# Question 2:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| A's play-safe strategy: find row minima: A1:$-3$, A2:$-5$, A3:$-2$, A4:$-3$; maximin $= -2$, so A plays 3 | M1 | Method of finding row minima and maximising |
| B's play-safe strategy: find column maxima: B1:$2$, B2:$5$, B3:$5$, B4:$4$; minimax $= 2$, so B plays 1 | A1 | |
| Both strategies correctly stated | A1 | |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| No stable solution because maximin $(-2) \neq$ minimax $(2)$ | B1 | Must state reason with values |

## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Column B4 is dominated by column B1 (every entry in B1 $\leq$ every entry in B4), so B4 is removed | B1 | |
| Row A2 is dominated by row A1 or A3, so A2 is removed | B1 | Must correctly identify which row/column is dominated with reason |

## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| Let $p_1, p_2, p_3$ be the probabilities that A plays 1, 2, 3 respectively | B1 | Variables defined |
| $p_1 + p_2 + p_3 = 1$, $p_1, p_2, p_3 \geq 0$ | B1 | |
| Maximise $V$ subject to: | M1 | Correct structure |
| $-3p_1 - 2p_2 + 2p_3 \geq V$ | A1 | |
| $2p_1 + 5p_2 - 3p_3 \geq V$ | A1 | |
| $5p_1 + 4p_2 - p_3 \geq V$ | A1 | |
| All constraints correct with normalisation | A1 | |

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