Can existence of pure Nash strategies simplify discovery of mixed ones?

Let's say I'm interested in two player games.

I'm aware of the complexities of finding mixed Nash equilibria. To really be sure we've found all we have to enumerate all possible combinations of row and column strategies and check for indifference (ignoring other simple tricks).

Although, let's assume some pure Nash equilibria exist. Does this say anything about mixed ones?

If pure strategies do exist can my search for mixed strategies include only those that contain at least one of the strategies from the pure Nash equilibrium in their support?

The answer is no, that is, if you have a $2$-player game with NE in pure strategies, not necessarily you can rely on the fact that the actions that are in the pure NE are also in the support of the mixed NE.

Here there is an example,

• where the actions of the Row player are Top, Middle, Down,
• and the actions of the Column player are Left, Center, Right.

$$\begin{bmatrix} 1,1 & 0,0 & 0,0 \\ 0,0 & 1,-1 & -1,1 \\ 0,0 & -1,1 & 1,-1 \\ \end{bmatrix}$$

[Notice that this is really like matching pennies, with the addition of the first row/column.]

There is one pure strategy NE, given by $(T, L)$, and a mixed strategy NE where players equally randomize between Middle/Down and Center/Right, much in the same spirit of matching pennies, and there is neither $T$ nor $L$ in the support of this mixed strategy NE equilibrium.

If pure Nash equilibria exists, it can be the case that there is no strictly mixed one: for example, consider a simple game where two players call an integer in range $0,\cdots,n$ independently, with the result $a-b$, where the first player called $a$ and tries to maximize the result and the second called $b$ and tries to minimize. It is obvious that for the first the most beneficial option is to call $n$, and for the second - $0$, independently of the choice of other. Thus this pure strategy pair is the only Nash equilibria.

We can also consider a vacuous game, where the players have many options, but the result score is always the same for both. Then we have all possible pure and mixed strategy pairs as Nash equilibria, which have no connections with each other at all.

• You write "If pure Nash equilibria exists, there can be no strictly mixed one". What about the Battle of the Sexes? Two pure NE and one in mixed strategies. – Kolmin Feb 19 '17 at 12:53
• Yes, that part needs clearing up. I have edited my last paragraph to make my question a bit clearer. – David Mitchell Feb 19 '17 at 13:26
• @Kolmin yes, that was a bad formulation, I edited. – Wolfram Feb 19 '17 at 14:47
• @Wolfram: I am not sure I understand your first example. That's how I read it, with set of players $N:= \{ 1,2 \}$. Thus, you write that $1$ wants to maximize, while $2$ wants to minimize (I guess you refer to the difference $a-b$, where $a$ is $1$'s choice, and $b$ is $2$'s choice). But then $(n , 0)$ is not a NE, because $2$ has an incentive to change his action to $n$. Notice also that $(n, n)$ is again not a NE, because this time is $1$ that has an incentive to change to $0$. As far as I can see, there is no pure NE in this game. – Kolmin Feb 19 '17 at 15:47