_b)\ \ \ \ \ \ \ \ \ \ & \mbox{else}\end{cases} $$

For $1/2 \le p \le 2/3$, consider the mixed strategy for $B$ where $x_b = 0$ with probability $2 - 3 p$, $1$ with probability $3p - 1$. Then $A$'s payoff is $(3/2) p (1-p)$ for $x_a = 0$ or $1$, and always less than that otherwise. So we have a symmetric Nash equilibrium with this mixed strategy.

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