2
Matrix Games – Domination
Contrary to Ferguson’s notation, we prefer to use the horizontal notation (
·
,
·
, . . . ,
·
) to represent
a
row
vector and hence (
·
,
·
, . . . ,
·
)
T
to denote a
column
vector.
2.1
Solve the game with matrix

1

3

2
2
¶
, that is find the value and an optimal (mixed) strategy
for both players.
Solution:
This matrix does not have a saddle point. Hence, the value of the game is
V
=

1
·
2

(

3)(

2)

1

(

3)+2

(

2)
=

8
6
=

4
3
=

1
1
3
, the probability with which Player I should choose the first row is
p
=
2

(

2)
6
=
2
3
, and the probability with which Player II should choose the first column is
q
=
2

(

3)
6
=
5
6
.
So the optimal mixed strategy for Player I is
(
2
3
,
1
3
)
and for Player II
(
5
6
,
1
6
)
.
2.2
Solve the game with matrix
0
2
t
1
¶
for an arbitrary real number
t
. Draw the graph of
v
(
t
)
,
the value of the game, as a function of
t
, for
∞
< t <
∞
.
Solution:
If
t <
0, then 0 is a saddle point and hence the value of the game would be 0 and the optimal
strategies are pure: Player I always chooses the first row and Player II always the first column.
If instead 0
< t
≤
1, then
t
is a saddle point and hence the value of the game would be
t
and the
optimal strategies are again pure: Player I always chooses the second row and Player II always
the first column.
If, however,
t
= 0, then both zeroelements in the first column are saddle
points, the value of the game is 0, and Player II still keeps its pure optimal strategy of choosing
the first column but Player I can now use an
arbitrary
mixed strategy of choosing between the
first and second row to be sure of a nonnegative average payoff. He might, however, prefer the
pure strategy of choosing the first row, since it dominates the second one. Finally, if
t >
1, there
is no saddle point. The value of the game would be
V
(
t
) =
0
·
1

2
t
0

2+1

t
=

2
t

1

t
=
2
t
t
+1
= 1 +
t

1
t
+1
,
the probability with which Player I should choose the first row is
p
=
1

t

1

t
=
t

1
t
+1
, and the
probability with which Player II should choose the first column is
q
=
1

2

1

t
=
1
t
+1
.
So an
optimal mixed strategy for Player I is then
‡
t

1
t
+1
,
2
t
+1
·
and for Player II
‡
1
t
+1
,
t
t
+1
·
. Finally,
note that lim
t
→∞
V
(
t
) = lim
t
→∞
2
1+
1
t
= 2.
2