0.18 Game theory  (Page 3/4)

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:

$R=\left[\begin{array}{cc}\text{.}\text{20}& \text{.}\text{80}\end{array}\right]$ and $C=\left[\begin{array}{c}\text{.}\text{70}\\ \text{.}\text{30}\end{array}\right]$ .

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

Since Robert chooses to play row 1 with $\text{.}\text{20}$ probability and Carol chooses to play column 1 with $\text{.}\text{70}$ probability, the move row 1, column 1 will be chosen with $\left(\text{.}\text{20}\right)\left(\text{.}\text{70}\right)=\text{.}\text{14}$ probability. The fact that this move has a payoff of 10 cents for Robert, Robert's expected payoff for this move is $\left(\text{.}\text{14}\right)\left(\text{.}\text{10}\right)=\text{.}\text{014}$ 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 $R=\left[\begin{array}{cc}\text{.}\text{20}& \text{.}\text{80}\end{array}\right]$ and Carol plays with the strategy $C=\left[\begin{array}{c}\text{.}\text{70}\\ \text{.}\text{30}\end{array}\right]$ , 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:

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

Let us suppose that the row player uses the strategy $R=\left[\begin{array}{cc}r& 1-r\end{array}\right]$ . Now if the column player plays column 1, the expected payoff $P$ for the row player is

$P\left(r\right)=1\left(r\right)+\left(-3\right)\left(1-r\right)=\mathrm{4r}-3$ .

Which can also be computed as follows:

$P\left(r\right)=\left[\begin{array}{cc}r& 1-r\end{array}\right]\left[\begin{array}{c}1\\ -3\end{array}\right]$ or $\mathrm{4r}-3$ .

If the row player plays the strategy $\left[\begin{array}{cc}r& 1-r\end{array}\right]$ and the column player plays column 2, the expected payoff $P$ for the row player is

$P\left(r\right)=\left[\begin{array}{cc}r& 1-r\end{array}\right]\left[\begin{array}{c}-2\\ 4\end{array}\right]=-\mathrm{6r}+4$ .

We have two equations

$P\left(r\right)=\mathrm{4r}-3$ and $P\left(r\right)=-\mathrm{6r}+4$

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

$r=7/\text{10}$ .

Therefore, the optimal strategy for the row player is $\left[\begin{array}{cc}\text{.}7& \text{.}3\end{array}\right]$ .

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 $r=\text{.}7$ , which gives us the optimal strategy $\left[\begin{array}{cc}\text{.}7& \text{.}3\end{array}\right]$ .

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

Suppose the column player's optimal strategy is represented by $\left[\begin{array}{c}c\\ 1-c\end{array}\right]$ . 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 $\left[\begin{array}{c}\text{.}6\\ \text{.}4\end{array}\right]$ .

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

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