## Question 5:

**(a)**

| Answer | Mark | Guidance |
|--------|------|----------|
| Romeo: row minima are $-4, -5, -2$; maximin $= -2$, so play-safe strategy is **C** | M1 A1 | M1 for finding row minima |
| Juliet: column maxima are $4, 1, 3$; minimax $= 1$, so play-safe strategy is **E** | A1 | (Juliet minimises, so looks at column maxima then takes minimum) |

**(b)**

| Answer | Mark | Guidance |
|--------|------|----------|
| Maximin $(-2) \neq$ minimax $(1)$, so no stable solution | B1 | Must state both values and conclude |

**(c)**

| Answer | Mark | Guidance |
|--------|------|----------|
| Strategy D is dominated by strategy F (or E): every entry in column D is greater than or equal to corresponding entry in column F (or E), so Juliet (minimiser) would never play D | B1 | |

**(d)(i)**

| Answer | Mark | Guidance |
|--------|------|----------|
| Since D is dominated/never played, remove column D. The matrix for Juliet is the negative of Romeo's pay-off matrix (zero-sum game), so negate remaining entries and transpose | M1 | |
| This gives the $2 \times 3$ matrix shown with rows E, F for Juliet against Romeo's strategies B, C | A1 | Accept clear explanation referencing negation of Romeo's values |

**(d)(ii)**

| Answer | Mark | Guidance |
|--------|------|----------|
| Let Juliet play E with probability $p$ and F with probability $(1-p)$ | M1 | |
| Against B (row 1): $5p + (-3)(1-p) = 8p - 3$ | M1 | Setting up expected values |
| Against C (row 2): $-p + 2(1-p) = 2 - 3p$ | A1 | |
| Set equal: $8p - 3 = 2 - 3p \Rightarrow 11p = 5 \Rightarrow p = \frac{5}{11}$ | DM1 A1 | |
| Optimal strategy: play E with probability $\frac{5}{11}$, play F with probability $\frac{6}{11}$ | A1 | |
| Check against row with $-1, 2$: verify this is optimal | A1 | |

**(d)(iii)**

| Answer | Mark | Guidance |
|--------|------|----------|
| Value $= 8\left(\frac{5}{11}\right) - 3 = \frac{40}{11} - \frac{33}{11} = \frac{7}{11}$ | B1 ft | Follow through from (d)(ii) |

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