<p>The payoff matrix in your question is not quite right (though the explanation of the payoffs in the paragraph above it is correct).</p>
<p>The payoff matrix should look like this:</p>
<p><span class="math-container">$$
    \begin{array}{ccccc}
   \qquad\qquad  &amp; \qquad\text{Player $2$}&amp;   \qquad\qquad &amp; \qquad\qquad &amp; \text{Player $2$} \\
    \end{array}\\
    \begin{array}{rr|c|c|crr|c|c|}
        &amp;            &amp; D  &amp; C &amp; &amp; &amp; &amp; D &amp; C    \\ 
\text{Player $1$}
        &amp; D             &amp;  -1,-1,-1 &amp; -1,0,-1  &amp; &amp; &amp;&amp; -1,-1,0 &amp; -4,-3,-3  \\ 
               &amp; C            &amp; 0,-1,-1 &amp;  -3,-3,-4&amp;&amp;&amp; &amp; -3,-4,-3 &amp;  -3,-3,-3 \\ 
    \end{array}\\
    \begin{array}{ccccc}
   \qquad\qquad  &amp; D&amp;   \qquad\qquad &amp; \qquad\qquad\qquad &amp; C \\
    \end{array}\\
    \begin{array}{cc}
   &amp;\qquad\qquad \qquad\text{Player $3$} \\
    \end{array}
$$</span></p>
<p><span class="math-container">$(D,D,D)$</span> cannot be a NE because any one player would be better off deviating to <span class="math-container">$C$</span> and saving the purification cost.</p>
<hr />
<p>To find the pure strategy NE, I have underlined the (payoffs from the) best responses:</p>
<p><span class="math-container">$$
    \begin{array}{ccccc}
   \qquad\qquad  &amp; \qquad\text{Player $2$}&amp;   \qquad\qquad &amp; \qquad\qquad &amp; \text{Player $2$} \\
    \end{array}\\
    \begin{array}{rr|c|c|crr|c|c|}
        &amp;            &amp; D  &amp; C &amp; &amp; &amp; &amp; D &amp; C    \\ 
\text{Player $1$}
        &amp; D             &amp;  -1,-1,-1 &amp; {\bf\underline{-1}},{\bf\underline{0}},{\bf\underline{-1}}  &amp; &amp; &amp;&amp; {\bf\underline{-1}},{\bf\underline{-1}},{\bf\underline{0}} &amp; -4,-3,-3  \\ 
               &amp; C            &amp; {\bf\underline{0}},{\bf\underline{-1}},{\bf\underline{-1}} &amp;  -3,-3,-4&amp;&amp;&amp; &amp; -3,-4,-3 &amp;  {\bf\underline{-3}},{\bf\underline{-3}},{\bf\underline{-3}} \\ 
    \end{array}\\
    \begin{array}{ccccc}
   \qquad\qquad  &amp; D&amp;   \qquad\qquad &amp; \qquad\qquad\qquad &amp; C \\
    \end{array}\\
    \begin{array}{cc}
   &amp;\qquad\qquad \qquad\text{Player $3$} \\
    \end{array}
$$</span></p>
<p>As you can see, there are four pure strategy Nash equilibria:</p>
<ul>
<li>One where all do not purify: <span class="math-container">$(C,C,C)$</span></li>
<li>Three where two purify (<span class="math-container">$D$</span>) and the other does not (<span class="math-container">$C$</span>)</li>
</ul>
<hr />
<p>Now consider mixed strategies. Since each player has a unique best response to any given actions of the two other players, there cannot be a mixed strategy NE where one player mixes and the other two do not.</p>
<p>Lets look for a mixed strategy NE where player <span class="math-container">$3$</span> mixes. Let <span class="math-container">$p_i$</span> denote the probability each player places on action <span class="math-container">$D$</span>. For <span class="math-container">$p_3\in(0,1)$</span>, player <span class="math-container">$3$</span> must be indifferent between <span class="math-container">$D$</span> and <span class="math-container">$C$</span>. The expected payoff from <span class="math-container">$D$</span> is:</p>
<p><span class="math-container">$$\begin{align*}
&amp;p_1p_2(-1)+p_1(1-p_2)(-1)+(1-p_1)p_2(-1)+(1-p_1)(1-p_2)(-4)\\
&amp;=[1-(1-p_1)(1-p_2)](-1)+(1-p_1)(1-p_2)(-4)\\
&amp;=-1+(1-p_1)(1-p_2)(-3)\end{align*}$$</span></p>
<p>The expected payoff from <span class="math-container">$C$</span> is</p>
<p><span class="math-container">$$\begin{align*}
&amp;p_1p_2(0)+p_1(1-p_2)(-3)+(1-p_1)p_2(-3)+(1-p_1)(1-p_2)(-3)\\
&amp;=(1-p_1p_2)(-3)\end{align*}$$</span></p>
<p>The two expected payoffs are the same if</p>
<p><span class="math-container">$$-1+(1-p_1)(1-p_2)(-3)=(1-p_1p_2)(-3)$$</span></p>
<p>which is equivalent to</p>
<p><span class="math-container">$$p_1+p_2-2p_1p_2=\frac{1}{3}\tag{IC3}$$</span></p>
<p>There are two cases to consider:</p>
<ul>
<li>Player <span class="math-container">$3$</span> and exactly one other player mix</li>
<li>All players mix</li>
</ul>
<hr />
<p>Can we have a NE where player <span class="math-container">$3$</span> and only one other player mix? Suppose player <span class="math-container">$1$</span> plays a pure strategy and player <span class="math-container">$2$</span> mixes. For player <span class="math-container">$2$</span> to mix it must be indifferent between <span class="math-container">$D$</span> and <span class="math-container">$C$</span> and so, given the symmetry of the game, must satisfy a similar condition to <span class="math-container">$(IC3)$</span>:</p>
<p><span class="math-container">$$p_1+p_3-2p_1p_3=\frac{1}{3}\tag{IC2}$$</span></p>
<ul>
<li>If <span class="math-container">$p_1=1$</span> (player <span class="math-container">$1$</span> plays <span class="math-container">$D$</span>), then <span class="math-container">$(IC2)$</span> and <span class="math-container">$(IC3)$</span> give <span class="math-container">$p_2=p_3=2/3$</span>. Then <span class="math-container">$1$</span>'s expected payoff from <span class="math-container">$D$</span> is <span class="math-container">$$(2/3)(2/3)(-1)+(2/3)(1/3)(-1)+(1/3)(2/3)(-1)+(1/3)(1/3)(-3)=(1-1/9)(-1)+(1/9)(-3)=-\frac{11}{9}$$</span> whereas the expected payoff from <span class="math-container">$C$</span> is <span class="math-container">$$(2/3)(2/3)(0)+(2/3)(1/3)(-3)+(1/3)(2/3)(-3)+(1/3)(1/3)(-3)=(1-4/9)(-3)=-\frac{15}{9}.$$</span> Since <span class="math-container">$1$</span>'s expected payoff from <span class="math-container">$D$</span> is higher than from <span class="math-container">$C$</span>, the mixed strategy profile <span class="math-container">$(p_1,p_2,p_3)=(1,2/3,2/3)$</span> constitutes a NE (as do permutations).</li>
<li>If <span class="math-container">$p_1=0$</span> (player <span class="math-container">$1$</span> plays <span class="math-container">$C$</span>), then <span class="math-container">$(IC2)$</span> and <span class="math-container">$(IC3)$</span> give <span class="math-container">$p_2=p_3=1/3$</span>. Then <span class="math-container">$1$</span>'s expected payoff from <span class="math-container">$D$</span> is <span class="math-container">$$(1/3)(1/3)(-1)+(1/3)(2/3)(-1)+(2/3)(1/3)(-1)+(2/3)(2/3)(-3)=(1-4/9)(-1)+(4/9)(-3)=-\frac{17}{9}$$</span> whereas the expected payoff from <span class="math-container">$C$</span> is <span class="math-container">$$(1/3)(1/3)(0)+(1/3)(2/3)(-3)+(2/3)(1/3)(-3)+(2/3)(2/3)(-3)=(1-1/9)(-3)=-\frac{24}{9}.$$</span> Since  <span class="math-container">$1$</span>'s expected payoff from <span class="math-container">$C$</span> is lower than from <span class="math-container">$D$</span>, there is no mixed strategy NE where player <span class="math-container">$1$</span> plays <span class="math-container">$C$</span> and the other two players mix.</li>
</ul>
<hr />
<p>Can we have a mixed strategy NE where all players mix? If so, then <span class="math-container">$(p_1,p_2,p_3)$</span> must satisfy the three equations <span class="math-container">$(IC2)$</span> and <span class="math-container">$(IC3)$</span> and <span class="math-container">$(IC1)$</span>:</p>
<p><span class="math-container">$$p_2+p_3-2p_2p_3=\frac{1}{3}\tag{IC1}.$$</span></p>
<p>Solving the system of three equations gives two solutions:</p>
<p><span class="math-container">$$p_1=p_2=p_3=\frac{(3-\sqrt(3)}{6}\approx 0.211 \qquad\text{and}\qquad  p_1=p_2=p_3=\frac{(3+\sqrt(3)}{6}\approx 0.789$$</span></p>
<hr />
<p>In summary, there are nine NE:</p>
<ul>
<li>One where all do not purify: <span class="math-container">$(C,C,C)$</span></li>
<li>Three where two purify (<span class="math-container">$D$</span>) and the other does not (<span class="math-container">$C$</span>)</li>
<li>Three where one player plays <span class="math-container">$D$</span> and the others place probability <span class="math-container">$2/3$</span> on <span class="math-container">$D$</span> and <span class="math-container">$1/3$</span> on <span class="math-container">$C$</span>.</li>
<li>Two where all players place positive probability on both actions.</li>
</ul>

CC BY-SA 4.0
2023-08-25T20:42:03.990Z
waow, thanks. Youre answer is very detailed and helpful to understand the problem. Thanks!
lasagna
CC BY-SA 4.0
2023-08-25T10:03:19.003Z
2023-08-25T10:33:54.900Z
2023-08-25T10:33:54.900Z
smcc
<p>I’m trying to understand the Nash equilibria in a game involving three firms that use water from a shared lake. Each firm can choose to purify the water before returning it to the lake or not purify it (and therefore pollute the lake). The cost of purifying the used water before returning it to the lake is 1. If two or more firms do not purify the water before returning it to the lake, all three firms incur a cost of 3 to treat the water before being able to use it. In summary:</p>
<p>If all three firms purify: -1 for each firm.
If two firms purify and one pollutes: -1 for each firm that purifies, 0 for the polluting firm.
If one firm purifies and two pollute: -4 for the firm that purifies, -3 for each polluting firm.
If all three firms pollute: -3 for each firm.
I represented this game in strategic form using two payoff matrices, one for each decision of the third firm. Each matrix represents the decisions of the first two firms, while the third firm is represented as a separate matrix. The available strategies for each firm are to purify (D) or not purify © the water before returning it to the lake.</p>
<p>If the third firm decides to purify the water, the payoff matrix would be:</p>
<p><a href="https://i.sstatic.net/KjUst.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/KjUst.png" alt="enter image description here" /></a>
I obtained two pure Nash equilibria: (C, C, C) and (D, D, D). However, I’m not sure if my reasoning is correct. Could someone help me understand if these are indeed the Nash equilibria of this game and if there are any mixed equilibria? Thank you!</p>

CC BY-SA 4.0
2023-08-24T15:46:59.620Z
D stands for defect and means not to purify in this context. C stands for cooperate, which means purify here. And your payoff matrices don't match your description: Firm 3's payoffs don't depend on the other firms' decisions in your matrices. Check again!
VARulle
CC BY-SA 4.0
2023-08-24T15:53:48.107Z
sorry, original problem was written in spanish. D is "depuran" that means pirify and C is "contaminar" that stands for "contaminate"
lasagna
<nash-equilibrium>
Nash Equilibria in a Game with Three Firms and a Shared Resource
lasagna
# Nash Equilibria in a Game with Three Firms and a Shared Resource

(Asked by: username_0 on 2023-08-24)

I’m trying to understand the Nash equilibria in a game involving three firms that use water from a shared lake. Each firm can choose to purify the water before returning it to the lake or not purify it (and therefore pollute the lake). The cost of purifying the used water before returning it to the lake is 1. If two or more firms do not purify the water before returning it to the lake, all three firms incur a cost of 3 to treat the water before being able to use it. In summary:

If all three firms purify: -1 for each firm.
If two firms purify and one pollutes: -1 for each firm that purifies, 0 for the polluting firm.
If one firm purifies and two pollute: -4 for the firm that purifies, -3 for each polluting firm.
If all three firms pollute: -3 for each firm.
I represented this game in strategic form using two payoff matrices, one for each decision of the third firm. Each matrix represents the decisions of the first two firms, while the third firm is represented as a separate matrix. The available strategies for each firm are to purify (D) or not purify © the water before returning it to the lake.

If the third firm decides to purify the water, the payoff matrix would be:

[![enter image description here](https://i.sstatic.net/KjUst.png)](https://i.sstatic.net/KjUst.png)
I obtained two pure Nash equilibria: (C, C, C) and (D, D, D). However, I’m not sure if my reasoning is correct. Could someone help me understand if these are indeed the Nash equilibria of this game and if there are any mixed equilibria? Thank you!

**Tags:** nash-equilibrium
> **Comments:**
> **username_1**: D stands for defect and means not to purify in this context. C stands for cooperate, which means purify here. And your payoff matrices don't match your description: Firm 3's payoffs don't depend on the other firms' decisions in your matrices. Check again!  
> **username_0**: sorry, original problem was written in spanish. D is "depuran" that means pirify and C is "contaminar" that stands for "contaminate"  

---

## 1 answers

### Answer by username_2 on 2023-08-25. Score: 0 | Accepted answer

The payoff matrix in your question is not quite right (though the explanation of the payoffs in the paragraph above it is correct).

The payoff matrix should look like this:

$$
\begin{array}{ccccc}
\qquad\qquad & \qquad\text{Player $2$}& \qquad\qquad & \qquad\qquad & \text{Player $2$} \\
\end{array}\\
\begin{array}{rr|c|c|crr|c|c|}
& & D & C & & & & D & C \\
\text{Player $1$}
& D & -1,-1,-1 & -1,0,-1 & & && -1,-1,0 & -4,-3,-3 \\
& C & 0,-1,-1 & -3,-3,-4&&& & -3,-4,-3 & -3,-3,-3 \\
\end{array}\\
\begin{array}{ccccc}
\qquad\qquad & D& \qquad\qquad & \qquad\qquad\qquad & C \\
\end{array}\\
\begin{array}{cc}
&\qquad\qquad \qquad\text{Player $3$} \\
\end{array}
$$

$(D,D,D)$ cannot be a NE because any one player would be better off deviating to $C$ and saving the purification cost.

---

To find the pure strategy NE, I have underlined the (payoffs from the) best responses:

$$
\begin{array}{ccccc}
\qquad\qquad & \qquad\text{Player $2$}& \qquad\qquad & \qquad\qquad & \text{Player $2$} \\
\end{array}\\
\begin{array}{rr|c|c|crr|c|c|}
& & D & C & & & & D & C \\
\text{Player $1$}
& D & -1,-1,-1 & {\bf\underline{-1}},{\bf\underline{0}},{\bf\underline{-1}} & & && {\bf\underline{-1}},{\bf\underline{-1}},{\bf\underline{0}} & -4,-3,-3 \\
& C & {\bf\underline{0}},{\bf\underline{-1}},{\bf\underline{-1}} & -3,-3,-4&&& & -3,-4,-3 & {\bf\underline{-3}},{\bf\underline{-3}},{\bf\underline{-3}} \\
\end{array}\\
\begin{array}{ccccc}
\qquad\qquad & D& \qquad\qquad & \qquad\qquad\qquad & C \\
\end{array}\\
\begin{array}{cc}
&\qquad\qquad \qquad\text{Player $3$} \\
\end{array}
$$

As you can see, there are four pure strategy Nash equilibria:

* One where all do not purify: $(C,C,C)$
* Three where two purify ($D$) and the other does not ($C$)

---

Now consider mixed strategies. Since each player has a unique best response to any given actions of the two other players, there cannot be a mixed strategy NE where one player mixes and the other two do not.

Lets look for a mixed strategy NE where player $3$ mixes. Let $p\_i$ denote the probability each player places on action $D$. For $p\_3\in(0,1)$, player $3$ must be indifferent between $D$ and $C$. The expected payoff from $D$ is:

$$\begin{align\*}
&p\_1p\_2(-1)+p\_1(1-p\_2)(-1)+(1-p\_1)p\_2(-1)+(1-p\_1)(1-p\_2)(-4)\\
&=[1-(1-p\_1)(1-p\_2)](-1)+(1-p\_1)(1-p\_2)(-4)\\
&=-1+(1-p\_1)(1-p\_2)(-3)\end{align\*}$$

The expected payoff from $C$ is

$$\begin{align\*}
&p\_1p\_2(0)+p\_1(1-p\_2)(-3)+(1-p\_1)p\_2(-3)+(1-p\_1)(1-p\_2)(-3)\\
&=(1-p\_1p\_2)(-3)\end{align\*}$$

The two expected payoffs are the same if

$$-1+(1-p\_1)(1-p\_2)(-3)=(1-p\_1p\_2)(-3)$$

which is equivalent to

$$p\_1+p\_2-2p\_1p\_2=\frac{1}{3}\tag{IC3}$$

There are two cases to consider:

* Player $3$ and exactly one other player mix
* All players mix

---

Can we have a NE where player $3$ and only one other player mix? Suppose player $1$ plays a pure strategy and player $2$ mixes. For player $2$ to mix it must be indifferent between $D$ and $C$ and so, given the symmetry of the game, must satisfy a similar condition to $(IC3)$:

$$p\_1+p\_3-2p\_1p\_3=\frac{1}{3}\tag{IC2}$$

* If $p\_1=1$ (player $1$ plays $D$), then $(IC2)$ and $(IC3)$ give $p\_2=p\_3=2/3$. Then $1$'s expected payoff from $D$ is $$(2/3)(2/3)(-1)+(2/3)(1/3)(-1)+(1/3)(2/3)(-1)+(1/3)(1/3)(-3)=(1-1/9)(-1)+(1/9)(-3)=-\frac{11}{9}$$ whereas the expected payoff from $C$ is $$(2/3)(2/3)(0)+(2/3)(1/3)(-3)+(1/3)(2/3)(-3)+(1/3)(1/3)(-3)=(1-4/9)(-3)=-\frac{15}{9}.$$ Since $1$'s expected payoff from $D$ is higher than from $C$, the mixed strategy profile $(p\_1,p\_2,p\_3)=(1,2/3,2/3)$ constitutes a NE (as do permutations).
* If $p\_1=0$ (player $1$ plays $C$), then $(IC2)$ and $(IC3)$ give $p\_2=p\_3=1/3$. Then $1$'s expected payoff from $D$ is $$(1/3)(1/3)(-1)+(1/3)(2/3)(-1)+(2/3)(1/3)(-1)+(2/3)(2/3)(-3)=(1-4/9)(-1)+(4/9)(-3)=-\frac{17}{9}$$ whereas the expected payoff from $C$ is $$(1/3)(1/3)(0)+(1/3)(2/3)(-3)+(2/3)(1/3)(-3)+(2/3)(2/3)(-3)=(1-1/9)(-3)=-\frac{24}{9}.$$ Since $1$'s expected payoff from $C$ is lower than from $D$, there is no mixed strategy NE where player $1$ plays $C$ and the other two players mix.

---

Can we have a mixed strategy NE where all players mix? If so, then $(p\_1,p\_2,p\_3)$ must satisfy the three equations $(IC2)$ and $(IC3)$ and $(IC1)$:

$$p\_2+p\_3-2p\_2p\_3=\frac{1}{3}\tag{IC1}.$$

Solving the system of three equations gives two solutions:

$$p\_1=p\_2=p\_3=\frac{(3-\sqrt(3)}{6}\approx 0.211 \qquad\text{and}\qquad p\_1=p\_2=p\_3=\frac{(3+\sqrt(3)}{6}\approx 0.789$$

---

In summary, there are nine NE:

* One where all do not purify: $(C,C,C)$
* Three where two purify ($D$) and the other does not ($C$)
* Three where one player plays $D$ and the others place probability $2/3$ on $D$ and $1/3$ on $C$.
* Two where all players place positive probability on both actions.

> **Comments:**
> **username_0**: waow, thanks. Youre answer is very detailed and helpful to understand the problem. Thanks!  

---

