Question

The pizza business in Little Town is split between two rivals, Tony and Joey. They are each investigating strategies to steal business away from the other. Joey is considering either lowering prices or cutting bigger slices. Tony is looking into starting up a line of gourmet pizzas, or offering outdoor seating, or giving free sodas at lunchtime. The effects of these various strategies are summarized in the following payoff matrix (entries are dozens of pizzas, Joey’s gain and Tony’s loss).

			TONY
		Gourmet	Seating	Freesoda
JOEY	Lower price	+2	0	-3
	BiggerSlices	_1	-2	+1

For instance, if Joey reduces prices and Tony goes with the gourmet option, then Tony will lose 2 dozen pizzas worth of nosiness to Joey.

What is the value of this game, and what are the optimal strategies for Tony and Joey?

Short answer

The value of this game is,$V=\frac{1}{5}$, The optimal strategies for Tony is$y-\left(\frac{2}{7},\frac{5}{7}\right)$ and for Joey is $x-\left(\frac{3}{5},\frac{2}{5}\right)$.

Step-by-step solution

Explain Payoff Matrix

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.

Calculate optimal strategies for Tony and Joey

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,

$\begin{array}{r}2x_1-x_2+z\le 0\\ -2x_2+z\le 0\\ -3x_1+x_2+z\le 0\end{array}$

Simplifying yields the following:

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

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,

$\begin{array}{r}2y_1-3y_3+w\ge 0\\ -1y_1-2y_2+y_3+w\ge 0\\ y_1+y_2+y_3=0\end{array}$

Simplifying yields the following:

$y_1,y_2,y_3\ge 0$.

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$=\frac{\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]\times G_{\mathrm{cof}}}{\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]\times G_{A\in fj}\times \left[\begin{array}{l}1\\ 1\end{array}\right]}$

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

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

The optimal strategy for Tony$=\frac{\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]\times G_{A\circ \delta j}}{\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]\times G_{A_{AJj}}\times \left[\begin{array}{l}1\\ 1\end{array}\right]}$

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 $x-\left(\frac{3}{5},\frac{2}{5}\right)$.

Calculate the value of the game.

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

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

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

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