Section 8.2

 (6) [ -1 3 ] 2 -6
When you check, you will see that this game is nonstrictly determined, so the formula applies.
D = (a+d) - (b+c) = (-1-6) - (2+3) = -12
The value of the game is (ad-bc)/D = ((-1)(-6) - (2)(3))/12 = 0, so the game is fair.
But for it to come out even, both R and C must play their optimal strategies.
R should play the strategy P* = [ (d-c)/D (a-b)/D ] = [8/12 4/12] = [2/3 1/3].
 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
This is a strictly determined game with a saddle value of 0.
Thus the value of the game is 0, and it is fair.
R should play the pure strategy P* = [1 0] (that is, always the first row).
 C should play the pure strategy Q* =[ 0 ] (that is always play the second column). 1

(10)
 [ 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

This game is nonstrictly determined, so our formula applies.
D = (a+d) - (b+c) = (4+3) - (-2-6) = 15
The value of the game is (ad-bc)/D = ((4)(3) - (-2)(-6))/15 = 0, so the game is fair.
But for it to come out even, both R and C must play their optimal strategies.

R should play the strategy P* = [ (d-c)/D (a-b)/D ] = [5/15 10/15] = [1/3 2/3].
But, remembering that the first and last rows are never played, this is P* = [0 1/3 2/3 0].
 C should play the strategy Q* =[ (d-b)/D ] = [ 9/15 ] = [ 3/5 ] (a-c)/D 6/15 2/5

(26 A)
 [ 0 2 -1 0 ] 1 0 -1 -2 2 3 -1 1 1 -2 0 0

R will never play row 1, for row 3 is always better than (or just as good as) than row 1.
Nor will R play row 2, for, again, row 3 is always better or just as good.
Thus, R ignores the first two rows, and gets the new game
 [ 2 3 -1 1 ] 1 -2 0 0

Now, C will never play column 1, for column 3 is better.
Nor will C play column 4, for column 3 is better.
Thus C ignores the first and last column, and the game now looks this way:
 [ 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

Remembering that C plays neither the first nor the last column, C's strategy is thus
 Q* =[ 0 ] 1/6 5/6 0

Thus Station C should do talk 1/6 of the time and sports 5/6 of the time. It should never do travel or soaps.