Suppose Robert is the row player, that is, he plays the rows, and Carol is a column player. If Robert shows a dime and Carol shows a dime, the sum will be less than thirty five cents and Robert will win ten cents. But, if Robert shows a dime and Carol shows a quarter, the sum will not be less than thirty five cents and Carol will win ten cents or Robert will lose ten cents. The following matrix depicts all four cases and their corresponding payoffs for Robert. Remember a negative value is a loss for Robert and a win for Carol.

Figure 1

The best strategy for Robert is to always show a dime because this way the worst he can do is lose ten cents. And, the best strategy for Carol is to always show a quarter because that way the worst she can do is to lose ten cents. If both Robert and Carol play their optimal strategies, Robert will lose ten cents each time. Therefore, the value of the game is negative ten cents.

We find the saddle point by placing an asterisk next to the minimum entry in each row, and by putting a box around the maximum entry in each column as shown below.

Figure 3

Since the second row, first column entry, which happens to be 10, has both an asterisk and a box, it is a saddle point. This implies that the value of the game is 10, and the optimal strategy for the row player is to always play row 2, and the optimal strategy for the column player is to always play column 1. If both players play their optimal strategies, the row player will win 10 units each time.

The row player's strategy is written as 0 1 0 1 indicating that he will play row 1 with a probability of 0 and row 2 with a probability of 1. Similarly the column player's strategy is written as 1 0 1 0 implying that he will play column 1 with a probability of 1, and column 2 with a probability of 0.

We write the payoff matrix for Robert as follows:

Figure 4

To determine whether the game is strictly determined, we look for a saddle point. Again, we place an asterisk next to the minimum value in each row, and a box around the maximum value in each column. We get

Figure 5

Since there is no entry that has both an asterisk and a box, the game does not have a saddle point, and thus it is non-strictly determined.

We wish to devise a strategy for Robert. If Robert consistently shows a dime, for example, Carol will see the pattern and will start showing a quarter, and Robert will lose. Conversely, if Carol repeatedly shows a quarter, Robert will start showing a quarter, thus resulting in Carol's loss. So a good strategy is to throw your opponent off by showing a dime some of the times and showing a quarter other times. Before we develop an optimal strategy for each player, we will consider an arbitrary strategy for each and determine the corresponding payoffs.

Let RR size 12{R} {} denote Robert's strategy and CC size 12{C} {} denote Carol's strategy.

Since Robert is a row player and Carol is a column player, their strategies are written as follows:

R=.20.80R=.20.80 size 12{R= left [ matrix { "." "20" {} # "." "80"{} } right ]} {} and C=.70.30C=.70.30 size 12{C= left [ matrix { "." "70" {} ## "." "30" } right ]} {}.

To find the expected payoff, we use the following reasoning.

Since Robert chooses to play row 1 with .20.20 size 12{ "." "20"} {} probability and Carol chooses to play column 1 with .70.70 size 12{ "." "70"} {} probability, the move row 1, column 1 will be chosen with .20.70=.14.20.70=.14 size 12{ left ( "." "20" right ) left ( "." "70" right )= "." "14"} {} probability. The fact that this move has a payoff of 10 cents for Robert, Robert's expected payoff for this move is .14.10=.014.14.10=.014 size 12{ left ( "." "14" right ) left ( "." "10" right )= "." "014"} {} cents. Similarly, we compute Robert's expected payoffs for the other cases. The table below lists expected payoffs for all four cases.

The above table shows that if Robert plays the game with the strategy R=.20.80R=.20.80 size 12{R= left [ matrix { "." "20" {} # "." "80"{} } right ]} {} and Carol plays with the strategy C=.70.30C=.70.30 size 12{C= left [ matrix { "." "70" {} ## "." "30" } right ]} {}, Robert can expect to lose 7.2 cents for every game.

Alternatively, if we call the game matrix GG size 12{G} {}, then the expected payoff for the row player can be determined by multiplying matrices RR size 12{R} {}, GG size 12{G} {} and CC size 12{C} {}. Thus, the expected payoff PP size 12{P} {} for Robert is as follows:

Which is the same as the one obtained from the table.

Let us suppose that the row player uses the strategy R=r1−rR=r1−r size 12{R= left [ matrix { r {} # 1 - r{} } right ]} {}. Now if the column player plays column 1, the expected payoff PP size 12{P} {} for the row player is

Pr=1r+−31−r=4r−3Pr=1r+−31−r=4r−3 size 12{P left (r right )=1 left (r right )+ left ( - 3 right ) left (1 - r right )=4r - 3} {}.

Which can also be computed as follows:

Pr=r1−r1−3Pr=r1−r1−3 size 12{P left (r right )= left [ matrix { r {} # 1 - r{} } right ] left [ matrix { 1 {} ## - 3 } right ]} {} or 4r−34r−3 size 12{4r - 3} {}.

If the row player plays the strategy r1−rr1−r size 12{ left [ matrix { r {} # 1 - r{} } right ]} {} and the column player plays column 2, the expected payoff PP size 12{P} {} for the row player is

Pr=r1−r−24=−6r+4Pr=r1−r−24=−6r+4 size 12{P left (r right )= left [ matrix { r {} # 1 - r{} } right ] left [ matrix { - 2 {} ## 4 } right ]= - 6r+4} {}.

We have two equations

Pr=4r−3Pr=4r−3 size 12{P left (r right )=4r - 3} {} and Pr=−6r+4Pr=−6r+4 size 12{P left (r right )= - 6r+4} {}

The row player is trying to improve upon his worst scenario, and that only happens when the two lines intersect. Any point other than the point of intersection will not result in optimal strategy as one of the expectations will fall short.

Solving for rr size 12{r} {} algebraically, we get

r=7/10r=7/10 size 12{r=7/"10"} {}.

Therefore, the optimal strategy for the row player is .7.3.7.3 size 12{ left [ matrix { "." 7 {} # "." 3{} } right ]} {} .

Alternatively, we can find the optimal strategy for the row player by, first, multiplying the row matrix with the game matrix as shown below.

And then by equating the two entries in the product matrix. Again, we get r=.7r=.7 size 12{r= "." 7} {}, which gives us the optimal strategy .7.3.7.3 size 12{ left [ matrix { "." 7 {} # "." 3{} } right ]} {} .

We use the same technique to find the optimal strategy for the column player.

Suppose the column player's optimal strategy is represented by c1−cc1−c size 12{ left [ matrix { c {} ## 1 - c } right ]} {}. We, first, multiply the game matrix by the column matrix as shown below.

And then equate the entries in the product matrix. We get

Therefore, the column player's optimal strategy is .6.4.6.4 size 12{ left [ matrix { "." 6 {} ## "." 4 } right ]} {} .

To find the expected value, VV size 12{V} {}, of the game, we find the product of the matrices RR size 12{R} {}, GG size 12{G} {} and CC size 12{C} {}.

That is, if both players play their optimal strategies, the row player can expect to lose .2.2 size 12{ "." 2} {} units for every game.

Since we have already determined that the game is non-strictly determined, we proceed to determine the optimal strategy for the game. We rewrite the game matrix.

Let R=r1−rR=r1−r size 12{R= left [ matrix { r {} # 1 - r{} } right ]} {} be Robert's strategy, and C=c1−cC=c1−c size 12{C= left [ matrix { c {} ## 1 - c } right ]} {} be Carol's strategy.

To find the optimal strategy for Robert, we, first, find the product RGRG size 12{ ital "RG"} {} as below.

By setting the entries equal, we get

35r−25=−35r+2535r−25=−35r+25 size 12{"35"r - "25"= - "35"r+"25"} {}
or r=5/7r=5/7 size 12{r=5/7} {}.

Therefore, the optimal strategy for Robert is 5/72/75/72/7 size 12{ left [ matrix { 5/7 {} # 2/7{} } right ]} {} .

To find the optimal strategy for Carol, we, first, find the following product.

We now set the entries equal to each other, and we get,

or c=1/2c=1/2 size 12{c=1/2} {}

Therefore, the optimal strategy for Carol is 1/21/21/21/2 size 12{ left [ matrix { 1/2 {} ## 1/2 } right ]} {} .

To find the expected value, VV size 12{V} {}, of the game, we find the product RGCRGC size 12{ ital "RGC"} {}.

If both players play their optimal strategy, the value of the game is zero. In such case, the game is called fair.

We first look for a saddle point and determine that none exist. Next, we try to reduce the matrix to a 2×22×2 size 12{2 times 2} {} matrix by eliminating the dominated row.

Since every entry in row 3 is larger than the corresponding entry in row 2, row 3 dominates row 2. Therefore, a rational row player will never play row 2, and we eliminate row 2. We get

Now we try to eliminate a column. Remember that the game matrix represents the payoffs for the row player and not the column player; therefore, the larger the number in the column, the smaller the payoff for the column player.

The column player will never play column 2, because it is dominated by both column 1 and column 3. Therefore, we eliminate column 2 and get the modified matrix, MM size 12{M} {}, below.

To find the optimal strategy for both the row player and the column player, we use the method learned in the Section 2.

Let the row player's strategy be R=r1−rR=r1−r size 12{R= left [ matrix { r {} # 1 - r{} } right ]} {}, and the column player's be strategy be C=c1−cC=c1−c size 12{C= left [ matrix { c {} ## 1 - c } right ]} {} .

To find the optimal strategy for the row player, we, first, find the product RMRM size 12{ ital "RM"} {} as below.

By setting the entries equal, we get

or r=1/3r=1/3 size 12{r=1/3} {}.

Therefore, the optimal strategy for the row player is 1/32/31/32/3 size 12{ left [ matrix { 1/3 {} # 2/3{} } right ]} {} , but relative to the original game matrix it is 1/302/31/302/3 size 12{ left [ matrix { 1/3 {} # 0 {} # 2/3{} } right ]} {} .

To find the optimal strategy for the column player we, first, find the following product.

We set the entries in the product matrix equal to each other, and we get,

or c=2/3c=2/3 size 12{c=2/3} {}

Therefore, the optimal strategy for the column player is 2/31/32/31/3 size 12{ left [ matrix { 2/3 {} ## 1/3 } right ]} {} , but relative to the original game matrix, the strategy for the column player is 2/301/32/301/3 size 12{ left [ matrix { 2/3 {} ## 0 {} ## 1/3 } right ]} {} .

To find the expected value, VV size 12{V} {}, of the game, we have two choices: either to find the product of matrices RR size 12{R} {}, MM size 12{M} {} and CC size 12{C} {}, or multiply the optimal strategies relative to the original matrix to the original matrix. We choose the first, and get

Therefore, if both players play their optimal strategy, the value of the game is zero.