| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Standard +0.8 This is a game theory problem requiring the simplex method or graphical solution to find mixed strategy Nash equilibrium in a 3×3 zero-sum game with no saddle point. While the technique is systematic once learned, it involves multiple steps (checking for saddle point, setting up and solving linear programming constraints, finding optimal mixed strategies) and is more procedurally complex than typical A-level content, though still within reach of Further Maths students who have studied this topic. |
| Spec | 7.08e Mixed strategies: optimal strategy using equations or graphical method |
| S plays 1 | S plays 2 | S plays 3 | |
| R plays 1 | 2 | 1 | 3 |
| R plays 2 | 1 | - 1 | 2 |
| R plays 3 | - 1 | 3 | - 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a) | \(R1\) dominates \(R2\), so deleted \(R2\) to give \(\begin{pmatrix} 2 & 1 & 3 \\ -1 & 3 & -3 \end{pmatrix}\) | B1 |
| Part (b) | If \(S\) plays 1; \(R\)'s gain is \(2p - (1-p) = 3p - 1\) | M1 |
| If \(S\) plays 2; \(R\)'s gain is \(p + 3(1-p) = 3 - 2p\) | ||
| If \(S\) plays 3; \(R\)'s gain is \(3p - 3(1-p) = 6p - 3\) | A1 | 1A1: CAO (condone incorrect simplification) |
| 2B1ft: Attempt at three lines (correct gradients and correct order of intersection with 'axes'), accept \(p > 1\) or \(p < 0\) here. Must be functions of \(p\). | ||
| 2M1: Finding their correct optimal point, must have three lines and three intersection points and set up an equation to find \(0 \leq p \leq 1\). Solving all three simultaneous equations only is M0. | ||
| 2A1: CSO (all previous marks must have been awarded) | ||
| 3B1ft: CAO \(0 \leq p \leq 1\), scale clear (or 1 line = 1), condone lack of labels. Rulers used. | ||
| 3A1ft: All three options listed must fit from their \(p\), check page 1 for \(R\) should never play 2. \(0 \leq\) probabilities \(\leq 1\). Dependent on both previous M marks being awarded. | ||
| Diagram showing \(3 - 2p = 3p - 1\) giving \(p = \frac{4}{5}\) | M1, A1 | |
| 4A1: CAO for the value of the game (\(\frac{7}{5}\)) | ||
| Robin should play \(R1\) with probability \(\frac{4}{5}\), \(R2\) never, \(R3\) with probability \(\frac{1}{5}\) | A1ft | |
| The value of the game is \(\frac{7}{5}\) to Robin | A1 | |
| 9 marks |
| **Part (a)** | $R1$ dominates $R2$, so deleted $R2$ to give $\begin{pmatrix} 2 & 1 & 3 \\ -1 & 3 & -3 \end{pmatrix}$ | B1 | 1B1: CAO |
|---|---|---|---|
| **Part (b)** | If $S$ plays 1; $R$'s gain is $2p - (1-p) = 3p - 1$ | M1 | |
| | If $S$ plays 2; $R$'s gain is $p + 3(1-p) = 3 - 2p$ | | |
| | If $S$ plays 3; $R$'s gain is $3p - 3(1-p) = 6p - 3$ | A1 | 1A1: CAO (condone incorrect simplification) |
| | | | 2B1ft: Attempt at three lines (correct gradients and correct order of intersection with 'axes'), accept $p > 1$ or $p < 0$ here. Must be functions of $p$. |
| | | | 2M1: Finding their correct optimal point, must have three lines and three intersection points and set up an equation to find $0 \leq p \leq 1$. Solving all three simultaneous equations only is M0. |
| | | | 2A1: CSO (all previous marks must have been awarded) |
| | | | 3B1ft: CAO $0 \leq p \leq 1$, scale clear (or 1 line = 1), condone lack of labels. Rulers used. |
| | | | 3A1ft: All three options listed must fit from their $p$, check page 1 for $R$ should never play 2. $0 \leq$ probabilities $\leq 1$. Dependent on both previous M marks being awarded. |
| | Diagram showing $3 - 2p = 3p - 1$ giving $p = \frac{4}{5}$ | M1, A1 | |
| | | | 4A1: CAO for the value of the game ($\frac{7}{5}$) |
| | Robin should play $R1$ with probability $\frac{4}{5}$, $R2$ never, $R3$ with probability $\frac{1}{5}$ | A1ft | |
| | The value of the game is $\frac{7}{5}$ to Robin | A1 | |
| | | **9 marks** | |
---4. Robin (R) and Steve (S) play a two-person zero-sum game which is represented by the following pay-off matrix for Robin.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& S plays 1 & S plays 2 & S plays 3 \\
\hline
R plays 1 & 2 & 1 & 3 \\
\hline
R plays 2 & 1 & - 1 & 2 \\
\hline
R plays 3 & - 1 & 3 & - 3 \\
\hline
\end{tabular}
\end{center}
Find the best strategy for Robin and the value of the game to him.\\
\hfill \mbox{\textit{Edexcel D2 2013 Q4 [9]}}