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.
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!
