DF: Game theory for zero-sum games — AQA A-Level Further Mathematics
Teaching note based on the AQA specification.
In AQA A-Level Further Mathematics, a two-person zero-sum game models conflict where one player's gain strictly equals the other's loss. Students learn to determine play-safe strategies, identify saddle points, apply dominance, and solve mixed-strategy games using graphical analysis and linear programming.
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Your focus
- Understand, interpret and construct pay-off matrices.
Which command word is this?
Show sufficient mathematical working or logical deduction to arrive at the solution. Stating the answer without showing row minima / column maxima or expected payoff equations will lose marks.
DF: Game theory for zero-sum games exam tips
Topic Overview
Game theory in AQA Further Mathematics focuses on two-person zero-sum games, where the total sum of gains and losses across both participants is always zero. The topic explores decision-making under competition, starting with pure strategies, play-safe criteria (maximin and minimax), and identifying stable equilibrium points known as saddle points.
When a pure strategy equilibrium does not exist, the topic expands to probabilistic decision-making via mixed strategies. Students master reduction techniques via dominance, solve 2 x n and n x 2 games geometrically using boundary envelopes, and formulate larger, complex pay-off matrices into linear programming models solvable via the Simplex algorithm.
Key Concepts
- →Zero-Sum Principle: The algebraic sum of payoffs to both players equals zero for every combination of strategies, meaning one player's gain is identically the other player's loss.
- →Play-Safe Criterion and Saddle Points: The row player chooses the maximum of row minimums (maximin), while the column player chooses the minimum of column maximums (minimax). If maximin equals minimax, the game has a stable solution (saddle point).
- →Dominance Rule: A strategy can be permanently deleted if another available strategy yields an equal or strictly better payoff under all opposing moves (higher payoffs for Row, lower payoffs for Column).
- →Mixed Strategy and Envelope Method: When no saddle point exists, players randomise choices with specific probabilities to maximise their minimum expected payoff, solved graphically using upper or lower boundary envelopes.
- →Linear Programming Conversion: Larger m x n games are converted into linear programming problems by defining non-negative scaled variables x_i = p_i / V after adding a constant k to ensure all payoffs are strictly positive.
Examiner Tips
- 💡Always check for a saddle point first before undertaking algebraic or graphical mixed-strategy methods; if maximin equals minimax, you can state the stable solution and game value immediately.
- 💡Clearly label axes (probability p from 0 to 1 on the horizontal, expected payoff on the vertical) and state line equations explicitly when sketching graphical solutions to guarantee full method marks.
- 💡When defining variables in Linear Programming formulations, always state the relationship between probabilities, game value, and variables clearly (e.g., x_i = p_i / V with p_i >= 0, sum p_i = 1).
Common Mistakes
- Assuming Column dominance follows the same rule as Row dominance: Students often forget that a pay-off matrix shows gains for Row and losses for Column. Therefore, Column B1 dominates Column B2 if all entries in B1 are less than or equal to B2.
- Selecting the upper envelope for the Row player in graphical methods: The Row player wants to maximise their worst-case outcome, so students must locate the highest point on the LOWER envelope, not the upper envelope.
- Forgetting to adjust the final game value after an LP transformation: If a constant k was added to the matrix to guarantee all entries are positive, the true value of the original game must be calculated as V = (1 / sum(x_i)) - k.
Revision Plan
- 1Day 1-2: Master pure strategies, calculate row minimums and column maximums, verify saddle points, and interpret stable solutions.
- 2Day 3-4: Practise dominance reductions for both row and column players, ensuring clear justification for each eliminated row or column.
- 3Day 5-7: Solve 2x2 and 2xn games using expected payoff equations and graphical lower/upper envelope analysis.
- 4Day 8-10: Work through Linear Programming formulations of 3x3 games, focusing on adding constant k, setting up constraints, and finding the value of the game.
- 5Day 11-14: Complete full AQA past-paper questions under timed conditions, checking against official mark schemes for correct terminology.
Exam Question Types
- 📋Pay-off Matrix Analysis & Saddle Point: Find row minimums and column maximums, deduce play-safe strategies, and determine whether a stable solution exists.
- 📋Dominance and Reduction: Reduce an m x n matrix by sequentially eliminating dominated rows and columns, giving clear justifications.
- 📋Graphical Solution (2 x n or m x 2): Formulate expected payoffs in terms of probability p, plot lines, highlight the envelope, and calculate optimal p and game value V.
- 📋Linear Programming Formulation: Formulate a game as an LP problem for the Simplex method, including shifting payoffs by constant k and defining variables.
Command Word Expectations (AQA)
Show sufficient mathematical working or logical deduction to arrive at the solution. Stating the answer without showing row minima / column maxima or expected payoff equations will lose marks.
Provide a complete, step-by-step algebraic or numerical proof reaching the given result without skipping intermediate lines of calculation.
Translate the game theory scenario into a fully specified mathematical model (e.g., LP problem). You must define all variables, state the objective function clearly, and list all constraints including non-negativity.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: Two players, A and B, play a zero-sum game with pay-off matrix for player A given by: Row 1: [2, -3], Row 2: [-1, 4]. Determine the play-safe strategies for both players, demonstrate that there is no stable solution, and find the optimal mixed strategy for player A.
- 1.Step 1: Determine row minimums and column maximums to find play-safe strategies. Row minimums: Row 1 = -3, Row 2 = -1. Maximin value = max(-3, -1) = -1 (Row 2). Column maximums: Column 1 = 2, Column 2 = 4. Minimax value = min(2, 4) = 2 (Column 1).
- 2.Step 2: Compare maximin and minimax. Maximin (-1) does not equal Minimax (2), so there is no saddle point and the game has no stable pure strategy solution.
- 3.Step 3: Define player A's mixed strategy. Let player A play Row 1 with probability p and Row 2 with probability (1 - p), where 0 <= p <= 1.
- 4.Step 4: Express expected payoffs against player B's choices. If B plays Col 1: E_1(p) = 2p - 1(1 - p) = 3p - 1. If B plays Col 2: E_2(p) = -3p + 4(1 - p) = 4 - 7p.
- 5.Step 5: Equate expected payoffs to find the optimal probability: 3p - 1 = 4 - 7p => 10p = 5 => p = 0.5.
- 6.Step 6: Calculate the value of the game to player A: V = 3(0.5) - 1 = 0.5.
Question: Player A chooses between strategies A1 and A2. Player B chooses between B1, B2, and B3. The pay-off matrix for A is: A1: [4, 1, 6], A2: [1, 5, 2]. Use a graphical method to determine the optimal mixed strategy for player A and find the value of the game.
- 1.Step 1: Let player A play A1 with probability p and A2 with probability (1 - p), with 0 <= p <= 1.
- 2.Step 2: Formulate expected payoffs against each of player B's strategies: Against B1: E_1 = 4p + 1(1 - p) = 3p + 1. Against B2: E_2 = 1p + 5(1 - p) = 5 - 4p. Against B3: E_3 = 6p + 2(1 - p) = 4p + 2.
- 3.Step 3: Sketch the lines E_1, E_2, E_3 for 0 <= p <= 1. At p=0: E_1=1, E_2=5, E_3=2. At p=1: E_1=4, E_2=1, E_3=6.
- 4.Step 4: Identify the lower boundary (the worst-case expected payoff for A for any choice of p). The lower envelope is formed by the intersection of E_1 (3p + 1) and E_2 (5 - 4p) because E_3 lies strictly above E_1 for all p in [0, 1] (B3 is dominated by B1 from B's perspective).
- 5.Step 5: Find the maximum point of this lower envelope by equating E_1 and E_2: 3p + 1 = 5 - 4p => 7p = 4 => p = 4/7.
- 6.Step 6: Calculate the game value: V = 3(4/7) + 1 = 12/7 + 7/7 = 19/7.