91Ó°ÊÓ

The payoff matrix had rows and columns that make a move for winning. Row and columns have a mixed strategy to win the game. By observing the opponent’s moves, strategies are predicted.

In the given problem, the strategies are investigated to steal business from others. The effects of the strategies are summarized in a payoff matrix.

There are two rivals, Joey and Tony, Joey represents the rows, and the Tony represents columns. Tony picks up any one option and Joey picks up an option from.

The payoff matrix is as follows:

$G=\begin{array}{ccc}+2& 0& −3\\ −1& −2& +1\end{array}$

Consider the Payoff matrix $G=\begin{array}{ccc}+2& 0& −3\\ −1& −2& +1\end{array}$. Let the rows and columns have a mixed strategy, specified by the vector $x=\left(x_1,x_2\right)\text{and}y=\left(y_1,y_2,y_3\right)$respectively. The sum of the vectors must equal one. The Row’s strategy is fixed; for the optimal column, move either Gourmet g , with payoff $\begin{array}{r}2x_1−1x_2\end{array}$or Seating s with payoff $\begin{array}{r}\\ −2x_2\\ \end{array}$or Free soda f with payoff $−3x_1+1x_2$.

Consider that Joey announces x before Tony. Pick $\left(x_1,x_2\right)$that maximizes from $min\left\{2x_1−x_2,−2x_2,−3x_1+x_2\right\}$. LP (Linear Programming) to pick $\left(x_1,x_2\right)$is $z=min\left\{2x_1−x_2,−2x_2,−3x_1+x_2\right\}$.

Joey needs to choose $\begin{array}{r}{x}_1\end{array}$and $\begin{array}{r}{x}_2\end{array}$to maximize the z as follows,

$\begin{array}{r}2x_1−x_2+z≤0\\ −2x_2+z≤0\\ −3x_1+x_2+z≤0\end{array}$

Simplifying yields the following:

$\begin{array}{r}{x}_1+x_2=3\\ x_1,x_2â‰\yen 0\end{array}$

Pick$(y_1,y_2,y_3)$that maximizes from .

LP (Linear Programming) to pick $y_1,y_2,y_3$is .

Tony needs to choose $y_1,y_2,$and $y_3$to minimize the w as follows,

$\begin{array}{r}2y_1−3y_3+wâ‰\yen 0\\ −1y_1−2y_2+y_3+wâ‰\yen 0\\ y_1+y_2+y_3=0\end{array}$

Simplifying yields the following:

$y_1,y_2,y_3â‰\yen 0$.

The LPs of Tony and Joey are not equal. Reduce the payoff matrix as follows,

$G=\begin{array}{ccc}+2& 0& −3\\ −1& −2& +1\end{array}$

Delete the dominant row or column to reduce the payoff matrix. Column 2has column three dominance; delete column 2 .

The optimal strategy for Joey$=\frac{\left[\begin{array}{l}1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots 1\end{array}\right]×G_{\mathrm{cof}}}{\left[\begin{array}{l}1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots 1\end{array}\right]×G_{A∈fj}×\left[\begin{array}{l}1\\ 1\end{array}\right]}$

The optimal strategy for Joey =$\begin{array}{r}\left[\begin{array}{ll}1& 1\end{array}\right]×\left[\begin{array}{ll}1& 1\\ 3& 2\end{array}\right]\\ \left[\begin{array}{ll}1& 1\end{array}\right]×\left[\begin{array}{ll}1& 3\\ 1& 2\end{array}\right]×\left[\begin{array}{c}1\\ 1\end{array}\right]\end{array}$

The optimal strategy for Joey$\left(\frac{3}{5},\frac{2}{5}\right)$

The optimal strategy for Tony$=\frac{\left[\begin{array}{l}1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots 1\end{array}\right]×G_{A∘\mathrm{δ}j}}{\left[\begin{array}{l}1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots 1\end{array}\right]×G_{A_{AJj}}×\left[\begin{array}{l}1\\ 1\end{array}\right]}$

The optimal strategy for Tony$\begin{array}{ccc}=(11& 1x\left|\begin{array}{cc}1& s\\ 1& 2\end{array}\right]& \\ |1& 1)& [\begin{array}{cc}1& 3\\ 1& 2\end{array}\left|\{1\begin{array}{c}1\\ 1\end{array}\right|\end{array}$

The optimal strategy for Tony$=\left(\frac{2}{7},\frac{5}{7}\right)$

Therefore, the optimal strategies for Tony are$y−\left(\frac{2}{7},\frac{5}{7}\right)$and for Joey is $x−\left(\frac{3}{5},\frac{2}{5}\right)$.

The value of the game is denoted by V. Consider the mixed strategy of Joey and Tony.

$V=\left[\begin{array}{cc}\frac{2}{7}& \frac{5}{7}\end{array}\right]×\left[\begin{array}{cc}2& −3\\ −1& 1\end{array}\right]×\left[\begin{array}{c}\frac{2}{5}\\ \frac{3}{5}\end{array}\right]$

$V=\frac{1}{5}$

Therefore, the value of the game is $\frac{1}{5}$