I've a doubt about the following game and I'm new with Game-Theory. We have two road managers that aim to maximize their profit. Let be $i=1,2$ the players and respectively $\omega_1,\omega_2\in [0,1]$ possible action. Let be $u_i$, profit of player $i$:

  1. $u_1(\omega_1,\omega_2)=\omega_1(1-\frac{2}{3}\max(0,\omega_1-\omega_2))$.
  2. $u_2(\omega_1,\omega_2)=\omega_2(\frac{2}{3}\max(0,\omega_1-\omega_2))$.

Where $\omega_j$ represents toll on player $i$'s road.

I've got to find Nash Equilibria of the game. I've found that $[1,1/2]$ is the unique Nash equilibrium, is that correct?

Thank in advance for your help

  • $\begingroup$ If player 1 plays 1, isn't player 2's best response to play $1\over2$? $\endgroup$
    – 3rdMoment
    Commented Jan 9, 2023 at 7:42
  • $\begingroup$ sure, I wrote wrong number, I've corrected it $\endgroup$
    – Lorenzo
    Commented Jan 9, 2023 at 18:24

1 Answer 1


If $\omega_1\le\omega_2$, we have $\max(0,\omega_1-\omega_2)=0$ and thus $u_1=\omega_1$, $u_2=0$. If $\omega_1\ge\omega_2$, we have $\max(0,\omega_1-\omega_2)=\omega_1-\omega_2$ and thus $u_1=\omega_1(1-\frac23(\omega_1-\omega_2))$ and $u_2=\frac23\omega_2(\omega_1-\omega_2)$.

Thus, the payoff for player $1$ increases with $\omega_1$ until $\omega_1=\omega_2$, so there can be no Nash equilibrium with $\omega_1\lt\omega_2$. The derivatives for $\omega_1\ge\omega_2$ are

$$ \frac{\partial u_1}{\partial\omega_1}=1-\frac43\omega_1+\frac23\omega_2 $$


$$ \frac{\partial u_2}{\partial\omega_2}=\frac23(\omega_1-2\omega_2)\;. $$

For a Nash equilibrium in the interior, $0\lt\omega_2\lt\omega_1\lt1$, these must both vanish. Solving the system of linear equations yields $\omega_1=1$ and $\omega_2=\frac12$. This doesn’t lie in the interior, but it’s a Nash equilibrium; both strategies are best responses to each other. The payoffs are $\frac23$ and $\frac16$, respectively.

Now we need to check the remaining boundary cases. Since we’ve excluded $\omega_1\lt\omega_2$ and we already found $\omega_1=1$ in solving the interior case, the only remaining boundary cases are $\omega_2=0$ and $\omega_2=\omega_1$. Neither of these yields a Nash equilibrium, so it seems you’ve correctly identified the one Nash equilibrium of the game.


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