Further Mathematics · OCR H245 · Option D · Topic 7.08

Game theory

Pay-off matrix, pure strategies, mixed strategies.

  • 3 sub-topics
  • Taught in Year 13
  • 68 practice questions
  • 3 animated lessons
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What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 7.08a-bPay-off matrixY13
  2. 7.08c-dPure strategiesY13
  3. 7.08e-fMixed strategiesY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 68 questions in this topic, free, no account needed.

Game theory

Not needed for Year 12Easy

Zero-sum games · Year 13 · 1 mark

  1. What makes a two-person game zero-sum?
    (1)

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  1. One pay-off matrix is therefore enough: the other player's pay-offs are its negatives.A: Whatever one player gains, the other loses

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Game theory

Not needed for Year 12Medium

Two-person zero-sum games · Year 13 · 5 marks

The pay-off matrix shows A's winnings in a zero-sum game between AA (rows A1–A3) and BB (columns B1–B3). (−14−53−26−2−3−4)\begin{pmatrix} -1 & 4 & -5 \\ 3 & -2 & 6 \\ -2 & -3 & -4 \end{pmatrix}
  1. (a)
    Explain what is meant by a zero-sum game.
    (1)
  2. (b)
    Find the play-safe strategy for each player.
    (3)
  3. (c)
    Write down the pay-off matrix for BB.
    (1)
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Part (a)1 mark

  1. Whatever one player wins, the other loses: the total winnings are zero.

Part (b)3 marks

  1. Row minima: -5, -2, -4; column maxima: 3, 4, 6.
  2. Maximise the row minimum; minimise the column maximum.A plays A2; B plays B1.

Part (c)1 mark

  1. Transpose and change every sign.(1−32−4235−64)\begin{pmatrix} 1 & -3 & 2 \\ -4 & 2 & 3 \\ 5 & -6 & 4 \end{pmatrix}

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Game theory

Not needed for Year 12Hard

Mixed strategies: graphical method · Year 13 · 6 marks

A zero-sum game has pay-off matrix for A (rows A1, A2; columns B1, B2, B3): (500−342)\begin{pmatrix} 5 & 0 & 0 \\ -3 & 4 & 2 \end{pmatrix}. A plays A1 with probability p and A2 with probability 1 − p.
  1. (a)
    Show that there is no stable solution.
    (1)
  2. (b)
    Find A's expected pay-off for each of B's strategies, in terms of pp.
    (2)
  3. (c)
    Use a graphical method to find the optimal value of pp and the value of the game.
    (3)

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Part (a)1 mark

  1. Max of row minima < min of column maxima.

Part (b)2 marks

  1. B1:8pB1: 8p−3, B2:{}- 3, \ B2:−4p{}-4p+4, B3:{}+ 4, \ B3:−2p{}-2p+2{}+ 2

Part (c)3 marks

  1. Draw the three lines for 0≤p≤10 \le p \le 1; A maximises the lowest line.
  2. The highest point of the lower boundary.p=12p = \frac{1}{2}
  3. value=1\text{value} = 1

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