| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2004 |
| Session | June |
| Marks | 14 |
| 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 procedures: checking for saddle points, solving a 2×3 game using linear programming or dominance/graphical methods, and understanding zero-sum relationships. While multi-step with 14 marks, it follows textbook algorithms without requiring novel insight. Slightly above average difficulty due to the algebraic manipulation involved in finding optimal mixed strategies. |
| Spec | 7.08c Pure strategies: play-safe strategies and stable solutions7.08e Mixed strategies: optimal strategy using equations or graphical method |
| Answer | Marks | Guidance |
|---|---|---|
| row min: \(-4\), \(-2\) ← max; Col. max: \(2\), \(1\), \(3\); ↑ min: \(-2 \neq 1\) ∴ not stable | M1 A1, A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: Let Emma play \(R_1\) with probability \(p\). If Freddie plays \(C_1\), Emma's winnings are \(-4p + 2(1-p) = 2 - 6p\). \(C_2\), Emma's winnings are \(-p + 1(1-p) = 1 - 2p\). \(C_3\), Emma's winnings are \(3p - 2(1-p) = -2 + 5p\) | M1 A1, A1 | 3 marks |
| Need intersection of \(2 - 6p\) and \(-2 + 5p\): \(2 - 6p = -2 + 5p\), \(4 = 11p\), \(p = \frac{4}{11}\) | M1, A1 | |
| Answer: So Emma should play \(R_1\) with probability \(\frac{4}{11}\), \(R_2\) with probability \(\frac{7}{11}\). The value of the game is \(-\frac{2}{11}\) to Emma | A1 ft | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: Value to Freddie \(\frac{2}{11}\), matrix \(\begin{pmatrix} 4 & -2 \\ 1 & -1 \\ -3 & 2 \end{pmatrix}\) | B1 ft B1, B1 | 3 marks |
### (a)
**Answer:**
$$\begin{pmatrix} -4 & -1 & 3 \\ 2 & 1 & -2 \end{pmatrix}$$
row min: $-4$, $-2$ ← max; Col. max: $2$, $1$, $3$; ↑ min: $-2 \neq 1$ ∴ not stable | M1 A1, A1 | 3 marks
### (b)
**Answer:** Let Emma play $R_1$ with probability $p$. If Freddie plays $C_1$, Emma's winnings are $-4p + 2(1-p) = 2 - 6p$. $C_2$, Emma's winnings are $-p + 1(1-p) = 1 - 2p$. $C_3$, Emma's winnings are $3p - 2(1-p) = -2 + 5p$ | M1 A1, A1 | 3 marks
Need intersection of $2 - 6p$ and $-2 + 5p$: $2 - 6p = -2 + 5p$, $4 = 11p$, $p = \frac{4}{11}$ | M1, A1 |
**Answer:** So Emma should play $R_1$ with probability $\frac{4}{11}$, $R_2$ with probability $\frac{7}{11}$. The value of the game is $-\frac{2}{11}$ to Emma | A1 ft | 3 marks
### (c)
**Answer:** Value to Freddie $\frac{2}{11}$, matrix $\begin{pmatrix} 4 & -2 \\ 1 & -1 \\ -3 & 2 \end{pmatrix}$ | B1 ft B1, B1 | 3 marks
---Emma and Freddie play a zero-sum game. This game is represented by the following pay-off matrix for Emma. $\begin{pmatrix} -4 & -1 & 3 \\ 2 & 1 & -2 \end{pmatrix}$
\begin{enumerate}[label=(\alph*)]
\item Show that there is no stable solution. [3]
\item Find the best strategy for Emma and the value of the game to her. [8]
\item Write down the value of the game to Freddie and his pay-off matrix. [3]
\end{enumerate}
(Total 14 marks)
\hfill \mbox{\textit{Edexcel D2 2004 Q4 [14]}}