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    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

    1. 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
    1. 1Day 1-2: Master pure strategies, calculate row minimums and column maximums, verify saddle points, and interpret stable solutions.
    2. 2Day 3-4: Practise dominance reductions for both row and column players, ensuring clear justification for each eliminated row or column.
    3. 3Day 5-7: Solve 2x2 and 2xn games using expected payoff equations and graphical lower/upper envelope analysis.
    4. 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.
    5. 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)
    Determine

    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.

    Show that

    Provide a complete, step-by-step algebraic or numerical proof reaching the given result without skipping intermediate lines of calculation.

    Formulate

    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)
    Pitfall: Confusing Row and Column player perspectives when identifying dominated strategies or optimal mixed strategies.
    ❌ Weak Answer (Loses Marks):Row 2 dominates Row 1 because 4 is bigger than -2, and Column 1 dominates Column 2 because 5 is bigger than 1.
    Example improved answer:For the row player, strategy R2 dominates R1 if every payoff in R2 is greater than or equal to the corresponding payoff in R1, with at least one strictly greater. For the column player, payoffs represent losses; therefore, strategy C1 dominates C2 only if every entry in C1 is strictly less than or equal to the corresponding entry in C2.
    Examiner Tip: Always remember that the standard pay-off matrix represents payouts to the Row player. The Row player seeks to maximise payoffs, whereas the Column player seeks to minimise them.
    Pitfall: Failing to add a constant to the pay-off matrix before setting up a Linear Programming (LP) formulation when elements are zero or negative.
    ❌ Weak Answer (Loses Marks):Let the value of the game be V. Maximise P = V subject to p1 + 2p2 >= V and p1 + p2 = 1.
    Example improved answer:Because simplex requires non-negative variables and a positive game value V, add a suitable constant k to all matrix elements such that all entries are strictly positive (a_ij + k > 0). Let V_new = V + k. Formulate the LP using standard non-negative variables x_i = p_i / V_new, and state clearly that the true value of the original game is V = V_new - k.
    Examiner Tip: AQA mark schemes strictly penalise LP formulations that do not guarantee a strictly positive value of V when probabilities are transformed into simplex variables.
    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. 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. 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. 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. 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. 5.Step 5: Equate expected payoffs to find the optimal probability: 3p - 1 = 4 - 7p => 10p = 5 => p = 0.5.
    6. 6.Step 6: Calculate the value of the game to player A: V = 3(0.5) - 1 = 0.5.
    Final Answer: Player A's play-safe strategy is Row 2 and Player B's is Column 1. Since maximin (-1) != minimax (2), there is no stable solution. Player A's optimal mixed strategy is to play Row 1 with probability 0.5 and Row 2 with probability 0.5, giving a game value of 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. 1.Step 1: Let player A play A1 with probability p and A2 with probability (1 - p), with 0 <= p <= 1.
    2. 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. 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. 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. 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. 6.Step 6: Calculate the game value: V = 3(4/7) + 1 = 12/7 + 7/7 = 19/7.
    Final Answer: Player A should play A1 with probability 4/7 and A2 with probability 3/7. The value of the game to Player A is 19/7 (or 2.71 to 2 d.p.).
    Active Recall Memory Test
    What mathematical condition indicates that a zero-sum game has a saddle point (stable solution)?
    Key Fact: The maximin value for the row player equals the minimax value for the column player.
    For the column player, when does Strategy C1 dominate Strategy C2 in a standard payoff matrix?
    Key Fact: When every payoff entry in Column 1 is less than or equal to the corresponding entry in Column 2, with at least one entry strictly less.
    In a 2 x n graphical solution for Player A, which envelope must you identify to find the maximin point?
    Key Fact: The lower envelope (the highest point of the lower boundary of all lines).
    Why must a constant k be added to matrix elements when converting a game into a standard Linear Program?
    Key Fact: To ensure all payoff values are strictly positive, guaranteeing that the game value V > 0 so that variables x_i = p_i / V are valid and strictly non-negative.
    Frequently Asked Questions
    What is the difference between a pure strategy and a mixed strategy?
    A pure strategy involves a player consistently choosing one specific action with 100% certainty every round. A mixed strategy involves randomising choices between two or more pure strategies according to a calculated probability distribution. Players adopt mixed strategies when no stable pure equilibrium (saddle point) exists, preventing their opponent from exploiting predictable moves.
    How do I choose between the upper and lower envelope in graphical game theory?
    If you are solving for Player A (the row player, who has 2 strategies), Player A wants to maximise their minimum payoff, so you identify the lower envelope of lines and pick its highest point (maximin). If you are solving for Player B (the column player, who has 2 strategies), Player B wants to minimise their maximum loss, so you look at the upper envelope and pick its lowest point (minimax).
    Can a zero-sum game have more than one saddle point?
    Yes, a game can have multiple saddle points. When this occurs, every saddle point will have the exact same pay-off value, ensuring that the value of the game remains completely unique and unambiguous. Both players can choose between the corresponding play-safe strategies without affecting their expected return.
    What does a negative value of the game mean?
    Because the pay-off matrix represents returns to the row player, a negative game value means that under optimal play by both sides, the row player will expect to lose that amount on average per round, while the column player makes a corresponding net gain. A game is called 'fair' if and only if its value is zero.
    Why do we divide probabilities by V when setting up an LP formulation?
    In linear programming, the objective is to maximise the value V subject to expected payoff inequalities like sum(a_ij * p_i) >= V. Dividing both sides by V turns the constraints into sum(a_ij * x_i) >= 1, where x_i = p_i / V. Because sum(p_i) = 1, summing all x_i gives sum(x_i) = 1 / V. Maximising V is mathematically identical to minimising sum(x_i), producing a standard linear program.