# Question 7:

## Part 7(a)(i):
| Writes transposed matrix of 3 rows and 4 columns with any twelve elements | M1 | |
| Fully correct pay-off matrix for Bex: $B_1$: $(-2, 4, 0, 3)$; $B_2$: $(1, 2, -1, -2)$; $B_3$: $(-3, -2, -1, 2)$ | A1 | |

## Part 7(a)(ii):
| $-3 < -2$, $-2 < 4$, $-1 < 0$, $2 < 3$, hence strategy $\mathbf{B_1}$ dominates strategy $\mathbf{B_3}$ | E1 | Must use game theory terminology |
| $-2 < 4$, $1 < 2$, $-3 < -2$, hence strategy $\mathbf{A_1}$ dominates strategy $\mathbf{A_2}$ | E1 | Must use game theory terminology |

## Part 7(b):
| Let Bex choose strategy $\mathbf{B_1}$ with probability $p$ and $\mathbf{B_2}$ with probability $1-p$ | B1 | Must introduce and define probability variable |
| Finds one (unsimplified) expected gain for Bex: $\mathbf{A_1}$: $-2p + 1(1-p) = -3p+1$ | M1 | |
| All three correct expected gains: $\mathbf{A_3}$: $-1(1-p) = p-1$; $\mathbf{A_4}$: $3p - 2(1-p) = 5p-2$ | A1 | |
| Draws graph with at least one vertical axis and three expected gains (one slip allowed) | M1 | |
| Identifies correct optimal intersection point and correct value of $p$: $p-1 = -3p+1 \Rightarrow p = \frac{1}{2}$ | A1 | |
| Value of game for Bex $= \frac{1}{2} - 1 = -\frac{1}{2}$; hence value of game for Ali $= \frac{1}{2}$ | B1F | Follow through from their $p$ |