Section 8.2
(6) [
|
-1
|
3
|
] |
2
|
-6
|
C should play the strategy
|
Q* =[
|
(d-b)/D
|
] = [ |
9/12
|
] = [ |
3/4
|
] |
(a-c)/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* =[
|
(d-b)/D
|
] = [ |
9/15
|
] = [ |
3/5
|
] |
(a-c)/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) - (-2-1) = 6.
The value of the game is (ad-bc)/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* = [ (d-c)/D (a-b)/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* =[
|
(d-b)/D
|
] = [ |
1/6
|
] |
(a-c)/D
|
5/6
|
Q* =[
|
0
|
] |
1/6
|
||
5/6
|
||
0
|