<p>If player <span class="math-container">$1$</span> mixes between <span class="math-container">$0$</span> and <span class="math-container">$1$</span>, and plays <span class="math-container">$0$</span> and <span class="math-container">$1$</span> with probability <span class="math-container">$p_0$</span> and <span class="math-container">$p_1=1-p_0$</span> respectively, then player <span class="math-container">$2$</span>'s expected payoff from playing</p>
<ul>
<li><span class="math-container">$0$</span> is <span class="math-container">$0$</span></li>
<li><span class="math-container">$1$</span> is <span class="math-container">$p_0(-c_1)+(1-p_0)(v-c_1)=(1-p_0)v-c_1$</span></li>
<li><span class="math-container">$2$</span> is <span class="math-container">$v-c_2$</span></li>
</ul>
<p>In order to make sure that player <span class="math-container">$2$</span> mixes between <span class="math-container">$1$</span> and <span class="math-container">$2$</span>, it must be the case that</p>
<ul>
<li><span class="math-container">$(1-p_0)v-c_1=v-c_2\geq 0$</span> i.e. <span class="math-container">$p_0=\dfrac{c_2-c_1}{v}$</span> and <span class="math-container">$v\geq c_2$</span>.</li>
</ul>
<p>If player <span class="math-container">$2$</span> mixes between <span class="math-container">$1$</span> and <span class="math-container">$2$</span>, and plays <span class="math-container">$1$</span> and <span class="math-container">$2$</span> with probability <span class="math-container">$q_1$</span> and <span class="math-container">$q_2=1-q_1$</span> respectively, then player <span class="math-container">$1$</span>'s expected payoff from playing</p>
<ul>
<li><span class="math-container">$0$</span> is <span class="math-container">$vq_2=v(1-q_1)$</span></li>
<li><span class="math-container">$1$</span> is <span class="math-container">$v-c_1$</span></li>
<li><span class="math-container">$2$</span> is <span class="math-container">$v-c_2$</span></li>
</ul>
<p>In order to make sure that player <span class="math-container">$1$</span> mixes between <span class="math-container">$0$</span> and <span class="math-container">$1$</span>, it must be the case that</p>
<ul>
<li><span class="math-container">$v(1-q_1)=v-c_1\geq v-c_2$</span> i.e. <span class="math-container">$q_1=\dfrac{c_1}{v}$</span> and <span class="math-container">$v&gt; c_1$</span>.</li>
</ul>
<p>Therefore, we have the following mixed strategy Nash equilibrium under the condition <span class="math-container">$v\geq c_2 &gt; c_1 &gt; 0$</span>:</p>
<ul>
<li><span class="math-container">$1$</span> plays mixed strategy <span class="math-container">$(p_0,p_1,p_2)=\left(\dfrac{c_2-c_1}{v}, \dfrac{v-(c_2-c_1)}{v}, 0\right)$</span>, where <span class="math-container">$p_i$</span> is the probability of playing action <span class="math-container">$i\in\{0,1,2\}$</span></li>
<li><span class="math-container">$2$</span> plays mixed strategy <span class="math-container">$(q_0,q_1,q_2)=\left(0, \dfrac{c_1}{v}, \dfrac{v-c_1}{v}\right) $</span>, where <span class="math-container">$q_i$</span> is the probability of playing action <span class="math-container">$i\in\{0,1,2\}$</span></li>
</ul>

CC BY-SA 4.0
2024-05-26T03:34:12.833Z
2024-05-26T03:34:12.833Z
Amit
<p>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.</p>
<p>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.</p>

CC BY-SA 4.0
2024-04-26T02:54:35.353Z
2024-04-26T02:54:35.353Z
ZZZ
<p>I'm a economics undergrad student currently studying the basics of Game Theory. I'm trying to solve the following mixed strategy game:</p>
<p>-Two players, Player 1 and Player 2</p>
<p>-Available actions: Each player can either make the contribution 0, 1, 2 (The costs of each contribution is increasing in the contribution, thus c2&gt;c1&gt;c0=0.</p>
<p>-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.</p>
<p>In summary, the payoff matrix looks like the following:</p>
<p><a href="https://i.sstatic.net/dCnGZ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dCnGZ.png" alt="enter image description here" /></a></p>
<p>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??</p>

<game-theory><nash-equilibrium>
Mixed Strategy Nash Equilibirum, Contributing to a Public Good
EconIsHard
# Mixed Strategy Nash Equilibirum, Contributing to a Public Good

(Asked by: username_0 on 2024-04-16)

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:

[![enter image description here](https://i.sstatic.net/dCnGZ.png)](https://i.sstatic.net/dCnGZ.png)

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??

**Tags:** game-theory, nash-equilibrium

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## 2 answers

### Answer by username_1 on 2024-05-26. Score: 1

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 $i\in\{0,1,2\}$


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### Answer by username_2 on 2024-04-26. Score: 0

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.


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