# Min and Max number of pure strategy Nash equilibria

Suppose we have a normal form game with two players and three strategies for each.

How can we find the minimum and maximum number of the pure strategy Nash equilibria for this particular game?

My thought was to write, given that $N=2$, the two strategy profiles like:

$$S_1=\{a_1,a_2,a_3\},S_2=\{b_1,b_2,b_3\}.$$

Next I thought to use the condition in the definition of Nash equilibrium, that is $u_i(s^*_i,s^*_{-i})\geq u_i(s^*_i,s^*_{-i})$ for all $s_i$ and for all $i$.

Now I am not sure how to formally derive the number of min and max. I think intuitively we could have no Nash equilibrium. Is that right?

As for the max, I am stuck. I would appreciate any help. Thank you.

• May you clarify your question: it's not clear what you mean by min/max number? If you have a specific $3 \times 3$ game, one can find the equilibria in pure strategy and count them. – mlc Sep 19 '18 at 18:50
• If you consider the class of all $3 \times 3$ games, there are games with zero equilibria in pure strategies and games with nine equilibria in pure strategies. So the min and the max for the class are respectively zero and nine. – mlc Sep 19 '18 at 18:51

First, let $N=2$ and $S_1=\{a_1,a_2\},S_2=\{b_1,b_2\}$. Can you construct a utility function $u_i$ on $S=S_1\times S_2$ s.t. the resulting game has no NE? Draw a 2x2 table, plug in some numbers, and confirm by definition that there are no NEs.
For the maximum, consider any constant utility function on the set of vectors of strategies $S=S_1\times\dotsb\times S_n$ of an $n$-player game. Confirm that every $s\in S$ is a NE, so the result is that any game can have a maximum of $\lvert S\rvert$ NE. In your case of a 3x3 game, the maximum is 9.