## Question 4(a):

**(i) Show stable solution:**

| Find row minima: A₁: $-6$, A₂: $-3$, A₃: $-5$, A₄: $-4$ | M1 | Method of finding row minima and column maxima |
|---|---|---|
| Maximin $= -3$ (row A₂) | A1 | |
| Find column maxima: B₁: $5$, B₂: $4$, B₃: $-3$ | | |
| Minimax $= -3$ (column B₃) | A1 | Since maximin = minimax $= -3$, stable solution exists |

**(ii) Play-safe strategies:**

| Adam plays $A_2$, Bill plays $B_3$ | B1 | Both required for mark |

**(iii) Value of game for Bill:**

| Value $= +3$ (since pay-off matrix is for Adam, Bill's value is $+3$) | B1 | Accept $3$ |

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## Question 4(b):

**(i) Strategy computer should never play:**

| $C_2$, since $C_1$ dominates $C_2$: every entry in $C_1$ is greater than corresponding entry in $C_2$ (i.e. $3>4$... ) | B1 | Must give reason: $C_1$ dominates $C_2$ since $3>-2$ wait — this is pay-off for Roza so computer wants to minimise; $C_2$ is dominated by $C_1$ since $4>3$ and $-1>-2$... Accept correct dominance argument |

**(ii) Expected gains for Roza:**

| If computer plays $C_1$: $E(C_1) = 3p + (-2)(1-p) = 5p - 2$ | B1 | |
|---|---|---|
| If computer plays $C_3$: $E(C_3) = -3p + 5(1-p) = 5 - 8p$ | B1 | |

**(iii) Value of $p$ to maximise expected gains:**

| Set $5p - 2 = 5 - 8p$ | M1 | Equating the two expressions |
| $13p = 7$, so $p = \frac{7}{13}$ | A1 | |

**(iv) Value of game for Roza:**

| $E = 5\left(\frac{7}{13}\right) - 2 = \frac{35}{13} - \frac{26}{13} = \frac{9}{13}$ | B1 | Follow through from their $p$ |