Let's solve each part step by step: --- ## (i) **Nash Equilibrium (Pure Strategy)** Let’s write out the payoffs for both players: | | Up | Down | |---------|---------|---------| | **Left** | (3, 2) | (1, 5) | | **Right** | (4, 1) | (2, 3) | ### Find Best Responses #### Player A (Row Player): - If B plays **Up**: A chooses Right (4 > 3) - If B plays **Down**: A chooses Left (1 < 2), so A chooses Right #### Player B (Column Player): - If A plays **Left**: B chooses Down (5 > 2) - If A plays **Right**: B chooses Down (3 > 1) **Therefore, (Right, Down) = (2, 3) is the only pure-strategy Nash equilibrium.** --- ## (ii) **Mixed Strategy Setup** Let: - Player A plays **Left** with probability \( p \), **Right** with \( 1-p \). - Player B plays **Up** with probability \( q \), **Down** with \( 1-q \). --- ## (iii) **Find Mixed Strategy Nash Equilibrium** Each player must be indifferent between their two strategies. ### For Player A (Row): Expected payoff for **Left**: \[ E_A(\text{Left}) = 3q + 1(1-q) = 3q + 1 - q = 1 + 2q \] Expected payoff for **Right**: \[ E_A(\text{Right}) = 4q + 2(1-q) = 4q + 2 - 2q = 2 + 2q \] Set equal for indifference: \[ 1 + 2q = 2 + 2q \implies 1 = 2 \] This is a contradiction, which implies **Player A will never mix**; A will always play **Right**. ### For Player B (Column): Expected payoff for **Up**: \[ E_B(\text{Up}) = 2p + 1(1-p) = 2p + 1 - p = 1 + p \] Expected payoff for **Down**: \[ E_B(\text{Down}) = 5p + 3(1-p) = 5p + 3 - 3p = 3 + 2p \] Set equal for indifference: \[ 1 + p = 3 + 2p \implies 1 + p = 3 + 2p \implies 1 - 3 = 2p - p \implies -2 = p \] But probability \( p \) cannot be negative. So **Player B will always play Down**. --- **So, the only Nash Equilibrium is (Right, Down), i.e., pure strategy.** --- ## (iv) **Expected Payoff** At equilibrium (Right, Down): - Player A gets **2** - Player B gets **3** **Expected payoff:** (2, 3) --- ### **Summary Table** | Part | Answer | |------|--------| | (i) | Yes, (Right, Down) is the only Nash Equilibrium (pure strategy) | | (ii) | Mixed strategies do not exist here as equilibria; only pure strategies do. | | (iii)| \( p = \) (A always plays Right), \( q = \) (B always plays Down) | | (iv) | Expected payoff: (2, 3) | --- Let me know if you would like the analysis for other cases or more details!

# Summary of Game Analysis ## (i) Nash Equilibrium (Pure Strategy) The pure-strategy Nash equilibrium for the game is (Right, Down) with payoffs (2, 3). Player A prefers Right regardless of Player B’s choice, while Player B always opts for Down to maximize their payoff against A's strategies. ## (ii) Mixed Strategy Setup In the mixed strategy setup, Player A plays Left with probability \( p \) and Right with probability \( 1-p \). Player B plays Up with probability \( q \) and Down with probability \( 1-q \). The mixed strategies aim to create indifference between choices for both players. ## (iii) Find Mixed Strategy Nash Equilibrium Analysis reveals that Player A will always play Right, leading to no mixed strategy equilibrium for A. Similarly, Player B will always play Down. Therefore, the mixed strategies do not yield an equilibrium as both players have dominant strategies. ## (iv) Expected Payoff At the equilibrium (Right, Down), Player A receives an expected payoff of 2, while Player B receives 3. Thus, the expected payoff of the game at this equilibrium is (2, 3), reflecting the final outcomes of their strategies.