D2 June 2013 (R) Q4
4. Robin (R) and Steve (S) play a two-person zero-sum game which is represented by the following pay-off matrix for Robin.
| S plays 1 | S plays 2 | S plays 3 | |
|---|---|---|---|
| R plays 1 | 2 | 1 | 3 |
| R plays 2 | 1 | –1 | 2 |
| R plays 3 | –1 | 3 | –3 |
Find the best strategy for Robin and the value of the game to him. (9)
| Scheme | Marks | ||||||
|---|---|---|---|---|---|---|---|
R1 dominates R2, so deleted R2 to give
| B1 | ||||||
| If S plays 1; R’s gain is \(2p - (1-p) = 3p - 1\) | M1 | ||||||
| If S plays 2; R’s gain is \(p + 3(1-p) = 3 - 2p\) | |||||||
| If S plays 3; R’s gain is \(3p - 3(1-p) = 6p - 3\) | A1 | ||||||
 | B1ft B1 | ||||||
| \(3 - 2p = 3p - 1\) giving \(p = \dfrac{4}{5}\) | M1 A1 | ||||||
| Robin should play R1 with probability \(\dfrac{4}{5}\) R2 never R3 with probability \(\dfrac{1}{5}\) | A1ft | ||||||
| The value of the game is \(\dfrac{7}{5}\) to Robin | A1 | ||||||
| 9 marks |
Notes
1B1: CAO
1M1: Setting up three probability expressions, implicit definition of ‘\(p\)’.
1A1: CAO (condone incorrect simplification)
2B1ft: Attempt at three lines (correct gradients and correct order of intersection with ‘axes’), accept \(p \gt 1\) or \(p \lt 0\) here. Must be functions of \(p\).
3B1: CAO \(0 \leqslant p \leqslant 1\), scale clear (or 1 line = 1), condone lack of labels. Rulers used.
2M1: Finding their correct optimal point, must have three lines and three intersection points and set up an equation to find \(0 \leqslant p \leqslant 1\). Dependent on the second B mark being earned. Solving all three simultaneous equations only is M0.
2A1: CSO (all previous marks must have been awarded)
3A1ft: All three options listed must ft from their \(p\), check page 1 for R should never play 2. \(0 \leqslant \text{probabilities} \leqslant 1\) Dependent on both previous M marks being awarded.
4A1: CAO for the value of the game \(\left(\dfrac{7}{5}\right)\)