D2 June 2018 Q2
2. 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 | −3 | 2 | 5 | −1 |
| A plays 2 | −5 | 3 | 1 | −1 |
| A plays 3 | −2 | 5 | 4 | 2 |
| A plays 4 | 2 | −3 | −1 | 4 |
| −3 | 2 | 5 |
| −2 | 5 | 4 |
| 2 | −3 | −1 |
| Scheme | Marks |
|---|---|
| Row minimum {–3, –5, –2, –3} Row maximin = –2 | M1 |
| Column maximum {2,5,5,4} Column minimax = 2 | A1 |
| So play safe for player A is 3 and play safe for player B is 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 six (of the eight) values stated correctly) – if they reduce the game to the given 3 by 3 matrix then they need five of the six values correct – working must be done in (a) – if correct play-safes for both players stated with no working then M1 only
a1A1: Correct Row maximin and Column minimax (dependent on all row mins and column maxs correct) – stated or clearly shown
a2A1: Correct play safes (A(3) and B(1)) for both players – not dependent on previous A mark
| Scheme | Marks |
|---|---|
| \(2 \neq -2\) so game is not stable | B1 |
| (1) |
Notes
b1B1: CAO (dependent on the first two marks in (a) or complete method seen in (b)) – states \(2 \neq -2\) or row(maximin) \(\neq\) col(minimax) as long as 2 and –2 are clearly identified + conclusion (not stable)
| Scheme | Marks |
|---|---|
| Column 1 dominates column 4 because –3 < –1, –5 < –1, –2 < 2 and 2 < 4 (so remove column 4) | B1 |
| Row 3 dominates row 2 because –2 > –5, 5 > 3, 4 > 1 (and 2 > –1) (so remove row 2) | B1 |
| (2) |
Notes
c1B1: Either correct domination stated for both rows and columns or one correct with correct justification
c2B1: CSO (correct domination and full justification) – for justification the minimum we will accept is that all values in C4 > C1 (although the C1 could be implied) and that all values in R3 > R2 (again R2 could be implied) – note strict inequalities
| Scheme | Marks |
|---|---|
| e.g. add at least 3 to each element e.g. \(\begin{pmatrix} 1 & 6 & 9 \\ 2 & 9 & 8 \\ 6 & 1 & 3 \end{pmatrix}\), \(\begin{pmatrix} 0 & 5 & 8 \\ 1 & 8 & 7 \\ 5 & 0 & 2 \end{pmatrix}\), etc. | B1 |
| Let \(p_1, p_3, p_4\) be the probability of (A) playing 1, 3, 4 respectively (where \(p_1, p_3, p_4 \geqslant 0\)) | B1 |
| Let \(V =\) value of the game (to player A) | B1 |
| Maximise \((P =)\, V\) | B1 |
| Subject to: e.g. \[\begin{aligned} V - p_1 - 2p_3 - 6p_4 &\leqslant 0 &\qquad V - p_3 - 5p_4 &\leqslant 0 \\ V - 6p_1 - 9p_3 - p_4 &\leqslant 0 &\qquad V - 5p_1 - 8p_3 &\leqslant 0 \\ V - 9p_1 - 8p_3 - 3p_4 &\leqslant 0 &\qquad V - 8p_1 - 7p_3 - 2p_4 &\leqslant 0 \\ p_1 + p_3 + p_4 &\leqslant 1 &\qquad p_1 + p_3 + p_4 &\leqslant 1 \end{aligned}\] | M1 A1 A1 |
| (7) | |
| 13 marks |
Notes
d1B1: Making all terms non-negative (any addition of at least 3 is acceptable) – must be shown (not just stated as a method), all values correct – could be implied in constraints
d2B1: Defining probability variables (allow throughout if defined using 1, 2 and 3 rather than 1, 3 and 4)
d3B1: Defining \(V\)
d4B1: Maximising + function/expression
d1M1: At least three inequalities (of the four) using column values in \((V)\), \(p_1, p_3, p_4\) (or equations using slack variables e.g. \(V - p_1 - 2p_3 - 6p_4 + r = 0\) but must be + \(r\)), all probability terms in the first three constraints having correct signs for their coefficients, must be \(V \leqslant \ldots\)
d1A1: The three inequalities in \(V, p_1, p_3, p_4\) CAO (must be expressed as inequalities)
d2A1: Probability sum inequality correct
If defining for the original 4 by 4 game (they would need to add at least 5) then first five marks available only in (d)