Math Problem Statement

Find the optimal strategies, P and Q, for the row and column players, respectively.

6 0

2 4

Solution

To find the optimal strategies for the row player (PP) and the column player (QQ) in a two-person zero-sum game, we apply the concept of mixed strategy equilibria. For the given payoff matrix:

6 & 0 \\ 2 & 4 \end{bmatrix}$$ ### Step 1: Formulate the problem Let $$P = [p_1, p_2]$$ be the row player's mixed strategy, where $$p_1 + p_2 = 1$$, and $$p_1, p_2 \geq 0$$. Similarly, let $$Q = [q_1, q_2]$$ be the column player's mixed strategy, where $$q_1 + q_2 = 1$$, and $$q_1, q_2 \geq 0$$. The expected payoff for the row player is: $$E = p_1 (6q_1 + 0q_2) + p_2 (2q_1 + 4q_2)$$ The column player minimizes this payoff by choosing $$q_1, q_2$$, while the row player maximizes it by choosing $$p_1, p_2$$. ### Step 2: Solve for the value of the game ($$V$$) Using linear programming or graphical methods, we solve for $$V$$, the game's value. #### Row Player's Constraints: 1. For column player's pure strategies, set payoffs equal: $$6q_1 + 0q_2 = 2q_1 + 4q_2$$ Solve: $$6q_1 = 2q_1 + 4q_2 \implies 4q_1 = 4q_2 \implies q_1 = q_2$$ Since $$q_1 + q_2 = 1$$, we have $$q_1 = q_2 = \frac{1}{2}$$. #### Column Player's Constraints: Similarly, for row player's mixed strategy, equate payoffs for pure strategies: $$6p_1 + 2p_2 = 0p_1 + 4p_2$$ Solve: $$6p_1 + 2p_2 = 4p_2 \implies 6p_1 = 2p_2 \implies 3p_1 = p_2$$ Since $$p_1 + p_2 = 1$$: $$p_1 + 3p_1 = 1 \implies 4p_1 = 1 \implies p_1 = \frac{1}{4}, \quad p_2 = \frac{3}{4}.$$ ### Step 3: Compute the value of the game ($$V$$) The value of the game $$V$$ is the expected payoff when both players use their optimal strategies: $$V = 6p_1q_1 + 0p_1q_2 + 2p_2q_1 + 4p_2q_2$$ Substitute $$p_1 = \frac{1}{4}, p_2 = \frac{3}{4}, q_1 = \frac{1}{2}, q_2 = \frac{1}{2}$$: $$V = 6 \cdot \frac{1}{4} \cdot \frac{1}{2} + 0 + 2 \cdot \frac{3}{4} \cdot \frac{1}{2} + 4 \cdot \frac{3}{4} \cdot \frac{1}{2}$$ Simplify: $$V = \frac{3}{4} + \frac{3}{4} + \frac{6}{4} = \frac{12}{4} = 3.$$ ### Final Answer: - **Row Player's Optimal Strategy:** $$P = \left[\frac{1}{4}, \frac{3}{4}\right]$$ - **Column Player's Optimal Strategy:** $$Q = \left[\frac{1}{2}, \frac{1}{2}\right]$$ - **Value of the Game:** $$V = 3$$ Would you like details or explanations for any step? Here are 5 relative questions to consider: 1. How are the optimal strategies related to the Nash equilibrium? 2. What happens if the payoff matrix is modified slightly? 3. Can these strategies be applied to non-zero-sum games? 4. How does the value of the game impact real-world decision-making? 5. Are there faster computational methods for solving larger payoff matrices? **Tip:** Always check if the game is symmetric; it can simplify computations.

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Math Problem Analysis

Mathematical Concepts

Game Theory
Mixed Strategy Equilibrium
Linear Programming

Formulas

Expected payoff: E = p1(a11q1 + a12q2) + p2(a21q1 + a22q2)
Probability constraints: p1 + p2 = 1, q1 + q2 = 1
Payoff equality for optimal strategies

Theorems

Nash Equilibrium
Minimax Theorem

Suitable Grade Level

Grades 11-12, College Level