# Nash Equilibrium for the prisoners dilemma when using mixed strategies

Consider the following game matrix $$\begin{array}{l|c|c} & \textbf{S} & \textbf{G} \\ \hline \textbf{S} & (-2,-2) & (-6, -1) \\ \hline \textbf{G} & (-1,-6) & (-4, -4) \end{array}$$ which corresponds to the well-known prisonder's dilemma. Now a Nash Equilibrium by using pure strategies would be (G,G) cause by choosing them neither can improve his outcome by unilaterally changing his strategy.

Now I wanted to calculate a Nash Equilibrium for mixed strategies using this payoff-matrix. (I am using an algorithm proposed at the German wikipedia Article on Nash-Equilibrium). So, I am looking for a mixed strategy for player II which makes player I indifferent regarding his strategy choices, and vice versa.

Let $q$ be the probability that player II chooses $S$ and accordingly $(1-q)$ that he chooses $G$, then the expected values for player I are \begin{align*} EV(I | S) & = (-2)q + (-6)(1-q) \qquad \textrm{if he chooses $S$} \\ EV(I | G) & = (-1)q + (-4)(1-q) \qquad \textrm{if he chooses $G$} \end{align*} (I use the notation $EV(I | S)$ to mean, expected value of player $I$ when he chooses $S$ and so on).

Equating them, to find the probablities for player II which make player I indifferent, I have to solve $$-2q - 6(1-q) = -q - 4(1-q)$$ which has the solution $q = 2$, which is impossible since q should be a probablity, so with the additional restriction $0 \le q \le 1$ it has no solution? But I heard that Nash-Equilibrium for mixed strategies always exists in finite games, so what did I wrong?

• pure strategy is a special case of mixed strategy Commented Mar 8, 2013 at 13:26
• yes, but then I should get $q = 0$ as a special solution, which I didn't... Commented Mar 8, 2013 at 13:28
• No. I do not read german, but I suppose that this is a standard algorithm, which works only for the set of strategies of non-zero probability. That is, there is an initial wrong assumption that $q>0$ Commented Mar 8, 2013 at 13:32
• www.coursera.org has Game Theory course where they explain how it works. Commented Mar 8, 2013 at 13:34

The reason this doesn't work is the same as the one I gave in this answer to this question of yours. Indifference must hold only between options that are assigned non-zero probabilities, and in the present case the strategies in the only Nash equilibrium assign zero probability to $S$, so there's no indifference between $S$ and $G$ at equilibrium.