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 19.0. 

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 19.0. 

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 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 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 19.0. 

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 19.0. 

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 R denote Robert's strategy and C denote Carol's strategy.

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

and .

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

Since Robert chooses to play row 1 with .20 probability and Carol chooses to play column 1 with .70 probability, the move row 1, column 1 will be chosen with (.20)(.70)=.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 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 and Carol plays with the strategy , Robert can expect to lose 7.2 cents for every game.

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

(19.1)

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

Let us suppose that the row player uses the strategy . Now if the column player plays column 1, the expected payoff P for the row player is

P(r)=1(r)+(−3)(1−r)=4r−3.

Which can also be computed as follows:

or 4r−3.

If the row player plays the strategy and the column player plays column 2, the expected payoff P for the row player is

.

We have two equations

P(r)=4r−3 and P(r)=−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 r algebraically, we get

r=7/10.

Therefore, the optimal strategy for the row player is .

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

(19.4)

And then by equating the two entries in the product matrix. Again, we get r=.7, which gives us the optimal strategy .

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

Suppose the column player's optimal strategy is represented by . We, first, multiply the game matrix by the column matrix as shown below.

(19.5)

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

Therefore, the column player's optimal strategy is .

To find the expected value, V, of the game, we find the product of the matrices R, G and C.

(19.8)

That is, if both players play their optimal strategies, the row player can expect to lose .2 units for every game.