D2 June 2013 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 | |
|---|---|---|---|
| A plays 1 | 5 | 4 | −6 |
| A plays 2 | −1 | −2 | 3 |
| A plays 3 | 1 | −1 | 2 |
| Scheme | Marks |
|---|---|
| \(\begin{bmatrix}4 & -6 \\ -2 & 3 \\ -1 & 2\end{bmatrix}\) column 2 dominates column 1 | B1 |
| (1) |
Notes
a1B1 CAO (accept reduced matrix or ‘column 2 dominates column 1’ or column crossed out). Allow recover in part (b)
| Scheme | Marks |
|---|---|
| \(\begin{bmatrix}-4 & 2 & 1 \\ 6 & -3 & -2\end{bmatrix}\) | B1 B1 |
| (2) |
Notes
b1B1 either \(3 \times 2\) matrix with correct values (including signs) or \(2 \times 3\) matrix with correct values (condone incorrect signs)
b2B1 CAO
| Scheme | Marks |
|---|---|
| Let \(p\) = probability that B plays new row 1 If A plays 1: B’s expected winnings \(= -4p + 6(1 - p) = 6 - 10p\) If A plays 2: B’s expected winnings \(= 2p - 3(1 - p) = -3 + 5p\) If A plays 3: B’s expected winnings \(= p - 2(1 - p) = -2 + 3p\) | 1M1 A1 |
 | B2, 1ft, 0 |
| \(6 - 10p = -2 + 3p\) \(8 = 13p\) \(p = \dfrac{8}{13}\) | 2M1 A1 |
| B should play 1: never, play 2 with probability \(\dfrac{8}{13}\) and play 3 with probability \(\dfrac{5}{13}\) The value of the game is \(-\dfrac{2}{13}\) to B | B1 B1 |
| (8) | |
| (11 marks) |
Notes
c1M1 Setting up three probability expressions, implicit definition of ‘p’.
c1A1 CAO (condone incorrect simplification)
c1B1ft Attempt at three lines (correct gradients and intersection with ‘axes’), accept \(p > 1\) or \(p < 0\) here. Must be functions of p.
c2B1 CAO \(0 \leqslant p \leqslant 1\), scale clear (or 1 line = 1), condone lack of labels. Rulers used.
c2M1 Finding their correct optimal point, must have three lines and set up an equation to find \(0 \leqslant p \leqslant 1\). Dependent on first B mark in part (c). Must have three intersection points. Solving all three simultaneous equations only is M0.
c2A1 CSO
c3B1 All three options listed must ft from their p, check page 1 for B should never play 1. \(0 \leqslant \text{probabilities} \leqslant 1\).
c4B1 \(-2/13\) CAO (accept awrt 0.154)
SC1: If column 2 deleted in (a) candidates can earn a maximum of (a) B0 (b) B1 B0 (c) M1 A0 B1 B0 M1 A0 B1 B1 (max. of 6) – the final B mark is for the value of the game being −4/3
SC2: If column 3 is deleted in (a) candidates can earn a maximum of (a) B0 (b) B1 B0 (c) M1 A0 B1 B0 M0 A0 B0 B0