# Mixed Strategy Nash Equilibirum, Contributing to a Public Good

I'm a economics undergrad student currently studying the basics of Game Theory. I'm trying to solve the following mixed strategy game:

-Two players, Player 1 and Player 2

-Available actions: Each player can either make the contribution 0, 1, 2 (The costs of each contribution is increasing in the contribution, thus c2>c1>c0=0.

-The preferences: are represented by the payoff function (v-c), where v is the value that each player attaches to the public good being provided, and c is the cost that each player pays.

In summary, the payoff matrix looks like the following:

I see that there are three pure strategy Nash Equilibria in this game, (0,2), (1,1), (2,0). But I have trouble determining the mixed strategy equilibria... specifically, I'm wondering whether there can be a mixed strategy equilibrium in which player 1 mixes between {0,1} and player 2 mixes between {1,2}, even though the payoff matrix is symmetric for the two players. Can you lend me a hand here??

If player $$1$$ mixes between $$0$$ and $$1$$, and plays $$0$$ and $$1$$ with probability $$p_0$$ and $$p_1=1-p_0$$ respectively, then player $$2$$'s expected payoff from playing

• $$0$$ is $$0$$
• $$1$$ is $$p_0(-c_1)+(1-p_0)(v-c_1)=(1-p_0)v-c_1$$
• $$2$$ is $$v-c_2$$

In order to make sure that player $$2$$ mixes between $$1$$ and $$2$$, it must be the case that

• $$(1-p_0)v-c_1=v-c_2\geq 0$$ i.e. $$p_0=\dfrac{c_2-c_1}{v}$$ and $$v\geq c_2$$.

If player $$2$$ mixes between $$1$$ and $$2$$, and plays $$1$$ and $$2$$ with probability $$q_1$$ and $$q_2=1-q_1$$ respectively, then player $$1$$'s expected payoff from playing

• $$0$$ is $$vq_2=v(1-q_1)$$
• $$1$$ is $$v-c_1$$
• $$2$$ is $$v-c_2$$

In order to make sure that player $$1$$ mixes between $$0$$ and $$1$$, it must be the case that

• $$v(1-q_1)=v-c_1\geq v-c_2$$ i.e. $$q_1=\dfrac{c_1}{v}$$ and $$v> c_1$$.

Therefore, we have the following mixed strategy Nash equilibrium under the condition $$v\geq c_2 > c_1 > 0$$:

• $$1$$ plays mixed strategy $$(p_0,p_1,p_2)=\left(\dfrac{c_2-c_1}{v}, \dfrac{v-(c_2-c_1)}{v}, 0\right)$$, where $$p_i$$ is the probability of playing action $$i\in\{0,1,2\}$$
• $$2$$ plays mixed strategy $$(q_0,q_1,q_2)=\left(0, \dfrac{c_1}{v}, \dfrac{v-c_1}{v}\right)$$, where $$q_i$$ is the probability of playing action $$i\in\{0,1,2\}$$

You could try to randomize the row. That is, let the probability of player 1 to choose 0, 1, 2 be p1, p2 and (1-p1-p2), respectively. In the equilibrium, player 2 must be indifferent to choices 0,1,2 as well, which means that the expected return should be exactly the same for the three opinions. By solving the equations, you should be able to find p1 and p2 in terms of v, c1 and c2.

In my personal opinion, I do not believe that there exists such an equilibrium where player 1 chooses {0,1} and {1,2} because, as you mention, the payoff matrix is symmetric.