Section 8.2
(6) [

1

3

] 
2

6

C should play the strategy

Q^{*} =[

(db)/D

] = [ 
9/12

] = [ 
3/4

] 
(ac)/D

3/12

1/4

(8) [

3

0

] 
1

4

C should play the pure strategy

Q^{*} =[

0

] (that is always play the second column). 
1

[

3

6

] 
4

6


2

3


3

0

R will never play row 1, for row 2 is just as good or better, regardless of
C's choice.
Nor will R play row 4, for row 3 is always the better bet, regardless of C's
choice.
Thus R plays only the second and third rows, so this game reduces to the folowing
matrix
[

4

6

] 
2

3

C should play the strategy

Q^{*} =[

(db)/D

] = [ 
9/15

] = [ 
3/5

] 
(ac)/D

6/15

2/5

[ 
0

2

1

0

] 
1

0

1

2


2

3

1

1


1

2

0

0

[ 
2

3

1

1

] 
1

2

0

0

[ 
3

1

] 
2

0

This is a nonstrictly determined 2 by 2 game, so our formula applies.
D = (a+d)  (b+c) = (3+0)  (21) = 6.
The value of the game is (adbc)/D = ((3)(0)  (2)(1))/6 = 1/3,
so the game is not fair. Station C will win in the long run.
R's optimal strategy is P^{*} = [ (dc)/D (ab)/D ] = [2/6 4/6] = [1/3
2/3].
Looling back at the original matrix, this becomes P^{*} = [0 0 1/3
2/3].
In words, Station R should do sit coms 1/3 of the time, and soaps 2/3 of the
time, and never do travel or news.
C should play the strategy

Q^{*} =[

(db)/D

] = [ 
1/6

] 
(ac)/D

5/6

Q^{*} =[

0

] 
1/6


5/6


0
