| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Zero-sum game optimal mixed strategy |
| Difficulty | Standard +0.3 This is a standard D2 game theory question requiring routine application of the mixed strategy algorithm for 2×3 games. Students follow a mechanical procedure: find row player's strategy by equating expected payoffs, then use the game value to find column player's strategy by solving simultaneous equations. While multi-step, it requires no novel insight—just careful arithmetic and following the textbook method. |
| 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 |
| \multirow{2}{*}{} | Collette | |||
| Strategy | \(\mathrm { C } _ { 1 }\) | \(\mathbf { C } _ { \mathbf { 2 } }\) | \(\mathrm { C } _ { 3 }\) | |
| \multirow{2}{*}{Roseanne} | \(\mathrm { R } _ { 1 }\) | -3 | 2 | 3 |
| \(\mathbf { R } _ { \mathbf { 2 } }\) | 2 | -1 | -4 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let Roseanne play \(R_1\) with probability \(p\) | M1 | Setting up equations |
| Against \(C_1\): \(-3p + 2(1-p) = 2 - 5p\) | A1 | |
| Against \(C_2\): \(2p - (1-p) = 3p - 1\) | A1 | |
| Against \(C_3\): \(3p - 4(1-p) = 7p - 4\) | A1 | |
| Set \(2-5p = 3p-1 \Rightarrow p = \frac{3}{8}\) | M1 | |
| Check \(C_3\): \(7(\frac{3}{8})-4 = -\frac{11}{8}\) which is less, so \(C_3\) not used | A1 | |
| Optimal: play \(R_1\) with prob \(\frac{3}{8}\), \(R_2\) with prob \(\frac{5}{8}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Value \(= 2 - 5(\frac{3}{8}) = 2 - \frac{15}{8} = -\frac{1}{2} = -0.5\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Probability of \(C_3 = 1 - p - q\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Against \(R_1\): \(-3p + 2q + 3(1-p-q) = 3 - 6p - q\) | M1 | |
| Against \(R_2\): \(2p - q - 4(1-p-q) = 6p + 3q - 4\) | A1 | |
| Set equal to \(-0.5\): \(3 - 6p - q = -0.5 \Rightarrow 6p + q = 3.5\) | M1 | |
| \(6p + 3q - 4 = -0.5 \Rightarrow 6p + 3q = 3.5\) | ||
| Solving: \(q = 0\), \(p = \frac{7}{12}\), prob \(C_3 = \frac{5}{12}\) | A1 |
# Question 3:
## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let Roseanne play $R_1$ with probability $p$ | M1 | Setting up equations |
| Against $C_1$: $-3p + 2(1-p) = 2 - 5p$ | A1 | |
| Against $C_2$: $2p - (1-p) = 3p - 1$ | A1 | |
| Against $C_3$: $3p - 4(1-p) = 7p - 4$ | A1 | |
| Set $2-5p = 3p-1 \Rightarrow p = \frac{3}{8}$ | M1 | |
| Check $C_3$: $7(\frac{3}{8})-4 = -\frac{11}{8}$ which is less, so $C_3$ not used | A1 | |
| Optimal: play $R_1$ with prob $\frac{3}{8}$, $R_2$ with prob $\frac{5}{8}$ | A1 | |
## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Value $= 2 - 5(\frac{3}{8}) = 2 - \frac{15}{8} = -\frac{1}{2} = -0.5$ | B1 | |
## Part (b)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Probability of $C_3 = 1 - p - q$ | B1 | |
## Part (b)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Against $R_1$: $-3p + 2q + 3(1-p-q) = 3 - 6p - q$ | M1 | |
| Against $R_2$: $2p - q - 4(1-p-q) = 6p + 3q - 4$ | A1 | |
| Set equal to $-0.5$: $3 - 6p - q = -0.5 \Rightarrow 6p + q = 3.5$ | M1 | |
| $6p + 3q - 4 = -0.5 \Rightarrow 6p + 3q = 3.5$ | |
| Solving: $q = 0$, $p = \frac{7}{12}$, prob $C_3 = \frac{5}{12}$ | A1 | |
---3 Two people, Roseanne and Collette, play a zero-sum game. The game is represented by the following pay-off matrix for Roseanne.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multirow{2}{*}{} & & \multicolumn{3}{|c|}{Collette} \\
\hline
& Strategy & $\mathrm { C } _ { 1 }$ & $\mathbf { C } _ { \mathbf { 2 } }$ & $\mathrm { C } _ { 3 }$ \\
\hline
\multirow{2}{*}{Roseanne} & $\mathrm { R } _ { 1 }$ & -3 & 2 & 3 \\
\hline
& $\mathbf { R } _ { \mathbf { 2 } }$ & 2 & -1 & -4 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find the optimal mixed strategy for Roseanne.
\item Show that the value of the game is - 0.5 .
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Collette plays strategy $\mathrm { C } _ { 1 }$ with probability $p$ and strategy $\mathrm { C } _ { 2 }$ with probability $q$. Write down, in terms of $p$ and $q$, the probability that she plays strategy $\mathrm { C } _ { 3 }$.
\item Hence, given that the value of the game is - 0.5 , find the optimal mixed strategy for Collette.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA D2 2008 Q3 [13]}}