Question

Find the value of the game specified by the following payoff matrix.

$\begin{array}{cccc}\mathbf{0}& \mathbf{0}& \mathbf{‐}\mathbf{1}& \mathbf{‐}\mathbf{1}\\ \mathbf{0}& \mathbf{1}& \mathbf{‐}\mathbf{2}& \mathbf{‐}\mathbf{1}\\ \mathbf{‐}\mathbf{1}& \mathbf{‐}\mathbf{1}& \mathbf{}\mathbf{}\mathbf{1}& \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\\ \mathbf{‐}\mathbf{1}& \mathbf{}\mathbf{}\mathbf{0}& \mathbf{}\mathbf{}\mathbf{0}& \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\\ \mathbf{1}& \mathbf{‐}\mathbf{2}& \mathbf{0}& \mathbf{‐}\mathbf{3}\\ \mathbf{1}& \mathbf{‐}\mathbf{1}& \mathbf{‐}\mathbf{1}& \mathbf{‐}\mathbf{1}\\ \mathbf{0}& \mathbf{‐}\mathbf{3}& \mathbf{2}& \mathbf{‐}\mathbf{1}\\ \mathbf{0}& \mathbf{‐}\mathbf{2}& \mathbf{1}& \mathbf{‐}\mathbf{1}\end{array}$

(Hint: Consider the mixed strategies $\left(\frac{1}{3},0,0,\frac{1}{2},\frac{1}{6,}0,0,0\right)$and )$\left(\frac{2}{3},0,0,\frac{1}{3}\right)$)

Short answer

The value of the game is$-1$ by reducing dominance. Otherwise, without a reduction solution for this payoff matrix is not possible.

Step-by-step solution

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. The average payoff is represented as follows,

$\mathbf{\sum }_{\mathbf{i}\mathbf{,}\mathbf{j}}{\mathit{G}}_{\mathbf{I}\mathbf{j}}\mathbf{.}\mathit{P}\left[Rows,column\right]\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{,}\mathbf{j}}{\mathit{G}}_{\mathbf{I}\mathbf{j}}{\mathit{x}}_{\mathbf{i}}{\mathit{y}}_{\mathbf{j}}$ ......(1)

Calculation of the game value

The given payoff matrix is more significant, with eight rows and four columns. The larger matrix is reduced by the dominance of the probability of rows and columns.

Sum the row and column values, and delete the rows and columns with the least values.

$$\begin{gathered} \left[\begin{array}{cccc}0& 0& ‐1& ‐1\\ 0& 1& ‐2& ‐1\\ ‐1& ‐1& 1& 1\\ ‐1& 0& 0& 1\\ 1& ‐2& 0& ‐3\\ 1& ‐1& ‐1& ‐1\\ 0& ‐3& 2& ‐1\\ 0& ‐2& 1& ‐1\end{array}\right]\begin{array}{c}‐2\\ ‐2\\ 0\\ 0\\ ‐4\\ ‐2\\ ‐2\\ ‐2\end{array} \\ 0‐80‐6 \end{gathered}$$

The reduced matrix after deleting the dominance is $\left[\begin{array}{cc}‐1& 1\\ ‐1& 0\end{array}\right]$

The row min max of the matrix $\left[\begin{array}{cc}‐1& 1\\ ‐1& 0\end{array}\right]$is -1, The column max min of the matrix is -1.

$$\begin{gathered} rowminmax=columnmaxmin \\ -1=-1 \end{gathered}$$

The above value is denoted as, $G_{ij}=-1$ ......(2)

From the given mixed strategies $\left(\frac{1}{3},0,0,\frac{1}{2},\frac{1}{6},0,0,0\right)$and $\left(\frac{2}{3},0,0,\frac{1}{3}\right),$

$$\begin{gathered} \frac{1}{3}+0+0+\frac{1}{2}+\frac{1}{6}+0+0+0=1 \\ \frac{2}{3}+0+0+\frac{1}{3}=1 \end{gathered}$$

Substitute the values of mixed strategies sums and the value of Equation (2) in Equation (1).

$$\begin{gathered} \sum _{\mathrm{i},\mathrm{j}}{\mathrm{G}}_{\mathrm{Ij}}.\mathrm{P}[\mathrm{Rows},\mathrm{column}]=\sum _{\mathrm{i},\mathrm{j}}{\mathrm{G}}_{\mathrm{Ij}}{\mathrm{x}}_{\mathrm{i}}{\mathrm{y}}_{\mathrm{j}} \\ \underset{}{\sum (-1)\left(\frac{1}{3}+0+0+\frac{1}{2}+\frac{1}{6}+0+0+0\right).\left(\frac{2}{3}+0+0+\frac{1}{3}=1\right)=}\sum _{\mathrm{i},\mathrm{j}}{\mathrm{G}}_{\mathrm{Ij}}{\mathrm{x}}_{\mathrm{i}}{\mathrm{y}}_{\mathrm{j}} \end{gathered}$$

Solving the above equation,

$\sum _{\mathrm{i},\mathrm{j}}{\mathrm{G}}_{\mathrm{Ij}}{\mathrm{x}}_{\mathrm{i}}{\mathrm{y}}_{\mathrm{j}}=-1$

Therefore, the value of the game is -1 by reducing dominance. Otherwise, without a reduction solution, a payoff matrix is not possible.