Let's solve part B step by step, **graphically** for the 2x3 game given (since part A already provides the solution). ### Payoff Matrix (Player A): | | B1 | B2 | B3 | |-------|----|-----|----| | **A1**| 2 | -3 | 8 | | **A2**| 3 | 3 | -6 | We are to find the optimal mixed strategy for Player A (let Player A play A1 with probability \( p \) and A2 with probability \( 1-p \)). --- ## **Step 1: Set up the expected payoff equations** Let \( p \) be the probability that Player A plays A1, and \( 1-p \) the probability for A2. Expected payoff for Player A if B plays: - **B1:** \( E_1 = 2p + 3(1-p) = 2p + 3 - 3p = 3 - p \) - **B2:** \( E_2 = -3p + 3(1-p) = -3p + 3 - 3p = 3 - 6p \) - **B3:** \( E_3 = 8p + (-6)(1-p) = 8p - 6 + 6p = 14p - 6 \) **Explanation:** We derived the expected payoff for each column as a function of \( p \), Player A's mixed strategy. --- ## **Step 2: Graph the payoff lines and find the optimal point** Now, we need to find the value of \( p \) that maximizes the minimum expected payoff (since Player A wants to maximize the minimum gain). Let's set up the three equations and look for intersection points (where two strategies give the same expected payoff): 1. \( 3 - p = 3 - 6p \) \( -p + 6p = 0 \implies 5p = 0 \implies p = 0 \) 2. \( 3 - p = 14p - 6 \) \( 3 + 6 = 14p + p \implies 9 = 15p \implies p = 0.6 \) 3. \( 3 - 6p = 14p - 6 \) \( 3 + 6 = 14p + 6p \implies 9 = 20p \implies p = 0.45 \) Evaluate each expected payoff at these \( p \) values: - **At \( p = 0 \):** - \( E_1 = 3 \) - \( E_2 = 3 \) - \( E_3 = -6 \) - **Minimum = -6** - **At \( p = 0.45 \):** - \( E_1 = 3 - 0.45 = 2.55 \) - \( E_2 = 3 - 6 \times 0.45 = 3 - 2.7 = 0.3 \) - \( E_3 = 14 \times 0.45 - 6 = 6.3 - 6 = 0.3 \) - **Minimum = 0.3** - **At \( p = 0.6 \):** - \( E_1 = 3 - 0.6 = 2.4 \) - \( E_2 = 3 - 6 \times 0.6 = 3 - 3.6 = -0.6 \) - \( E_3 = 14 \times 0.6 - 6 = 8.4 - 6 = 2.4 \) - **Minimum = -0.6** So, the **maximum of the minimum expected payoffs** is at \( p = 0.45 \), with minimum payoff **0.3**. **Explanation:** By equating the expected payoffs and evaluating the minimum at each intersection, we found the optimal probability where Player A maximizes their minimum gain. --- ### **Final Answer:** - **Player A should play A1 with probability \( 0.45 \) and A2 with probability \( 0.55 \)** - **The value of the game is \( 0.3 \)** --- If you need a graphical representation or want to see the lines plotted, let me know!