# Infinite number of nash equilibria in 3x3 game

In the simple game below, there are two strategies, one for each player, that are weakly dominated (another strategy has greater than or equal payoff). If we remove these strategies then I can find a single mixed Nash equilibrium. But if we leave these strategies in, I think there are then an infinite number of nash equilibria. This seems odd to me, so I wanted to check for views.

The game is represented in the following matrix

$$\begin{pmatrix} & A & B & C \newline A & (3,1) & (3,2) & (3,0) \newline B & (4,1) & (0,0) & (4,0) \newline C & (0,1) & (0,2) & (0,0) \end{pmatrix}$$

We can see that strategy $$C$$ is weakly dominated for both the row and the column player.

If we remove it, we get the following game:

$$\begin{pmatrix} & A & B \newline A & (3,1) & (3,2) \newline B & (4,1) & (0,0) \end{pmatrix}$$

There are clearly two pure Nash equilibria for this game: $$(A,B)$$ and $$(B,A)$$. We can also see that there is a mixed Nash equilibrium too.

Lets denote the probability that the row player plays $$A$$ be $$p_1$$ and the probability that the row player plays $$B$$ be $$p_2$$, and assume that $$p_1, p_2 \geq 0$$ and $$p_1+p_2 = 1$$. At the same time, let's also assume that the probability that the column player plays $$A$$ be $$q_1$$ and the probability that the column player plays $$B$$ be $$q_2$$. We'll make the parallel assumptions that $$q_1,q_2 \geq 0$$ and $$q_1 +q_2=1$$.

We know that at the nash equilibria, the row player needs to make the column player indifferent between their choice of $$A$$ or $$B$$, so we must have:

$$p_1+ p_2 = 2p_1$$

$$\therefore p_1+(1-p_1)=2p_1$$

$$\therefore p_1 =0.5$$ and $$p_2 = 0.5$$

Similarly for the column player to make the row player indifferent we know we must have:

$$3q_1 + 3q_2 = 4q_1$$

$$\therefore q_1 = 0.75$$ and $$q_2 = 0.25$$

So for the smaller game, where we have removed the weakly dominated strategies, we have two pure Nash equilibria and one mixed.

But if we return to the full 3x3 version of the game, and carry the convention on with $$p_3$$ being the probability that the row player plays $$C$$ and $$q_1$$ the probability the column player plays $$C$$, I get an odd result.

If we first solve for the mixed Nash equilibria for the column player as follows:

$$3q_1 + 3q_2 +3 q_3 = 4 q_1 +4 q_3$$

$$\therefore 3q_2 = q_1 + q_3$$, noting that $$q_2 = 1-q_1-q_3$$ we have:

$$3(1-q_1-q_3) = q_1 + q_3$$

$$\therefore 0.75 = q_1 + q_3$$

As such, it appears to me that we have an infinite number of solutions where $$q_3 = 0.25$$ and our choice of $$q_1$$ and $$q_3$$ lie anywhere on the line $$q_1 = 0.75 - q_3$$.

A similar approach to the row player gives an infinite number of Nash equilibria at $$p_2 =0.5$$ and any $$p_1, p_2$$ that fall on the line $$p_1 = 0.5 - p_2$$.

This all feels fairly straightforward. But the conclusion seems really counterintuitive: in the 3x3 game the mixed nash equilibirium involves both players playing strategies that are are weakly dominated some of the time.

Have I made a mistake anywhere here?

Any help greatly appreciated.

The $$C$$ strategies are in fact strictly dominated by the $$A$$ strategies for both players. They can never appear in any Nash equilibrium.
• Thanks for this. Two quick follow-ups: 1) I'm not following on the strictly dominated point. My reasoning is that if the row player plays strategy $B$, then strategies $B$ and $C$ give the same outcome for the column player, so $C$ is not strictly dominated? Commented Apr 11 at 6:02
• Second follow-up 2) if strategy $C$ can never feature in a Nash equilibrium, do you know what I've done wrong to derive non-zero probabilities associated with it in the mixed Nash? Commented Apr 11 at 6:06