3 The Principle of Indiﬀerence
3.1
Consider the game with matrix

2 2

1
1 1
1
3 0
1
.
(a) Note that this game has a saddle point.
(b) Show that the inverse of the matrix exists.
(c) Show that II has an optimal strategy giving positive weight to each of his columns.
(d) Why then, don’t equations (16) give an optimal strategy for II?
Solution:
(a) Denote the above matrix by
A
:= (
a
ij
)
1
≤
i,j
≤
3
. Then
a
23
= 1 is the minimum of the
elements in its row and the maximum of the elements in its column, hence a saddle point.
(b) We ﬁnd that the determinant of
A
, expanding on its middle column, is
det(
A
) =
±
±
±
±
±
±

2 2

1
1 1
1
3 0
1
±
±
±
±
±
±
=

2
·
±
±
±
±
1 1
3 1
±
±
±
±
+
±
±
±
±

2

1
3
1
±
±
±
±
=

2(1
·
1

1
·
3) + (

2
·
1

(

1)
·
3)
= 5
,
so det(
A
)
6
= 0 and hence its inverse exists.
(c) Since
a
23
= 1 is a saddle point, we know the value of the game to be 1. Hence, any optimal
strategy
q
for Player II must guarantee this value. i.e.

2
q
1
+ 2
q
2

q
3
≤
1
q
1
+
q
2
+
q
3
≤
1
3
q
1
+
q
3
≤
1
Since
q
1
+
q
2
+
q
3
= 1, we have equality in the second equation. Hence
q
3
= 1

q
1

q
2
,
which we can substitute in the other two equations to eliminate one variable. This gives
us

q
1
+ 3
q
2

1
≤
1
2
q
1

q
2
+ 1
≤
1
,
i.e.
q
2
≤
1
3
q
1
+
2
3
, and
q
2
≥
2
q
1
. Finally, we require that 0
≤
q
1
≤
1, 0
≤
q
2
≤
1 and also
0
≤
q
3
≤
1 which is equivalent to 0
≤
q
1
+
q
2
≤
1. You can make a diagram of this and
see that the set of optimal strategies of Player II is exactly
²
q
∈
R
3
:
q
2
≥
0
,q
2
≤
1
3
q
1
+
2
3
,q
2
≤
1

q
1
,q
2
≥
2
q
1
,
and
q
3
= 1

q
1

q
2
³
.
The optimal strategies giving positive weight to each of his columns are those for which
q
1
>
0,
q
2
>
0, and
q
3
>
0, i.e. (see diagram)
²
q
∈
R
3
:
q
2
>
0
,q
2
<
1
3
q
1
+
2
3
,q
2
<
1

q
1
,q
2
>
2
q
1
,
and
q
3
= 1

q
1

q
2
³
.
2