Question

Solve the following game by graphical method.

\begin{bmatrix} -2&0 \\ 3&-1 \\ -3&2 \\ 5&-4 \end{bmatrix}

14 Feb 2018
Answer :
Word Count : 601
We are asked to solve the 4×2 game with payoff matrix: $$ A = \begin{bmatrix} -2 & 0 \\ 3 & -1 \\ -3 & 2 \\ 5 & -4 \end{bmatrix} $$ using the graphical method. --- Step 1: Let player A use mixed strategies. Let the probabilities for player A choosing rows $R_1, R_2, R_3, R_4$ be $x_1, x_2, x_3, x_4 \ge 0$, with $$ x_1 + x_2 + x_3 + x_4 = 1 $$ Let the probability for player B choosing columns $C_1, C_2$ be $y$ for $C_1$ and $1-y$ for $C_2$. --- Step 2: Express expected payoff for player A. If A plays row $i$, expected payoff against B’s mixed strategy $y$ is: $$ E(R_i) = \text{payoff in column 1} \cdot y + \text{payoff in column 2} \cdot (1-y) $$ So: * $R_1: E_1 = -2y + 0(1-y) = -2y$ * $R_2: E_2 = 3y + (-1)(1-y) = 3y -1 + y = 4y -1$ * $R_3: E_3 = -3y + 2(1-y) = -3y +2 -2y = -5y +2$ * $R_4: E_4 = 5y + (-4)(1-y) = 5y -4 +4y = 9y -4$ --- Step 3: Plot these lines vs $y$ and find upper envelope. We can find the intersection points of these lines to determine optimal mixed strategy for B (minimizing A’s payoff). Step 3a: Find intersections $E_i = E_j$ 1. $E_1 = E_2: -2y = 4y -1 \implies 6y = 1 \implies y = 1/6$
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