| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Discrete (Further AS Paper 2 Discrete) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Non-group structures |
| Difficulty | Challenging +1.2 This is a standard game theory question requiring dominance identification (part a) and optimal mixed strategy calculation (part b). While it involves multiple steps and some algebraic manipulation, the techniques are routine for Further Maths students who have studied this topic. The dominance argument is straightforward, and finding the optimal mixed strategy follows a well-practiced algorithm with no novel insight required. |
| Spec | 7.08a Pay-off matrix: zero-sum games7.08b Dominance: reduce pay-off matrix7.08e Mixed strategies: optimal strategy using equations or graphical method |
| Lui | ||||
| \cline { 2 - 5 } | Strategy | \(\mathbf { L } _ { \mathbf { 1 } }\) | \(\mathbf { L } _ { \mathbf { 2 } }\) | \(\mathbf { L } _ { \mathbf { 3 } }\) |
| \(\mathrm { Kez } \quad \mathbf { K } _ { \mathbf { 1 } }\) | 4 | 1 | - 2 | |
| \(\mathbf { K } _ { \mathbf { 2 } }\) | - 4 | - 2 | 0 | |
| \(\mathbf { K } _ { \mathbf { 3 } }\) | - 2 | - 1 | 2 | |
| Answer | Marks |
|---|---|
| \(-4 < -2,\ -2 < -1,\ 0 < 2\), hence strategy \(K_3\) dominates strategy \(K_2\); therefore Kez should never play strategy \(K_2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Let Kez choose strategy \(K_1\) with probability \(p\) and \(K_3\) with probability \(1-p\) | B1 | PI |
| \(L_1\): expected gain for Kez \(= 4p - 2(1-p) = 6p - 2\) | M1 | Uses model to find one expected gain |
| Answer | Marks |
|---|---|
| All three correct | A1 |
| Uses graph with straight lines, at least one vertical axis, one line sketched correctly | M1 |
| Identifies optimal intersection point: \(2p - 1 = -4p + 2 \Rightarrow p = 0.5\) | A1 |
| \(0.5 \times 20 = 10\); Kez plays strategy \(K_3\) 10 times out of 20 | B1F |
## Question 7(a):
$-4 < -2,\ -2 < -1,\ 0 < 2$, hence strategy $K_3$ dominates strategy $K_2$; therefore Kez should never play strategy $K_2$ | B1 |
---
## Question 7(b):
Let Kez choose strategy $K_1$ with probability $p$ and $K_3$ with probability $1-p$ | B1 | PI
$L_1$: expected gain for Kez $= 4p - 2(1-p) = 6p - 2$ | M1 | Uses model to find one expected gain
$L_2$: expected gain $= p - (1-p) = 2p - 1$
$L_3$: expected gain $= -2p + 2(1-p) = -4p + 2$
All three correct | A1 |
Uses graph with straight lines, at least one vertical axis, one line sketched correctly | M1 |
Identifies optimal intersection point: $2p - 1 = -4p + 2 \Rightarrow p = 0.5$ | A1 |
$0.5 \times 20 = 10$; Kez plays strategy $K_3$ **10 times** out of 20 | B1F |7 Kez and Lui play a zero-sum game. The game does not have a stable solution.
The game is represented by the following pay-off matrix for Kez.
\begin{center}
\begin{tabular}{ l | c | c | c | c | }
& \multicolumn{3}{c}{Lui} & \\
\cline { 2 - 5 }
& Strategy & $\mathbf { L } _ { \mathbf { 1 } }$ & $\mathbf { L } _ { \mathbf { 2 } }$ & $\mathbf { L } _ { \mathbf { 3 } }$ \\
\hline
$\mathrm { Kez } \quad \mathbf { K } _ { \mathbf { 1 } }$ & 4 & 1 & - 2 & \\
\hline
$\mathbf { K } _ { \mathbf { 2 } }$ & - 4 & - 2 & 0 & \\
\hline
& $\mathbf { K } _ { \mathbf { 3 } }$ & - 2 & - 1 & 2 \\
\hline
\end{tabular}
\end{center}
7
\begin{enumerate}[label=(\alph*)]
\item State, with a reason, why Kez should never play strategy $\mathbf { K } _ { \mathbf { 2 } }$
7
\item $\quad$ Kez and Lui play the game 20 times.\\
Kez plays their optimal mixed strategy.\\
Find the expected number of times that Kez will play strategy $\mathbf { K } _ { \mathbf { 3 } }$\\
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Discrete 2022 Q7 [7]}}