# Nash Equilibria in a Game with Three Firms and a Shared Resource

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:

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!

• 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! Aug 24, 2023 at 15:46
• sorry, original problem was written in spanish. D is "depuran" that means pirify and C is "contaminar" that stands for "contaminate" Aug 24, 2023 at 15:53

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.
• waow, thanks. Youre answer is very detailed and helpful to understand the problem. Thanks! Aug 25, 2023 at 20:42