# Question 3:

## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| The gain of one player equals the loss of the other player | B1 | Must mention both gain and loss |
| The sum of the payoffs (to both players) is zero | B1 | Accept equivalent statement |

## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Colum wishes to minimise the pay-off to Roz, so looks at maximum in each column | M1 | Must show column maxima: $C_1: -2$, $C_2: 3$, $C_3: -1$ |
| Play-safe strategy is $C_1$ (choosing minimum of these maxima = $-2$) | A1 | Must give reason: minimax = $-2$ |

## Part (c)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Row minima: $R_1: -6$, $R_2: -6$, $R_3: -4$ | M1 | |
| $R_2$ is dominated by $R_3$ (every entry in $R_3 \geq$ every entry in $R_2$): $-3>-5$, $3>2$, $-4>-6$ | A1 | Accept $R_1$ dominated by $R_3$: $-3>-2$ fails — must show $R_2$ dominated by $R_3$ |

## Part (c)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Let Roz play $R_1$ with probability $p$ and $R_3$ with probability $(1-p)$ | M1 | |
| Expected payoff against $C_1$: $-2p + (-3)(1-p) = p - 3$ | A1 | |
| Expected payoff against $C_2$: $-6p + 3(1-p) = 3 - 9p$ | A1 | |
| Expected payoff against $C_3$: $-p + (-4)(1-p) = 3p - 4$ | A1 | |
| Setting $C_1 = C_2$: $p - 3 = 3 - 9p \Rightarrow p = \frac{3}{5}$ | M1 | |
| Check $C_3$ at $p = \frac{3}{5}$: $3(\frac{3}{5}) - 4 = -\frac{11}{5}$, which is less than $-\frac{12}{5}$ so $C_3$ not relevant | A1 | |
| Optimal strategy: play $R_1$ with probability $\frac{3}{5}$, $R_3$ with probability $\frac{2}{5}$, value $= -\frac{12}{5}$ | A1 | |

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