D2 June 2017 Q3
3. 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 | 0 | –2 | 6 |
| A plays 2 | 3 | 4 | 1 |
| A plays 3 | –1 | 1 | –3 |
| Scheme | Marks |
|---|---|
| Row minima: -2, 1, -3 max is 1 Column maxima: 3, 4, 6 min is 3 | M1 A1 |
| Play safe is A plays 2 and B plays 1 | A1 |
| (3) |
Notes
a1M1: Clear attempt to find the Row maximin and Column minimax (either the Row minimums or Column maximums correct or at least four (of the six) values stated correctly) – some candidates are removing row 3 at this stage which is fine – they will therefore need to find at least four (of the five) correct values for this mark to be awarded
a1A1: Correct Row maximin and Column minimax (dependent on all row mins and column maxs correct) – these could either be stated or clearly shown
a2A1: Correct play safe for A (2) and B (1) – not dependent on the previous A mark
| Scheme | Marks |
|---|---|
| Row maximin (1) \(\neq\) Column minimax (3) so not stable | B1 |
| (1) |
Notes
b1DB1: CAO (dependent on all rowmins and colmaxs correct) states \(1 \neq 3\) (or row (maximin) \(\neq\) col (minimax) as long as 1 is clearly identified as the row maximin and 3 as the column minimax) and draws the correct conclusion
| Scheme | Marks |
|---|---|
| Row 2 dominates row 3 so delete row 3 | B1 |
| Let A play 1 with probability \(p\) and 2 with probability \(1-p\) | |
| If B plays 1 A’s expected winnings are \(3(1-p) = -3p+3\) If B plays 2 A’s expected winnings are \(-2p+4(1-p) = -6p+4\) If B plays 3 A’s expected winnings are \(6p+(1-p) = 5p+1\) | M1 A1 |
 | M1 A1 |
| \(5p+1 = 3-3p \;\Rightarrow\; p = \dfrac{1}{4}\) | DM1 A1 |
| A should play row 1 with probability \(\frac{1}{4}\), row 2 with probability \(\frac{3}{4}\) and row 3 never | A1 |
| (8) |
Notes
c1B1: CAO row 2 dominates row 3 (maybe implied by later working) – accept reduced matrix or ‘row 2 dominates row 3’ or row 3 crossed out
c1M1: Setting up three probability expressions (allow \(p-1\)), implicit definition of ‘\(p\)’
c1A1: CAO (condone incorrect simplification)
c2M1: Attempt at their three lines (correct slant direction and relative intersection with ‘axes’), accept \(p \gt 1\) or \(p \lt 0\) here but must go from axis to axis (give bod if close). Must be functions of \(p\)
c2A1: CAO \(0 \leqslant p \leqslant 1\), scaling correct and clear (expect to see 1 line = 1, although other scalings are acceptable eg 1 line = 2), condone lack of labels. Rulers used
c3DM1: Finding their correct optimal point, must have three 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
c3A1: CAO
c4A1: CSO (must have scored all previous marks in (c)) – all three options listed, check page 1 for A should never play 3
SC1: If row 2 is deleted in (c) candidates can earn a maximum in (c) and (d) of
(c) B0 M1 A0 M1 A0 M1 A0 A1 (d) B1 (max. of 5) – the final A mark is for A should play 2 never, play 1 and 3 with probability \(\frac{1}{2}\). The B mark in (d) is for \(\frac{1}{2}\)
SC2: If row 1 is deleted in (c) candidates can earn a maximum in (c) and (d) of
(c) B0 M1 A0 M1 A0 M0 A0 A0 (d) B0 (max. of 2)
If candidates remove a column then send to review
| Scheme | Marks |
|---|---|
| Value of the game to player B is \(-\dfrac{9}{4}\) | B1 |
| (1) | |
| 13 marks |
Notes
d1B1: CAO