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



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> **Is there (always) a symmetric (mixed-strategies) equilibrium for the game? Can we find one?**
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> 
> **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 $\{0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1\}$).**
> 
> 
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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?
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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 $p=\frac{2}{3}$. 



How to find equilibrium of the following game?
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.


