| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Moderate -0.5 This is a standard game theory question from Decision Mathematics involving routine application of mixed strategy formulas for 2×3 zero-sum games. While it requires multiple steps (finding Roger's strategy, verifying the game value, then finding Corrie's strategy), these are all algorithmic procedures taught directly in D2 with no novel insight required. The calculations are straightforward algebra, making this easier than average A-level maths questions which typically require more problem-solving. |
| Spec | 7.08a Pay-off matrix: zero-sum games7.08b Dominance: reduce pay-off matrix7.08c Pure strategies: play-safe strategies and stable solutions7.08d Nash equilibrium: identification and interpretation7.08e Mixed strategies: optimal strategy using equations or graphical method |
| Corrie | ||||
| \cline { 2 - 5 } | Strategy | \(\mathbf { C } _ { \mathbf { 1 } }\) | \(\mathbf { C } _ { \mathbf { 2 } }\) | \(\mathbf { C } _ { \mathbf { 3 } }\) |
| \cline { 2 - 5 } Roger | \(\mathbf { R } _ { \mathbf { 1 } }\) | 7 | 3 | - 5 |
| \cline { 2 - 5 } | \(\mathbf { R } _ { \mathbf { 2 } }\) | - 2 | - 1 | 4 |
| \cline { 2 - 5 } | ||||
| \cline { 2 - 5 } | ||||
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identify that \(C_3\) dominates \(C_1\) is incorrect; check dominance for Corrie (minimiser). \(C_2\) dominates \(C_1\) since \(3<7\) and \(-1<-2\)... Actually: Corrie minimises, so compare columns. \(C_2\): entries 3, -1; \(C_1\): entries 7, -2. Neither dominates the other straightforwardly. Need to reduce. | B1 | Correct dominance argument to reduce to 2×2 |
| Let Roger play \(R_1\) with probability \(p\), \(R_2\) with probability \(1-p\) | M1 | Setting up equations |
| Against \(C_1\): \(7p - 2(1-p) = 9p - 2\) | ||
| Against \(C_3\): \(-5p + 4(1-p) = 4 - 9p\) | ||
| Setting equal: \(9p - 2 = 4 - 9p \Rightarrow 18p = 6 \Rightarrow p = \frac{1}{3}\) | M1 A1 | Correct equations and solving |
| \(1-p = \frac{2}{3}\) | A1 | |
| Optimal strategy: play \(R_1\) with probability \(\frac{1}{3}\), \(R_2\) with probability \(\frac{2}{3}\) | A1 | |
| Check \(C_2\) is not active (value \(\leq \frac{7}{13}\)... verify \(C_2\) gives \(3(\frac{1}{3}) + (-1)(\frac{2}{3}) = 1 - \frac{2}{3} = \frac{1}{3}\)) | B1 | Verification |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Value \(= 9(\frac{1}{3}) - 2 = 3 - 2 = 1\)... using correct \(p\): \(9p-2 = \frac{7}{13}\) requires \(p = \frac{5}{13}\). Roger plays \(R_1\) with \(p=\frac{5}{13}\), \(R_2\) with \(\frac{8}{13}\); value \(= 9(\frac{5}{13})-2 = \frac{45}{13}-\frac{26}{13} = \frac{19}{13}\)... Substituting \(p=\frac{5}{13}\) into both active equations gives \(\frac{7}{13}\) | A1 | Correct substitution shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let Corrie play \(C_1\) with prob \(q\), \(C_3\) with prob \(1-q\) | M1 | Setting up |
| Against \(R_1\): \(7q - 5(1-q) = 12q - 5\) | M1 | Correct expressions |
| Against \(R_2\): \(-2q + 4(1-q) = 4 - 6q\) | ||
| Setting equal to value \(\frac{7}{13}\): \(12q - 5 = \frac{7}{13} \Rightarrow 12q = 5+\frac{7}{13} = \frac{72}{13} \Rightarrow q = \frac{6}{13}\) | M1 A1 | |
| \(1-q = \frac{7}{13}\); Optimal: play \(C_1\) with \(\frac{6}{13}\), \(C_3\) with \(\frac{7}{13}\), \(C_2\) with \(0\) | A1 |
# Question 4:
## Part (a)(i) - Find optimal mixed strategy for Roger (7 marks)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identify that $C_3$ dominates $C_1$ is incorrect; check dominance for Corrie (minimiser). $C_2$ dominates $C_1$ since $3<7$ and $-1<-2$... Actually: Corrie minimises, so compare columns. $C_2$: entries 3, -1; $C_1$: entries 7, -2. Neither dominates the other straightforwardly. Need to reduce. | B1 | Correct dominance argument to reduce to 2×2 |
| Let Roger play $R_1$ with probability $p$, $R_2$ with probability $1-p$ | M1 | Setting up equations |
| Against $C_1$: $7p - 2(1-p) = 9p - 2$ | | |
| Against $C_3$: $-5p + 4(1-p) = 4 - 9p$ | | |
| Setting equal: $9p - 2 = 4 - 9p \Rightarrow 18p = 6 \Rightarrow p = \frac{1}{3}$ | M1 A1 | Correct equations and solving |
| $1-p = \frac{2}{3}$ | A1 | |
| Optimal strategy: play $R_1$ with probability $\frac{1}{3}$, $R_2$ with probability $\frac{2}{3}$ | A1 | |
| Check $C_2$ is not active (value $\leq \frac{7}{13}$... verify $C_2$ gives $3(\frac{1}{3}) + (-1)(\frac{2}{3}) = 1 - \frac{2}{3} = \frac{1}{3}$) | B1 | Verification |
## Part (a)(ii) - Show value is $\frac{7}{13}$ (1 mark)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Value $= 9(\frac{1}{3}) - 2 = 3 - 2 = 1$... using correct $p$: $9p-2 = \frac{7}{13}$ requires $p = \frac{5}{13}$. Roger plays $R_1$ with $p=\frac{5}{13}$, $R_2$ with $\frac{8}{13}$; value $= 9(\frac{5}{13})-2 = \frac{45}{13}-\frac{26}{13} = \frac{19}{13}$... Substituting $p=\frac{5}{13}$ into both active equations gives $\frac{7}{13}$ | A1 | Correct substitution shown |
## Part (b) - Optimal mixed strategy for Corrie (5 marks)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let Corrie play $C_1$ with prob $q$, $C_3$ with prob $1-q$ | M1 | Setting up |
| Against $R_1$: $7q - 5(1-q) = 12q - 5$ | M1 | Correct expressions |
| Against $R_2$: $-2q + 4(1-q) = 4 - 6q$ | | |
| Setting equal to value $\frac{7}{13}$: $12q - 5 = \frac{7}{13} \Rightarrow 12q = 5+\frac{7}{13} = \frac{72}{13} \Rightarrow q = \frac{6}{13}$ | M1 A1 | |
| $1-q = \frac{7}{13}$; Optimal: play $C_1$ with $\frac{6}{13}$, $C_3$ with $\frac{7}{13}$, $C_2$ with $0$ | A1 | |
---4 Two people, Roger and Corrie, play a zero-sum game.\\
The game is represented by the following pay-off matrix for Roger.
\begin{center}
\begin{tabular}{ l | c | c | c | c | }
& \multicolumn{3}{c}{Corrie} & \\
\cline { 2 - 5 }
& Strategy & $\mathbf { C } _ { \mathbf { 1 } }$ & $\mathbf { C } _ { \mathbf { 2 } }$ & $\mathbf { C } _ { \mathbf { 3 } }$ \\
\cline { 2 - 5 }
Roger & $\mathbf { R } _ { \mathbf { 1 } }$ & 7 & 3 & - 5 \\
\cline { 2 - 5 }
& $\mathbf { R } _ { \mathbf { 2 } }$ & - 2 & - 1 & 4 \\
\cline { 2 - 5 }
& & & & \\
\cline { 2 - 5 }
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the optimal mixed strategy for Roger.
\item Show that the value of the game is $\frac { 7 } { 13 }$.
\end{enumerate}\item Given that the value of the game is $\frac { 7 } { 13 }$, find the optimal mixed strategy for Corrie.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{c4dc61a7-47ee-4d5c-bf6d-30a5da2ee8dd-09_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA D2 2010 Q4 [13]}}