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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 $p=0.5$ then $x_a=x_b=0.5$ is a symmetric equilibrium .
• 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 $p=\frac{2}{3}$.

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 $p=\frac{2}{3}$.

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 $p=0.5$ then $x_a=x_b=0.5$ is a symmetric equilibrium .
• 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?

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 $p=0.5$ then $x_a=x_b=0.5$ is a symmetric equilibrium .
• 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 $p=\frac{2}{3}$.

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 $p=0.5$ then $x_a=x_b=0.5$ is a symmetric equilibrium .
• 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?

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 $p=0.5$ then $x_a=x_b=0.5$ is a symmetric equilibrium .
• 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?