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# ZERO-SUM GAMES

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1. Introduction
So far we have analyzed normal form games in a broad sense. Nonetheless,
soetime we may be interested in analyzing specific classes of games where some
condition is satisfied. For instance, two-player games have properties that do not
necessarily hold when we increase the number of players.
Here we analyze two-player zero-sum games. A zero-sum game is a game where,
for each pure strategy profile, the sum of the payoffs of all the players is equal to 0.
While zero-sum games have importance of their own, this lecture is mainly useful
to introduce some important concepts and terminology.
2. Zero-sum game
Let us start giving a formal definition.
Definition 1 (zero-sum Games). A game G = {N, {Si }iN , {ui }iN } is a zero-sum
game if for every s = (s1 , . . . , sn ) S
u1 (s) + u2 (s) + + un (s) = 0.
Notice that the definition also implies that for every mixed strategy =
(1 , . . . , n ) we have that U1 () + U2 () + + Un () = 0.
Most of the study of zero-sum games has been devoted to two-player zero-sum
games.
Definition 2 (Two-player zero-sum game). A two-player zero-sum game G =
{S1 , S2 , u1 , u2 } is a two-player game where for every (s1 , s2 ) S1 S2
u1 (s1 , s2 ) = u2 (s1 , s2 ).
The previous definition also implies that for every mixed strategy profile we
have that U1 () = U2 ().
Example 1 (The Matching Pennies Game). The typical example of a zero-sum
game is the matching game.
heads
tails

heads
1, 1
1, 1

tails
1, 1
1, 1

For every pure strategy profile s = (s1 , s2 ), it holds that u1 (s1 , s2 ) = u2 (s1 , s2 ).
Example 2 (Rock, Paper, Scissors). The well known rock, raper, scissors is another illustrative example of zero-sum game.
1

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Rock
Paper
Scissors

Rock
0, 0
1, 1
1, 1

Paper
1, 1
0, 0
1, 1

Scissors
1, 1
1, 1
0, 0

## Again, every pure strategy profile s = (s1 , s2 ) satisfies u1 (s1 , s2 ) = u2 (s1 , s2 ).

The theory of two-person zero-sum games was developed by von Neumann and
Morgenstern in their book Theory of Games and Economic Behavior (1944) and
constitutes the foundation of the work of Nash. John Nash constructed the general
framework that was covered in the past lectures.

3. Minmax Strategies.
The specific formulation of two-player zero-sum games facilitates their study.
In the next section, we will see that players can guarantee themselves their Nash
equilibrium payoff by choosing their minmax strategy. Loosely speaking, a players
minmax strategy is a strategy that minimizes the maximum damage that the opponents could inflict him. Formally,
Definition 3 (Minmax Strategy). Given a game G, i is a minmax strategy of
player i if


0
i arg max
min
U
(
,
s
)
.
i i i
0
i

si

It is always easier to understand the idea behind the concept if we only consider
pure strategies.
Example 3. Let us compute the minmax (pure) strategies of the following game.
T
B

L
3, 5
2, 3

R
1, 2
4, 4

If player 1 plays T the worst thing that could happen to him is that player 2
plays R which would give him a payoff equal to 1. If player 1 plays T the worst
thing that could happen to him is that player 2 plays L which would give him a
payoff equal to 2. Player 1s minmax (pure) strategy is, therefore, B. You can
check that player 2s minmax (pure) strategy is L.
However, we know that we cannot prevent players from randomizing. If player 1
played 21 T + 12 R the worst thing (in fact, the only thing) that could happen to him
is that he obtains an expected payoff equal to 25 which is larger than 2. Therefore,
player 2s minmax strategy is 12 T + 12 R. An analogous argument can be made for
player 2 that shows that his minmax strategy is 12 L + 12 R.
We now present the Minmax Theorem. It receives this name because the strategies that the theorem pins down are minmax strategies.

ZERO-SUM GAMES

4. Minmax Theorem
Theorem 1 (Minmax Theorem). Every two-player zero-sum game has a unique
value V and optimal strategies for both players 1 and 2 such that
U1 (1 , 2 ) V for every 2 2
U2 (1 , 2 ) V for every 1 1 .
The reason why the strategies are called optimal is that if player 1 (respectively,
player 2) plays 1 (resp. 2 ) he guarantees himself an expected payoff equal to V
(resp. equal to V ) independently of what player 2 (resp. player 1) plays. Let us
proceed to prove the Theorem.
Proof. Consider an arbitrary two-player zero-sum game G. We know that every
game has a Nash equilibrium. Let
= (
1 ,
2 ) be one Nash equilibrium of G.
Since
is a Nash equilibrium we have that
U1 (
1 ,
2 ) U1 (1 ,
2 ) for every 1 1 .
In a two-player zero-sum game U1 () = U2 (). Therefore, we can write the
previous inequality as
U2 (
1 ,
2 ) U2 (1 ,
2 ) for every 1 1 ,
and multiplying both sides by (1) yields
U2 (
1 ,
2 ) U2 (1 ,
2 ) for every 1 1 .
Hence, if player 2 plays the equilibrium strategy
2 he can guarantee himself his
equilibrium payoff U2 (
1 ,
2 ) whatever is the strategy played by player 1. Therefore,
let us assign U2 (
1 ,
2 ) = V and
2 = 2 .
We can now write the Nash equilibrium conditions for player 2,
U2 (
1 ,
2 ) U2 (
1 , 2 ) for every 2 2
V U2 (
1 , 2 ) for every 2 2 .
Since U2 (
1 , 2 ) = U1 (
1 , 2 ) we have that
V U1 (
1 , 2 ) for every 2 2 .
Multiplying both sides by (1) we obtain
V U1 (
1 , 2 ) for every 2 2 .
Which implies that
1 is the optimal strategy 1 for player 1.
It remains to prove that the value V is unique. Suppose that we have two Nash
equilibria
and 0 that yield, respectively, a expected payoff to player 1 equal to
0
V and V . Without loss of generality let us assume the V > V 0 . Therefore:
U1 (
1 ,
2 ) > U1 (10 , 20 ).
Since (10 , 20 ) is a Nash equilibrium it must hold that player 1 does not want to
deviate to
1 , i.e. U1 (10 , 20 ) U1 (
1 , 20 ). This, together with the last inequality
implies that
U1 (
1 ,
2 ) > U1 (
1 , 20 ).
We use again the fact that for every , U1 () = U2 (), and write the previous
inequatily in terms of player 2s payoffs
U2 (
1 ,
2 ) > U2 (
1 , 20 ),

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## multiplying both sides by (1) yields

U2 (
1 ,
2 ) < U2 (
1 , 20 ).
Which contradicts the fact that (
1 ,
2 ) is a Nash equilibrium because player 2
would have an incentive to deviate to 20 . Consequently, the value V is unique. 
One consequence of the previous theorem is that the set of Nash equilibria in a
two-player zero-sum game is interchangeable. That is, if (
1 ,
2 ) and (10 , 20 ) are
0
0
Nash equilibria, then (
1 , 2 ) and (1 ,
2 ) are also Nash equilibria.
(Nash defined a game as solvable if the set of Nash equilibria is interchangeable.
Therefore, using the terminology of Nash, we obtained that every two-player zerosum game is solvable.)
5. Constant Sum Games
A constant sum game is game G with the property that for every s =
(s1 , . . . , sn ) S we have that u1 (s) + + un (s) = c for some constant c.
Every constant sum game is equivalent to a zero-sum game (see affine transformations in the lecture devoted to Decision Theory). Consequently there is no
added generality from considering values of c different from 0.
Appendix A. Tutorial Questions
Exercise 1. Find all minmax pure strategies of the following game
F
G
H
I
J

A
2, 3
7, 3
2, 1
1, 8
0, 5

B
6, 1
8, 2
9, 5
4, 3
2, 6

C
7, 0
4, 2
3, 2
1, 0
2, 3

D
7, 2
2, 0
1, 8
5, 9
4, 6

E
1, 2
9, 4
3, 5
8, 2
7, 3

Exercise 2. Find all Nash equilibria in pure strategies of the game in Exercise 1.
Exercise 3. Find the unique equilibrium of the following game. Compare the
game with the matching pennies game. Why do they have the same set of Nash
equilibria?
L
2, 3
0, 1

T
B

R
0, 1
2, 3

Exercise 4. Find the set of Nash equilibria of the following game. Compare the
game with the Battle of the Sexes game. Why do they have the same set of Nash
equilibria?
T
B

L
9, 1
1, 0

R
1, 0
3, 4

Exercise 5. Find one zero-sum game (there are many) that has the same set of
Nash equilibria than the following game.

ZERO-SUM GAMES

3, 4

5, 2

4, 3

7, 0

1, 6

7 7
2, 2
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