| Exam Board | AQA |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| 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 textbook application of game theory requiring dominance identification and solving a system of linear equations with the game value given. Part (a) is straightforward comparison of payoffs (strategy A dominates B), and part (b) involves routine algebraic manipulation to find probabilities that sum to 1 and yield the given value—no novel insight required, just methodical application of learned techniques. |
| Spec | 7.08a Pay-off matrix: zero-sum games7.08b Dominance: reduce pay-off matrix7.08c Pure strategies: play-safe strategies and stable solutions7.08e Mixed strategies: optimal strategy using equations or graphical method |
| Owen | ||||
| \cline { 2 - 5 }\cline { 2 - 5 } | Strategy | D | E | F |
| A | 4 | 1 | - 1 | |
| \cline { 2 - 5 } Mark | B | 3 | - 2 | - 2 |
| \cline { 2 - 5 } | C | - 2 | 0 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Strategy B is dominated by strategy A (every entry in row A is greater than corresponding entry in row B), so Mark should never play B | B1 | Must reference domination explicitly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Reduce to 2×3 matrix (delete row B): rows A and C, columns D, E, F | M1 | |
| Let Owen play D with prob \(p\), E with prob \(q\), F with prob \(1-p-q\) | M1 | Setting up equations |
| For Mark playing A: \(4p + q - (1-p-q) = 0.6 \Rightarrow 5p + 2q = 1.6\) | A1 | |
| For Mark playing C: \(-2p + 0(q) + 3(1-p-q) = 0.6 \Rightarrow -5p - 3q = -2.4\) | A1 | |
| Solving: \(q = 0.8\), \(p = 0\) | A1 | |
| \(1-p-q = 0.2\), so \(F\) probability \(= 0.2\) | A1 | |
| Owen's optimal strategy: play D with prob \(0\), E with prob \(\frac{4}{5}\), F with prob \(\frac{1}{5}\) | A1 | cao |
# Question 5:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Strategy B is dominated by strategy A (every entry in row A is greater than corresponding entry in row B), so Mark should never play B | B1 | Must reference domination explicitly |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Reduce to 2×3 matrix (delete row B): rows A and C, columns D, E, F | M1 | |
| Let Owen play D with prob $p$, E with prob $q$, F with prob $1-p-q$ | M1 | Setting up equations |
| For Mark playing A: $4p + q - (1-p-q) = 0.6 \Rightarrow 5p + 2q = 1.6$ | A1 | |
| For Mark playing C: $-2p + 0(q) + 3(1-p-q) = 0.6 \Rightarrow -5p - 3q = -2.4$ | A1 | |
| Solving: $q = 0.8$, $p = 0$ | A1 | |
| $1-p-q = 0.2$, so $F$ probability $= 0.2$ | A1 | |
| Owen's optimal strategy: play D with prob $0$, E with prob $\frac{4}{5}$, F with prob $\frac{1}{5}$ | A1 | cao |
---5 Mark and Owen play a zero-sum game. The game is represented by the following pay-off matrix for Mark.
\begin{center}
\begin{tabular}{ l | c | c | c | c | }
& \multicolumn{4}{c}{Owen} \\
\cline { 2 - 5 }\cline { 2 - 5 }
& Strategy & D & E & F \\
\hline
A & 4 & 1 & - 1 & \\
\cline { 2 - 5 }
Mark & B & 3 & - 2 & - 2 \\
\cline { 2 - 5 }
& C & - 2 & 0 & 3 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Explain why Mark should never play strategy B.
\item It is given that the value of the game is 0.6 . Find the optimal strategy for Owen.\\
(You are not required to find the optimal mixed strategy for Mark.)\\[0pt]
[7 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA D2 2014 Q5 [8]}}