1. **Problem Statement:** Simplify the given payoff matrix for Player A using the Rule of Dominance and find the optimal strategies for both players. 2. **Given Payoff Matrix:** $$\begin{array}{c|cccc} & B1 & B2 & B3 & B4 \\ \hline A1 & 4 & 6 & 2 & 4 \\ A2 & 6 & 8 & 6 & 7 \\ A3 & 5 & 6 & 5 & 5 \end{array}$$ 3. **Rule of Dominance:** - A strategy dominates another if it yields a better or equal payoff in every column and strictly better in at least one. - Dominated strategies can be eliminated to simplify the game. 4. **Check for Dominated Rows (Player A's strategies):** - Compare A1 and A3: - A1: (4,6,2,4) - A3: (5,6,5,5) - A3 is better or equal in every column and strictly better in columns 1,3,4. - So, A3 dominates A1. Eliminate A1. - Compare A2 and A3: - A2: (6,8,6,7) - A3: (5,6,5,5) - A2 is better in every column. So, A3 does not dominate A2. 5. **Check for Dominated Columns (Player B's strategies):** - Compare B3 and B4: - B3: (2,6,5) - B4: (4,7,5) - B4 is better or equal in every row and strictly better in rows 1 and 2. - So, B4 dominates B3. Eliminate B3. - Compare B1 and B2: - B1: (4,6,5) - B2: (6,8,6) - B2 is better in every row. So, B1 is dominated by B2. Eliminate B1. 6. **Reduced Matrix:** $$\begin{array}{c|cc} & B2 & B4 \\ \hline A2 & 8 & 7 \\ A3 & 6 & 5 \end{array}$$ 7. **Find Optimal Strategies:** - Now the game is 2x2. Use mixed strategies. - Let Player A play A2 with probability $p$ and A3 with $1-p$. - Let Player B play B2 with probability $q$ and B4 with $1-q$. - Expected payoff for Player A when Player B plays B2: $$E_A(B2) = 8p + 6(1-p) = 6 + 2p$$ - Expected payoff for Player A when Player B plays B4: $$E_A(B4) = 7p + 5(1-p) = 5 + 2p$$ - Player A wants to maximize the minimum expected payoff, so set these equal: $$6 + 2p = 5 + 2p$$ This is impossible unless payoffs are equal, so Player A will choose $p=0$ or $p=1$. - Check payoffs at $p=0$ (only A3): min payoff = $\min(6,5) = 5$ - Check payoffs at $p=1$ (only A2): min payoff = $\min(8,7) = 7$ - So Player A prefers $p=1$ (pure strategy A2). - For Player B, expected payoffs when Player A plays A2: $$E_B(A2) = 8q + 7(1-q) = 7 + q$$ - When Player A plays A3: $$E_B(A3) = 6q + 5(1-q) = 5 + q$$ - Player B wants to minimize Player A's payoff, so set equal: $$7 + q = 5 + q$$ No solution, so Player B chooses pure strategy with minimum payoff for Player A. - For B2 ($q=1$), payoff to A is $8$ or $6$. - For B4 ($q=0$), payoff to A is $7$ or $5$. - Player B prefers $q=0$ (pure strategy B4) to minimize payoff. 8. **Conclusion:** - Optimal strategies: Player A plays A2 pure, Player B plays B4 pure. - Value of the game: payoff at (A2, B4) = 7. **Final answer:** - Player A's optimal strategy: A2 - Player B's optimal strategy: B4 - Value of the game: 7