# Question 6:

## Part (a)

| Answer | Marks | Guidance |
|--------|-------|----------|
| Strategy $B$ is dominated by strategy $C$ (every entry in row $C$ is greater than corresponding entry in row $B$) | B1 | Must refer to domination |

## Part (b) - Kate's Optimal Mixed Strategy

| Answer | Marks | Guidance |
|--------|-------|----------|
| Reduce to $2 \times 3$ game using rows $A$ and $C$ only | M1 | Removing row $B$ |
| Let Kate play $A$ with probability $p$, $C$ with probability $1-p$ | M1 | Setting up expressions |
| Expected payoff vs $D$: $-2p + 4(1-p) = 4 - 6p$ | A1 | |
| Expected payoff vs $E$: $0p + 1(1-p) = 1-p$ | A1 | |
| Expected payoff vs $F$: $3p + (-1)(1-p) = 4p - 1$ | A1 | |
| Setting $4-6p = 4p-1 \Rightarrow p = \frac{1}{2}$ | M1 | Equating appropriate lines |
| Check: vs $D$: $1$, vs $E$: $\frac{1}{2}$, vs $F$: $1$; optimal $p = \frac{1}{2}$ | A1 | |
| Kate plays $A$ with probability $\frac{1}{2}$, $C$ with probability $\frac{1}{2}$, value of game $= 1$ | A1 | |

## Part (c) - Pippa's Optimal Mixed Strategy

| Answer | Marks | Guidance |
|--------|-------|----------|
| Strategy $E$ dominated; Pippa mixes $D$ and $F$ only | M1 | |
| Let Pippa play $D$ with probability $q$, $F$ with probability $1-q$ | M1 | |
| vs $A$: $-2q + 3(1-q) = 3 - 5q$; vs $C$: $4q + (-1)(1-q) = 5q-1$ | A1 | |
| Setting equal: $3 - 5q = 5q - 1 \Rightarrow q = \frac{2}{5}$ | A1 | |
| Pippa plays $D$ with probability $\frac{2}{5}$, $F$ with probability $\frac{3}{5}$ | A1 | Condone no mention of $E$ |