| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Easy -1.8 This is a standard Decision Mathematics game theory question requiring routine application of textbook algorithms: finding row minima, checking for saddle points, eliminating dominated strategies, and solving a 2×2 game using the standard mixed strategy formula. All steps are mechanical with no novel insight required, making it significantly easier than average A-level maths questions. |
| 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 |
| \cline { 2 - 5 } | Colleen | |||
| \cline { 2 - 5 } Strategy | \(\mathbf { C } _ { \mathbf { 1 } }\) | \(\mathbf { C } _ { \mathbf { 2 } }\) | \(\mathbf { C } _ { \mathbf { 3 } }\) | |
| \cline { 2 - 5 } Rhona | \(\mathbf { R } _ { \mathbf { 1 } }\) | 2 | 6 | 4 |
| \cline { 2 - 5 } | \(\mathbf { R } _ { \mathbf { 2 } }\) | 3 | - 3 | - 1 |
| \cline { 2 - 5 } | \(\mathbf { R } _ { \mathbf { 3 } }\) | \(x\) | \(x + 3\) | 3 |
| \cline { 2 - 5 } | ||||
| \cline { 2 - 5 } | ||||
| Answer | Marks | Guidance |
|---|---|---|
| Row minima are: \(2, -3, x\) | B1 | All three correct (accept in table) |
| Answer | Marks | Guidance |
|---|---|---|
| Column maxima are: \(3, 6, 4\) | M1 | Attempting column maxima |
| Maximin \(= 2\), minimax \(= 3\) | A1 | Both correct |
| Since maximin \(\neq\) minimax, there is no stable solution | A1 | Conclusion stated |
| Answer | Marks | Guidance |
|---|---|---|
| Every entry in \(R_3\) is less than or equal to the corresponding entry in \(R_1\) (since \(x < 2\) means \(x < 2\) and \(x+3 < 6\) and \(3 < 4\)) | B1 | \(R_3\) is dominated by \(R_1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Let Rhona play \(R_1\) with probability \(p\) and \(R_2\) with probability \(1-p\) | M1 | Setting up expected values |
| \(E(C_1) = 2p + 3(1-p) = 3 - p\) | A1 | |
| \(E(C_2) = 6p - 3(1-p) = 9p - 3\) | A1 | |
| Setting \(3 - p = 9p - 3\) | M1 | Equating two expressions |
| \(6 = 10p\) | ||
| \(p = \frac{3}{5}\) | A1 | |
| Optimal strategy: play \(R_1\) with probability \(\frac{3}{5}\), \(R_2\) with probability \(\frac{2}{5}\) | A1 | Both probabilities stated |
| Check \(E(C_3)\): \(4(\frac{3}{5}) + (-1)(\frac{2}{5}) = \frac{10}{5} = 2\) | A1 | Verify \(C_3\) gives lower value so not relevant |
| Answer | Marks | Guidance |
|---|---|---|
| Value of game \(= 3 - \frac{3}{5} = \frac{12}{5}\) | B1ft | Follow through their \(p\) |
## Question 3:
**(a)(i)**
Row minima are: $2, -3, x$ | B1 | All three correct (accept in table)
**(a)(ii)**
Column maxima are: $3, 6, 4$ | M1 | Attempting column maxima
Maximin $= 2$, minimax $= 3$ | A1 | Both correct
Since maximin $\neq$ minimax, there is no stable solution | A1 | Conclusion stated
**(b)**
Every entry in $R_3$ is less than or equal to the corresponding entry in $R_1$ (since $x < 2$ means $x < 2$ and $x+3 < 6$ and $3 < 4$) | B1 | $R_3$ is dominated by $R_1$
**(c)(i)**
Let Rhona play $R_1$ with probability $p$ and $R_2$ with probability $1-p$ | M1 | Setting up expected values
$E(C_1) = 2p + 3(1-p) = 3 - p$ | A1 |
$E(C_2) = 6p - 3(1-p) = 9p - 3$ | A1 |
Setting $3 - p = 9p - 3$ | M1 | Equating two expressions
$6 = 10p$ |
$p = \frac{3}{5}$ | A1 |
Optimal strategy: play $R_1$ with probability $\frac{3}{5}$, $R_2$ with probability $\frac{2}{5}$ | A1 | Both probabilities stated
Check $E(C_3)$: $4(\frac{3}{5}) + (-1)(\frac{2}{5}) = \frac{10}{5} = 2$ | A1 | Verify $C_3$ gives lower value so not relevant
**(c)(ii)**
Value of game $= 3 - \frac{3}{5} = \frac{12}{5}$ | B1ft | Follow through their $p$
---3 Two people, Rhona and Colleen, play a zero-sum game. The game is represented by the following pay-off matrix for Rhona.
\begin{center}
\begin{tabular}{ l | c | c | c | c | }
\multicolumn{5}{c}{} \\
\cline { 2 - 5 }
& \multicolumn{4}{c}{Colleen} \\
\cline { 2 - 5 }
Strategy & $\mathbf { C } _ { \mathbf { 1 } }$ & $\mathbf { C } _ { \mathbf { 2 } }$ & $\mathbf { C } _ { \mathbf { 3 } }$ & \\
\cline { 2 - 5 }
Rhona & $\mathbf { R } _ { \mathbf { 1 } }$ & 2 & 6 & 4 \\
\cline { 2 - 5 }
& $\mathbf { R } _ { \mathbf { 2 } }$ & 3 & - 3 & - 1 \\
\cline { 2 - 5 }
& $\mathbf { R } _ { \mathbf { 3 } }$ & $x$ & $x + 3$ & 3 \\
\cline { 2 - 5 }
& & & & \\
\cline { 2 - 5 }
\end{tabular}
\end{center}
It is given that $x < 2$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the three row minima.
\item Show that there is no stable solution.
\end{enumerate}\item Explain why Rhona should never play strategy $R _ { 3 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the optimal mixed strategy for Rhona.
\item Find the value of the game.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA D2 2011 Q3 [13]}}