We’re talking about zero-sum 2-player normal form games. Last time we saw that in a Nash equilibrium for a game like this, both players must use a maximin strategy. Now let’s try to prove the converse!

In other words: let’s try to prove that if both players use a maximin strategy, the result is a Nash equilibrium.

Today we’ll only prove this is true *if* a certain equation holds. It’s the cool-looking equation we saw last time:

Last time we showed this cool-looking equation is true whenever our game has a Nash equilibrium. In fact, this equation is always true. In other words: it’s true for *any* zero-sum two-player normal form game. The reason is that *any* such game has a Nash equilibrium. But we haven’t showed that yet.

So, let’s do what we can easily do.

### Maximin strategies give Nash equilibria… sometimes

Let’s start by remembering some facts we saw in Part 16 and Part 17.

We’re studying a zero-sum 2-player normal form game. Player A’s payoff matrix is , and player B’s payoff matrix is

We saw that a pair of mixed strategies one for player A and one for player B, is a Nash equilibrium if and only if

1) for all

and

2) for all

We saw that is a maximin strategy for player A if and only if:

We also saw that is a maximin strategy for player B if and only if:

With these in hand, we can easily prove our big result for the day. We’ll call it Theorem 4, continuing with the theorem numbers we started last time:

**Theorem 4.** Suppose we have a zero-sum 2-player normal form game for which

holds. If is a maximin strategy for player A and is a maximin strategy for player B, then is a Nash equilibrium.

**Proof.** Suppose that is a maximin strategy for player A and is a maximin strategy for player B. Thus:

and

But since ★ holds, the right sides of these two equations are equal. So, the left sides are equal too:

Now, since a function is always less than or equal to its maximum value, and greater than or equal to its minimum value, we have

But ★★ says the quantity at far left here equals the quantity at far right! So, the quantity in the middle must equal both of them:

By the definition of minimum value, the first equation in ★★★:

says that

for all This is condition 2) in the definition of Nash equilibrium. Similarly, by the definition of maximum value, the second equation in ★★★:

says that

for all This is condition 1) in the definition of Nash equilibrium. So, the pair is a Nash equilibrium. █

[...] Let’s remember what we’ve proved in Part 16 and Part 18: [...]