Foundations of Game Theory: Mathematical Perspective
Section 3: Minimax Theorem and the value of a matrix game
Summary
This section follows on from Section 2 in describing mixed strategies and the expected
payoff function for a matrix game.
The von Neumann minimax theorem will be
introduced, but the main focus will be on its application, rather than rigorous proof.
Some general properties of the expected payoff function will be described along with
selective propositions which follow from the minimax theorem and Section 2. We will
then show how these ideas can be applied to find mixed strategy solutions and the value
of mx2 and 2xn matrix games.
Mixed Strategies
“Matching Pennies”
Players 1,2 choose heads or tails on a coin, each without revealing his choice to his
opponent. Their choices are compared and player 1 receives 1 unit from player 2 if both
have the same choice, otherwise player 2 gets 1 unit from player 1.
The normalized form of the game is given as
1 y



1
x
1
1
row min
1 x 1
1  1
1
1

y
col max
We observe that max min aij  min max aij and so no pure strategy solution exists. No
i
optimal pure strategies.
j
j
i
Instead, suppose that each player chooses pure strategies by some random device. Let
player 1 choose pure strategy 1 with probability x and player 2 choose pure strategy 2
with probability y. The expected payoff E(x,y) to player 1 is
E ( x, y )  xy *1  x(1  y )  (1  x) y  (1  x)(1  y )  (2 x  1)( 2 y  1)
By inspection, E(x,0.5)=0=E(0.5,y), for all x in [0,1], and for all y in [0,1] and
E(0.5,0.5)=0. So we may consider the pair of mixed strategies (0.5,0.5) as a saddle point
of this game and the value of the game is zero. Theoretically, if both choose a payoff of
0.5, they can’t lose.
Example
y 1 y 

x
1
3 1 row min
1 x 4
2 2
4
3



col max
The game has no saddle point, so we assume the game to be played with mixed
strategies:
E ( x, y )  xy *1  3x(1  y )  (1  x)4 y  2(1  x)(1  y )  4 xy  x  2 y  2 
1
1 5
 4( x  )( y  ) 
2
4 2
From this expression, it is clear that player 1 can ensure his expectation is at least 5/2 (by
choosing x=0.5). And he cannot be sure of more than 5/2 for by taking y=1/4, player 2
can ensure that player 1’s expectation id exactly 5/2. Thus player 1 may as well settle for
5/2 and play x=1/2, so as to gain this amount. Similarly, player 2 may as well reconcile
himself to get -5/2 and play y=1/4 so as to gain it.
Now we observe that
E ( x,0.25)  E (0.5,0.25)  E (0.5, y )
and so (0.5,0.25) is a saddle point of the game. The value of the game is 5/2.
Definition
If a player of a matrix game has k pure strategies then a mixed strategy for the player is a
k-tuple
~
x  ( x1 , x2 ,, xk ) of real numbers such that xi  0 for i=1,2,…,k and
k
x
i 1
i
 1.
A vector space on which an inner product is defined is called a Euclidean space, for
example  3 produces a 3-D space. The real k dimensional space is denoted by  k .
Each mixed strategy for a player is a point if a Euclidean space. To denote the set of all
possible mixed strategies for a player put
S k  {( x1 , x 2 , , x k )  R k : xi  0
i  1,..., k
k
and
x
i 1
i
 1}
Definition
For the game with payoff matrix A=(aij) where i=1,…,m and j=1,…,n, the expected
payoff function E is defined on SmxSn by
m
n
E(~
x, ~
y )   aij xi y j
i 1 j 1
x  ( x1 , x2 ,, xm )  S m and ~
y  ( y1 , y 2 ,, y n )  S n .
where ~
It turns out (proof omitted) that
max min E( ~
x, ~
y )  min max E( ~
x, ~
y)
xSm yS n
yS m xS m
for every matrix game and so there always exist optimal mixed strategies.
This is the content of the famous von Neumann minimax theorem. From this it will
follow that there exist
~
x*  Sm , ~
y *  S n such that
E(~
x, ~
y *)  E ( ~
x *, ~
y *)  E ( ~
x *, ~
y)
~
x  Sm , ~
y  Sn
Consider the matrix game with payoff matrix (aij) of order mxn. Sets of mixed strategies
for players 1,2 are respectively Sm, Sn where
k


S k  ( z1 , z 2 ,, z k   k : z i  0,  z i  1
i 1


A mixed strategy for a player having at least two non zero components is called a proper
mixed strategy.
The expected payoff function is the real valued function defined on SmxSn by
m
n
E ( x, y)   aij xi y j
i 1 j 1
The object of player 1 is to choose x to maximize E ( x, y ) . The object of player 2 is to
choose y to minimize E ( x, y ) . If player 1 chooses x  S m , the payoff to player 1 will be
at least min E ( x, y) , hence player 1 can expect a payoff of at least max min E ( x, y) . If
xS m yS n
yS n
player 2 uses y  S n then the expected payoff to player 1 is at most max E ( x, y) and so
xS m
player 2 can ensure the expected payoff to player 1 is at most min max E ( x, y) . The
yS n xS m
principal aim is to determine if min max E ( x, y) = max min E ( x, y) .
yS n xS m
xS m yS n
Let i be a pure strategy for player 1 and let y  S n be a mixed strategy for player 2. Then
n
E (i, y)   aij y j
j 1
Now let x  S m be a mixed strategy for player 1. Then
m
n
m
n
m
i 1
j 1
i 1
E ( x, y)   aij xi y j   xi  aij y j  xi E (i, y)
i 1 j 1
Similarly, let j be a pure strategy for player 2 and let x  S m be a mixed strategy for
player 1. Then
m
E ( x, j )   aij xi
i 1
Now let y  S n be a mixed strategy for player 2, then
m
n
n
m
n
j 1
i 1
j 1
E ( x, y)   aij xi y j   y j  aij xi  y j E ( x, j )
i 1 j 1
Lemma 3.1
If E is the expected payoff function of a matrix game with an mxn payoff matrix and
y  S n , x  S m then
max E ( x, y )  max E (i, y )
xS m
1i  m
min E ( x, y )  min E ( x, j )
yS n
1 j  n
Proof
Let v  max E ( x, y) and w  max E (i, y ) . Since v is taken over Sm which ensures at least
1i m
xS m
m possible strategies, and w is taken over i≤m, it follows that w≤v. If x  S m , then
m
m
m
i 1
i 1
i 1
E ( x, y )   xi E (i, y )   xi w  w xi  w since
m
x
i 1
x  S m and so v≤w, from which it follows that v=w.
Proposition 3.2 (von Neumann minimax theorem)
i
 1 . Thus E(x,y)≤w for all
■
If E is the expected payoff function of a matrix game with an mxn payoff matrix then
min max E ( x, y) = max min E ( x, y)
yS n xS m
xS m yS n
Proof
Omitted
A consequence of the minimax theorem and our discussion in Section 2 is that E has a
saddle point, where E is the payoff function of any mxn matrix game. Thus, there exists a
pair ( ~
x *, ~
y *)  S m  S n such that
E(~
x, ~
y *)  E ( ~
x *, ~
y *)  E ( ~
x *, ~
y)
( ~
x, ~
y )  Sm  Sn
Definition 3.3
Let E be the expected payoff function of a matrix game. The real number
v  min max E( ~
x, ~
y ) = max min E( ~
x, ~
y)
ySn xS m
xS m ySn
x *, ~
y *)  S m  S n
is called the value of the game and a mixed strategy solution is a pair ( ~
such that
E(~
x, ~
y *)  E ( ~
x *, ~
y *)  E ( ~
x *, ~
y)
( ~
x, ~
y )  Sm  Sn
y *  S n such
A mixed strategy ~
x * for player 1 is called optimal if there exists a strategy ~
~
~
that ( x *, y *) is a mixed strategy solution. An analogous definition applies for an optimal
strategy for player 2.
Remark
x *, ~
y *) is a mixed strategy
By the second proposition in Section 2, we know that if ( ~
~
~
solution of a game with value v, then E ( x *, y *)  v . Moreover, if ~
x * and ~y * are
x *, ~
y *) is a mixed strategy
optimal mixed strategies for players 1,2 respectively then ( ~
solution.
Proposition 3.4
If (i0,j0) is a pure strategy solution of a matrix game with payoff matrix a ij then the value
of the matrix game is a i 0 j 0 and (i0,j0) is a mixed strategy solution.
Proof
Let E be the expected payoff function. Suppose that (aij) has order mxn. If 1≤i≤m and
1≤j≤n, then E (i, j )  aij and for i=1,…,m, j=1,…,n,
aij0  ai0 j0  ai0 j
But by Lemma 3.1,
max E ( x, j 0 )  max E (i, j 0 )  max aij0  ai0 j0
1i  m
xS
1 i  m
m
and so E ( x, j0 )  E (i0 , j0 ) x  S m .
We can similarly show that
min E(i0 , y)  min E(i0 , j )  min ai0 j  ai0 j0 and so E (i0 , y)  E (i0 , j0 )
yS n
1 j n
1 j n
y  S n .
From these inequalities, it follows that
E ( x, j0 )  E (i0 , j0 )  E (i0 , y)
x, y  S m  S n and so (i0,j0) is a mixed strategy
solution. The value of the game is E (i0 , j 0 )  ai0 j0
■
Proposition 3.5
If E is the expected payoff function of a matrix game with an mxn payoff function, then
max min E ( x, j )  min max E (i, y )
yS n
x  S 1 j n
1 i  m
m
Proof
max E ( x, y )  max E (i, y )
xS m
1i  m
min E ( x, y )  min E ( x, j )
yS n
1 j  n
Now, min max E( x, y)  min max E(i, y)
ySn xS m
ySn 1i m
and
max min E( x, y)  max min E( x, j )
xSm ySn
xS m 1 j n
and by von Neumann’s minimax theorem 3.2, the result follows automatically. ■
Proposition 3.6
If the value of a game with an mxn payoff matrix is equal to v and ~
x * is an element of
~
S m such that E( x *, j )  v
j  1,..., n where E is the expected payoff function, then ~
x*
is optimal.
Proof
It follow from Lemma 3.1 that min E( x*, y)  min E( x*, j ) and so
yS n
E ( x*, y)  v
1 j n
y  S n .
Let ( x' , y' )  S m  S n be a mixed strategy solution of the game, then
E ( x, y' )  E ( x' , y' )  E ( x' , y)
( x, y)  S m  S n and E ( x' , y ' )  v
Therefore, E( x*, y' )  E( x' , y' )  v ie E( x*, y' )  v , but we also showed that
E ( x*, y)  v y  S n , hence E ( x*, y ' )  E ( x' , y ' )  v.
This shows that if ( x, y)  S m  S n then,
E ( x, y ' )  E ( x*, y ' )  v  E ( x*, y )
and so (x*,y’) is a mixed strategy solution by Definition 3.3. Hence x* is optimal.
■
Corollary 3.7
Let E be the expected payoff function of a matrix game with an mxn payoff matrix and
the value of the game is v, then an element ~
x * of S m is an optimal mixed strategy for
player 1 if and only if
min E ( ~
x *, j )  v
1 j  n
Proof
If min E ( ~
x *, j )  v
1 j  n
then
E(~
x *, j )  v for j=1,…,n and so ~
x * is optimal by
y*  S n
Proposition 3.6. Conversely, if ~
x * is optimal for player 1 then there exists a ~
~
~
such that ( x *, y *) is a mixed strategy solution. Then
E(~
x *, ~
y *)  v
and
E(~
x *, ~
y *)  E ( ~
x *, y) y  S n . Thus, min E ( ~
x *, y)  v and
yS n
so by Lemma 3.1, we have min E ( ~
x *, j )  v
1 j  n
Proposition 3.8
■
Let E be the expected payoff function of a matrix game with an mxn payoff matrix and
value equal to v. Let y* be an element of Sn. Then the following statements are
equivalent
i.
ii.
y* is an optimal strategy for player 2
max E (i, y*)  v
iii.
E(i, y*)  v
1i  m
for
i  1,...m
Proof
No proof will be given, since 3.6 and 3.7 are analogous.
Proposition 3.9
Let E be the expected payoff function of a matrix game with an mxn payoff matrix. If
x*  S m , y*  S n , and
v   s.t. E (i, y*)  v  E ( x*, j ) for i=1,…,m and
j=1,…,n, then v is the value of the game and (x*,y*) is a mixed strategy solution.
Proof
We need to prove that E ( x, y*)  E ( x*, y*)  E ( x*, y ).
Since E (i, y*)  v for i=1,…,m, it follows from Lemma 3.1 that E ( x, y*)  v x  S m
Similarly, E ( x*, y)  v y  S n . Thus, E ( x*, y*)  v  E ( x*, y*) and so
E ( x*, y*)  v. Now E ( x, y*)  E ( x*, y*)  E ( x*, y)
( x, y)  S m  S n . Hence
(x*,y*) is a mixed strategy solution and the value of the game is E ( x*, y*)  v . ■
Proposition 3.10
Let E be the expected payoff function of a matrix game with an mxn payoff matrix and
value equal to v. If x*=(x1*,x2*,…,xm*) and y* are optmal strategies for players 1 and 2
respectively, then xi*=0 for every i such that E(i,y*)<v. A similar property applies with
regard to player 2’s optimal strategies, ie yj*=0 for every j such that E(x*,j)>v.
Proof (by contradiction)
We have E (i, y*)  v for every i. Suppose there exists an r  {1,..., m} such that
E(r, y*)  v and
xr *  0 . Then E(r, y*) xr *  vxr * and so
m
m
m
i 1
i 1
i 1
E ( x*, y*)   E (i, y*) xi *   vxi *  v  xi *  v
However, this contradicts the fact that E ( x*, y*)  v since (x*,y*) are optimal. Therefore,
we must have xi *  0 for every i such that E (i, y*)  v as required.
■
Note that a similar property applies with regard to player 2’s optimal strategies, ie
yj*  0


E ( x*, j )  v 
This proposition is required for solving 2xn and mx2 matrix games which is described
next.
Solving Matrix Games:
2xn and mx2 matrix games
Consider the game with payoff matrix
2 3 11
7 5 2 


If x  ( ,1   ) is a mixed strategy solution for player 1 we have
E(x,1)=2α+7(1- α)=7-5 α
E(x,2)=3α+5(1- α)=5-2α
E(x,3)=11α+2(1- α)=2+9α
In the α-β plane we draw the graphs of the lines corresponding to
β=7-5 α
β=5-2α
β=2+9α
7
β
β=2+9α
5
β=5-2α
β=7-5α
2
0
β
1
β
α
Figure 1: Graphical solution of matrix game.
By Proposition 3.5, the value of the mxn matrix game with payoff function E is
x  S m is optimal if and only if min E( ~
max min E ( ~
x , j ) , and by Corrollary 3.7, ~
x , j)  v .
xS m 1 j  n
1 j  n
x , j ) is a function value of j. Hence, min E ( ~
In our example, E ( ~
x , j ) means the minimum
1 j n
of the 1≤j≤n coordinates at α of each of the three lines, and max min E ( ~
x , j ) means the
xS m 1 j  n
maximum of the ordinates at α taken over all α between 0 and 1. This occurs when β=52α intersects with β=2+9α, i.e. 11α=3 or α=3/11.
Hence, the unique optimal mixed strategy for player 1 is (3/11,8/11)=x* and the value of
the game is
E(~
x *,2)  E ( ~
x *,3)  49 / 11
Now, E(x*,1)=7-5*3/11=62/11>v which implies j=0, and so by Proposition 3.10, an
optimal mixed strategy for player 2 must have the form (0,γ,1- γ).
Since E(1,y*)=3 γ+11(1- γ)=11-8γ, but we know the value of the game is 49/11 so we
must have 11-8γ=49/11 so γ=9/11. The unique optimal mixed strategy for player 2 is
thus y*=(0,9/11,2/11).
Note, we could have used as an alternative E(2,y*)=5 γ+2(1- γ)=3 γ+2=49/11 so γ=9/11.
Summary of Graphical Method
Consider the matrix game with 2xn payoff matrix:
a1
b
 1
a2  an 
b2  bn 
Mixed strategy solutions for Player 1 have the form ( ,1   ),0    1. If E is the
expected payoff function and x  ( ,1   ) for j=1,…,n, then
E ( x, j )  a j  (1   )b j  (a j  b j )  b j
and if v is the value of the matrix game, then
v  max min [(a j  b j )  b j ]
0 1 1 j  n
In the αβ plane draw the graphs of the lines   (a j  b j )  b j . For each value of α
satisfying 0≤α≤1, let v(α) be the minimum of the ordinates at α of the n lines, i.e.
v( )  min [(a j  b j )  b j ]
1 j  n
Then, v  max min [(a j  b j )  b j ]  v( 0 ) , say, and so x  ( 0 ,1   0 ) is optimal for
0 1 1 j n
player 1.
Remark
For games having mx2 payoff matrices, the corresponding results are
v  min max E(i, y) and y* in Sn is optimal if and only if v  max E (i, y*)
yS n 1i m
1 i m