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A raider picks which of two flanks to hit. The defender picks one flank to guard. Every point the raider takes is a point the defender loses. There is no shared good outcome to hope for, only the other side’s mistake.

Zero-sum games and minimax

In a strictly competitive game, whatever you win the other player loses. That sounds bleak, but it makes the game unusually clean: there is a single number, the value, that a careful player can guarantee no matter how cleverly the opponent plays. The way to guarantee it is to mix.

From earlier lessons

Two quick ones before we start.

In Rock-Paper-Scissors the equilibrium mix is...

Adding the same constant to every payoff in a game...

Predict, then play

You are the raider. The strong flank is worth more if you get through, but the defender knows that too. The slider sets how often you hit the strong flank. Each line shows what you can guarantee if the defender guards one flank or the other. A guaranteed payoff is the worse of the two, so you want to lift that floor.

Why did that happen?

Take a guess, then reveal the reason

If you commit to one flank, the defender guards it and you lose. If you mix, the defender cannot cover both, and there is one mix where guarding either flank leaves you the same expected gain. At that mix the defender has no good reply, so the payoff you get there is one you can guarantee whatever they do. That guaranteed number is the value of the game.

The math

Definition, zero-sum and constant-sum games

A two-player game is zero-sum if the two payoffs add to zero in every outcome. It is constant-sum if they add to the same constant in every outcome. Subtracting that constant from one player’s payoff converts a constant-sum game into a strategically equivalent zero-sum game, because the shift does not change that player’s ranking of any strategies.

Here are the raid’s four payoffs, written to the raider. The defender earns the negative of each, so every cell sums to zero.

Exhibit 2 Payoff to the raider

You raid \ Defender guards	Strong	Weak
Strong flank	-1	+2
Weak flank	+1	-1

Read it as a raider. Hit a flank they are guarding and you lose a point. Hit an open one and you gain: $+1$ through the weak flank, $+2$ through the strong one, because there is more behind it.

Now mix. Play the strong flank with probability $p$ and the weak flank with probability $1-p$ . The defender has to commit to one flank, so fix their choice and average down that column of the table:

\underbrace{p(-1) + (1-p)(+1)}_{\text{they guard strong}} = 1 - 2p, \qquad \underbrace{p(+2) + (1-p)(-1)}_{\text{they guard weak}} = 3p - 1.

That is where the two lines come from: each is your own pair of payoffs in that column, weighted by how often you go each way. Sanity-check the ends. At $p=0$ you always take the weak flank, so you score $+1$ when they guard strong and $-1$ when they guard weak, which is what the two expressions give. At $p=1$ they read $-1$ and $+2$ , the strong-flank row.

The defender will pick whichever is worse for you, so your floor is $\min(1-2p,\,3p-1)$ . That minimum is highest where the two lines meet:

3p - 1 = 1 - 2p \quad\Rightarrow\quad p = \tfrac{2}{5}, \qquad \text{value} = 3\cdot\tfrac{2}{5} - 1 = \tfrac{1}{5}.

Definition, von Neumann's minimax theorem

In any finite two-player zero-sum game, the most the row player can guarantee (the maximin) equals the least the column player can hold them to (the minimax). That common number is the value of the game, and the mixes that achieve it form a Nash equilibrium. So here minimax is not pessimism: it is the equilibrium.

The defender solves the mirror problem and mixes $\tfrac35$ on guarding the strong flank, holding you to $\tfrac15$ from above. Maximin from below and minimax from above meet at the value.

Work it out

Find the value and the optimal mix

Write the raider’s guarantee against each pure defense as a line in $p$ : guarding the strong flank yields $-p + (1-p) = 1 - 2p$ ; guarding the weak flank yields $2p + (-1)(1-p) = 3p - 1$ . One line rises in $p$ , the other falls, so their crossing is the highest floor.

Set $3p - 1 = 1 - 2p$ to get $p = \tfrac25$ , and substitute back for a value of $\tfrac15$ . Because the game is zero-sum, the defender’s optimal mix guarantees the same value from the other side, so $\tfrac15$ is the value of the game, not just a floor.

A note from the real world

Reality check

Minimax mixing sounds like a blackboard ideal, too demanding for anyone under pressure. Yet it shows up in the field. Studying thousands of penalty kicks, Palacios-Huerta (2003) found professional kickers and keepers mix left and right at rates almost exactly matching the equilibrium, and in no readable order. The Behavioral unit returns to where this holds and where it breaks.

Key idea

In a zero-sum game there is one number you can guarantee, the value, and the mix that guarantees it is the equilibrium. Minimax play is not cautious here; it is optimal.

Check yourself

Check yourselfrecognise

What does it mean for a game to be zero-sum?

Check yourselfapply

In this raid game, what does mixing 2/5 on the strong flank get you?

Check yourselfbrings back: mixed strategies

Why is minimax play in a zero-sum game not merely defensive pessimism?

Sources

von Neumann, J. (1928). Zur Theorie der Gesellschaftsspiele. Mathematische Annalen, 100, 295-320.
Osborne, M. J. (2004). An Introduction to Game Theory. Oxford University Press.
Palacios-Huerta, I. (2003). Professionals play minimax. Review of Economic Studies, 70(2), 395-415.