Let's solve the given game graphically in **2 steps**, with explanations after each step, and then give the **final answer**. ### **Step 1: Set up Player A's mixed strategy** Let Player A play \(A_1\) with probability \(p\) and \(A_2\) with probability \(1-p\). The expected payoffs for Player A against each of Player B's strategies are: - **Against \(B_1\):** \[ 2p + 3(1-p) = 2p + 3 - 3p = 3 - p \] - **Against \(B_2\):** \[ -3p + 3(1-p) = -3p + 3 - 3p = 3 - 6p \] - **Against \(B_3\):** \[ 8p + (-6)(1-p) = 8p - 6 + 6p = 14p - 6 \] **Explanation:** We express Player A’s expected payoffs as functions of \(p\) (the probability of choosing \(A_1\)). These are the lines we will plot on a graph for \(0 \leq p \leq 1\). --- ### **Step 2: Graphically identify the optimal strategy** Now, plot the following three lines for \(0 \leq p \leq 1\): - \(y_1 = 3 - p\) - \(y_2 = 3 - 6p\) - \(y_3 = 14p - 6\) We must find the value of \(p\) where the **minimum** of these three payoffs is **maximized** (maximin). This point is where two of these lines intersect and are both lowest at that point. #### Find intersection points: - Set \(3 - 6p = 14p - 6\): \[ 3 - 6p = 14p - 6 \implies 3 + 6 = 14p + 6p \implies 9 = 20p \implies p = \frac{9}{20} \] At \(p = 9/20\): \[ y_2 = 3 - 6 \times \frac{9}{20} = 3 - \frac{54}{20} = 3 - 2.7 = 0.3 \] \[ y_3 = 14 \times \frac{9}{20} - 6 = \frac{126}{20} - 6 = 6.3 - 6 = 0.3 \] - Set \(3 - p = 3 - 6p\): \[ 3 - p = 3 - 6p \implies -p = -6p \implies 5p = 0 \implies p = 0 \] At \(p = 0\): both are 3. - Set \(3 - p = 14p - 6\): \[ 3 - p = 14p - 6 \implies 3 + 6 = 14p + p \implies 9 = 15p \implies p = \frac{3}{5} \] At \(p = 3/5\): \[ y_1 = 3 - \frac{3}{5} = 2.4 \] \[ y_3 = 14 \times \frac{3}{5} - 6 = 8.4 - 6 = 2.4 \] Now, **maximin** happens at the intersection of \(y_2\) and \(y_3\), which is at \(p = 9/20\). **Explanation:** By finding where the lowest expected payoff is highest, we determine the optimal mixed strategy. The intersection point where two lines (corresponding to B’s strategies) cross and are both lowest is the equilibrium. --- ## **Final Answer** - Player A should play \(A_1\) with probability \(\boxed{\frac{9}{20}}\) and \(A_2\) with probability \(\boxed{\frac{11}{20}}\). - The **value of the game** is \(\boxed{0.3}\). --- **Summary:** - **Step 1:** Write expected payoffs as functions of \(p\). - **Step 2:** Find the \(p\) where the minimum payoff is maximized (intersection of lowest lines). - **Final Answer:** Optimal strategy is \(A_1: \frac{9}{20}\), \(A_2: \frac{11}{20}\), with game value \(0.3\).