| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Discrete (Further Paper 3 Discrete) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Zero-sum game optimal mixed strategy |
| Difficulty | Challenging +1.2 This is a standard Further Maths game theory question requiring systematic application of the minimax theorem to find optimal mixed strategies. While it involves multiple steps (checking for dominance, setting up probability equations, solving linear equations), the procedure is algorithmic and well-practiced in Further Maths courses. It's harder than typical A-level pure maths due to the specialized topic, but represents a routine application of taught techniques rather than requiring novel insight. |
| Spec | 7.08a Pay-off matrix: zero-sum games7.08e Mixed strategies: optimal strategy using equations or graphical method |
| \multirow{2}{*}{} | Danielle | |||
| Strategy | \(\boldsymbol { X }\) | \(Y\) | \(\boldsymbol { Z }\) | |
| \multirow{3}{*}{John} | \(A\) | 2 | 1 | -1 |
| B | -3 | -2 | 2 | |
| \(\boldsymbol { C }\) | -3 | -4 | 1 | |
| Answer | Marks |
|---|---|
| \(-3 = -3,\ -4 < -2,\ 1 < 2\), hence strategy \(B\) dominates strategy \(C\); strategy \(C\) is dominated | B1 (AO2.2a) |
| Let John choose strategy \(A\) with probability \(p\) and strategy \(B\) with probability \(1-p\); introduce probability variable | M1 (AO3.3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(Z\): expected gain for John \(= -p + 2(1-p) = 2 - 3p\) | A1 (AO1.1b) | All three expressions correct |
| Construct graph with at least one vertical axis, plotting expected gains correctly | M1 (AO1.1a) | |
| Identify optimal point of intersection from graph: \(3p - 2 = 2 - 3p \Rightarrow 6p = 4 \Rightarrow p = \frac{2}{3}\) | A1 (AO1.1b) | |
| John should play strategy \(A\) with probability \(\frac{2}{3}\) and strategy \(B\) with probability \(\frac{1}{3}\) (and \(C\) with probability 0) | E1 (AO3.2a) | Correct contextual interpretation required |
## Question 8:
$-3 = -3,\ -4 < -2,\ 1 < 2$, hence strategy $B$ dominates strategy $C$; strategy $C$ is dominated | B1 (AO2.2a) |
Let John choose strategy $A$ with probability $p$ and strategy $B$ with probability $1-p$; introduce probability variable | M1 (AO3.3) |
$X$: expected gain for John $= 2p - 3(1-p) = 5p - 3$
$Y$: expected gain for John $= p - 2(1-p) = 3p - 2$
$Z$: expected gain for John $= -p + 2(1-p) = 2 - 3p$ | A1 (AO1.1b) | All three expressions correct
Construct graph with at least one vertical axis, plotting expected gains correctly | M1 (AO1.1a) |
Identify optimal point of intersection from graph: $3p - 2 = 2 - 3p \Rightarrow 6p = 4 \Rightarrow p = \frac{2}{3}$ | A1 (AO1.1b) |
John should play strategy $A$ with probability $\frac{2}{3}$ and strategy $B$ with probability $\frac{1}{3}$ (and $C$ with probability 0) | E1 (AO3.2a) | Correct contextual interpretation required
**Total: 6 marks**8 John and Danielle play a zero-sum game which does not have a stable solution. The game is represented by the following pay-off matrix for John.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multirow{2}{*}{} & & \multicolumn{3}{|c|}{Danielle} \\
\hline
& Strategy & $\boldsymbol { X }$ & $Y$ & $\boldsymbol { Z }$ \\
\hline
\multirow{3}{*}{John} & $A$ & 2 & 1 & -1 \\
\hline
& B & -3 & -2 & 2 \\
\hline
& $\boldsymbol { C }$ & -3 & -4 & 1 \\
\hline
\end{tabular}
\end{center}
Find the optimal mixed strategy for John.\\
\hfill \mbox{\textit{AQA Further Paper 3 Discrete Q8 [6]}}