# Pure strategy Nash equilibria

I'm given a matrix:

$$\begin{bmatrix} (4,3) & (5,1) & (6,2) \\ (2,1) & (8,4) & (3,6\\ (3,0) & (9,6) & (2,8) \end{bmatrix}$$

I'm supposed to find the pure strategy Nash equilibrium and Pareto efficiency in this game. I know how to approach when I need to find a mixed strategy using probability distributions, but I do not know how to approach this one. Any help?

Suppose the left elements from the pairs are the payoffs for player $1$, and the right the ones for player $2$, then you underline the optimal choice for player $i$ ($i=1,2$), given the choice of the other player. So for your example you would have

\begin{bmatrix} (\underline{4},\underline{3}) & (5,1) & (\underline{6},2) \\ (2,1) & (8,4) & (3,\underline{6})\\ (3,0) & (\underline{9},6) & (2,\underline{8}) \end{bmatrix}

The strategy $(1,1)$ yields a pure strategy Nash equilibrium, since the strategy $1$ is optimal for both players given the choice of the other. Can you see why this works in general for $2$ player games with a finite amount of choices?

• I see. And as far as I can understand, there're no pareto efficient outcomes from this game? – kjanko May 29 '18 at 18:45
• There are several Pareto efficient outcomes for this game... Do you know the definition of Pareto efficiency? – Václav Mordvinov May 29 '18 at 18:48
• Improve the score of one player without ruining the score of the other player. Since there's a (2, 8) outcome, wouldn't this mean that whenever player 1 chooses an outcome that has a <8 gain for the second player, he instantly ruins the second player's outcome? – kjanko May 29 '18 at 18:49
• There are $5$ Pareto efficient outcomes, can you see which ones? You have to look for outcomes for which we cannot increase the score of one player without strictly decreasing the other player's score. Consider en.wikipedia.org/wiki/Pareto_efficiency – Václav Mordvinov May 29 '18 at 18:51
• I think I misunderstand the concept – kjanko May 29 '18 at 18:52