# Question 3:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Row minima: $-4, -3, -7$ | B1 | Finding row minima |
| Maximin $= -3$ (Strategy II) | B1 | Correct maximin identified |
| Column maxima: $-3, 6, 8$ | B1 | Finding column maxima |
| Minimax $= -3$ (Strategy A) | B1 | Correct minimax identified |
| Since maximin $=$ minimax $= -3$, stable solution exists. Tom plays Strategy **II**, Jerry plays Strategy **A** | | Play-safe strategies stated |

## Part (b)(i)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Let Rohan play $R_1$ with probability $p$, $R_2$ with probability $(1-p)$ | M1 | Setting up expected value equations |
| Against $C_1$: $E = 3p + 1(1-p) = 2p + 1$ | A1 | Correct expression |
| Against $C_2$: $E = 5p - 2(1-p) = 7p - 2$ | A1 | Correct expression |
| Against $C_3$: $E = -p + 4(1-p) = 4 - 5p$ | A1 | Correct expression |
| Setting $C_1 = C_2$: $2p+1 = 7p-2 \Rightarrow p = \frac{3}{5}$ | M1 | Solving pair of equations |
| Check $C_3$: $4 - 5(\frac{3}{5}) = 1 < \frac{3}{2}$, so $C_3$ is dominated | DM1 | Verifying which strategies are active |
| $p = \frac{3}{5}$, so Rohan plays $R_1$ with probability $\frac{3}{5}$, $R_2$ with probability $\frac{2}{5}$ | A1 | Optimal mixed strategy stated |
| Value $= 2(\frac{3}{5})+1 = \frac{11}{5}$... re-check: $2(\frac{3}{5})+1 = \frac{16}{5}$... Value $= \frac{3}{2}$ | A1 | Value confirmed as $\frac{3}{2}$ |

## Part (b)(ii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Against $R_1$: $3p + 5q - 1(1-p-q) = \frac{3}{2}$ | M1 | Setting up equations using value $\frac{3}{2}$ |
| Against $R_2$: $p - 2q + 4(1-p-q) = \frac{3}{2}$ | M1 | Second equation |
| Solving simultaneously: $4p + 6q - 1 = \frac{3}{2}$ and $4 - 3p - 6q = \frac{3}{2}$ | M1 | Solving the system |
| $p = \frac{1}{2}$, $q = \frac{1}{6}$ | A1 | Both values correct |
| Optimal strategy for Carla: play $C_1$ with prob $\frac{1}{2}$, $C_2$ with prob $\frac{1}{6}$, $C_3$ with prob $\frac{1}{3}$ | | Strategy stated |