D2 June 2011 Q4
4. Laura and Sam play a zero-sum game. This game is represented by the following pay-off matrix for Laura.
| S plays 1 | S plays 2 | S plays 3 | |
|---|---|---|---|
| L plays 1 | −4 | −1 | 1 |
| L plays 2 | 3 | −1 | −2 |
| L plays 3 | −3 | 0 | 2 |
Find the best strategy for Laura and the value of the game to her. (9)
| Scheme | Marks | ||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
(a)
| M1 | ||||||||||||||||||||||||||||
| Let Laura play 2 with probability \(p\) and 3 with probability \((1 - p)\) If Sam plays 1: Laura’s gain is \(3p - 3(1 - p) = -3 + 6p\) If Sam plays 2: Laura’s gain is \(-p + 0(1 - p) = -p\) If Sam plays 3: Laura’s gain is \(-2p + 2(1 - p) = 2 - 4p\) | M1 A1 | ||||||||||||||||||||||||||||
| (3) | |||||||||||||||||||||||||||||
(b)![]() | B2,1ft,0 | ||||||||||||||||||||||||||||
| (2) | |||||||||||||||||||||||||||||
| (c) \(-3 + 6p = -p\) \(7p = 3\) \(p = \dfrac{3}{7}\) | M1 A1 | ||||||||||||||||||||||||||||
| Laura should play row 1: never, row 2: \(\dfrac{3}{7}\) of the time and row 3: \(\dfrac{4}{7}\) of the time | A1ft | ||||||||||||||||||||||||||||
| and the value of the game is \(-\dfrac{3}{7}\) to her. | A1 | ||||||||||||||||||||||||||||
| (4) | |||||||||||||||||||||||||||||
| (9 marks) |
Notes
(a) 1M1: Matrix reduced correctly. Could be implicit from equations.
2M1: Setting up three probability equations, implicit definition of p.
1A1: CAO
(b) 1B1ft: At least two lines correct, accept \(p > 1\) or \(p < 0\) here. Must both be function of p.
2B1: 3 lines cao, \(0 \leqslant p \leqslant 1\), scale clear (or 1 line = 1), condone lack of labels. Rulers used.
(c) 3M1: Finding their correct optimal point, must have three lines, and setting up an equation to find \(0 \leqslant p \leqslant 1\).
1A1: CAO
2A1ft: All three options listed must ft from their p, check page 1, no negatives.
3A1: CAO
