D2 June 2014 Q4
4. A two-person zero-sum game is represented by the following pay-off matrix for player A.
| B plays 1 | B plays 2 | B plays 3 | B plays 4 | |
|---|---|---|---|---|
| A plays 1 | 2 | –1 | 1 | –3 |
| A plays 2 | –3 | 2 | –2 | 1 |
| Scheme | Marks |
|---|---|
| Row mins {−3, −3} Column max {2, 2, 1, 1} | M1 |
| Row maximin (−3) \(\neq\) column minmax (1) so not stable | A1 |
| (2) |
Notes
a1M1: Finding row minimums and column maximums – condone one error.
a1A1: CAO states \(-3 \neq 1\) (or row (maximin) \(\neq\) col (minimax)) and draws the conclusion.
| Scheme | Marks | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
Column 4 dominates column 2 so delete column 2 or if B plays 2 A’s expected winnings are \(-p + 2(1-p)\ (= 2 - 3p)\)
| B1 | ||||||||||||
| Let A play 1 with probability \(p\) and 2 with probability \(1-p\) | B1 | ||||||||||||
| If B plays 1 A’s expected winnings are \(2p - 3(1-p) = 5p - 3\) If B plays 3 A’s expected winnings are \(p - 2(1-p) = 3p - 2\) If B plays 4 A’s expected winnings are \(-3p + (1-p) = 1 - 4p\) | M1 A1 | ||||||||||||
 | M1 A1 | ||||||||||||
| \(5p - 3 = 1 - 4p\) | M1 | ||||||||||||
| \(p = \dfrac{4}{9}\) | A1 | ||||||||||||
| A should play row 1 with probability \(\dfrac{4}{9}\) and row 2 with probability \(\dfrac{5}{9}\) | A1 | ||||||||||||
| (9) | |||||||||||||
| 11 marks |
Notes
b1B1: CAO Col 4 dominates Col 2 (maybe implied by later working) or correctly stating the expression for A’s expected winnings if B plays 2 \((2 - 3p)\).
b2B1: Defines \(p\). Allow those who only define that A plays 1 with prob. \(p\) – no incorrect statements be generous.
b1M1: Setting up three probability equations, implicit definition of \(p\).
b1A1: CAO (condone incorrect simplification).
b2M1: Either attempt at three lines (correct slant direction and relative intersection with ‘axes’) or four lines if no earlier domination, accept \(p \gt 1\) or \(p \lt 0\) here. Must be functions of \(p\).
b2A1: CAO \(0 \leqslant p \leqslant 1\), scaling correct and clear (or 1 line = 1), condone lack of labels. Rulers used.
b3DM1: Finding their correct optimal point, must have three (or four) lines and set up an equation to find \(0 \leqslant p \leqslant 1\). Dependent on previous M mark. Must have at least three intersection points. Solving all three simultaneous equations and stating incorrect \(p\) is M0.
b3A1: CAO (must have scored all marks except b2B1 (define \(p\) mark) in this part).
b4A1: CAO
SC1: If column 4 is deleted in (b) candidates can earn a maximum of B0 B1 M1 A0 M1 A0 M1 A0 A1 (max. of 5 out of 9 in part b). The final A mark is for ‘A should play row 1 with prob. \(\frac{2}{3}\) and row 2 with prob. \(\frac{1}{3}\)’.
SC2: If column 1 or 3 is deleted in (b), candidates can earn a maximum of B0 B1 M1 A0 M1 A0 M0 A0 A0 (max. of 3 out of 9 in part b)