Does the payoff matrix have 1 or 4 Nash Equilibria?

Question

I'm confused. Does every grid in the following payoff matrix constitute a Nash Equilibrium? ie. Are there 4 Nash Equilibria in this game?

\begin{array}{c|c|c|} & {1} & {2} \\ \hline 1 & (10,10) & (-1,10) \\ 2 & (10,-1) & (-1,-1) \\ \end{array}

Answer 1

Let's begin with the definition of Nash equilibrium. A Nash equilibrium is a set of strategies for each player such that each player's strategy is a best response to the other players' strategies. This includes mixed strategies, where we assign a probability distribution over the options.

Let's say Player 1 (the row player) has a strategy to randomize between picking Option 1 with probability $p_{1 1}$ and Option 2 with probability $1 - p_{1 1}$. Similarly, let's say Player 2 (the column player) has a strategy to pick Option 1 with probability $p_{2 1}$ and Option 2 with probability $1 - p_{2 1}$.

Note that the payoff for Player 1 is $10 p_{2 1} + (-1) (1 - p_{2 1}) = 11 p_{2 1} - 1$. This is notably not a function of $p_{1 1}$, which is to say, all strategies are best responses for Player 1. If you perform the same analysis for Player 2's payoff, which is $11 p_{1 1} - 1 \ne f (p_{2 1})$, you will find that all strategies are best responses for Player 2 as well.

You are correct that all four pure strategy sets $\{(1, 1), (1, 2), (2, 1), (2, 2)\}$ are Nash equilibria for this game. But there are an infinite number of mixed equilibria as well, as we have demonstrated.