| Exam Board | Edexcel |
|---|---|
| Module | FD2 AS (Further Decision 2 AS) |
| Session | Specimen |
| Marks | 12 |
| 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 exercise in game theory requiring routine application of the minimax theorem and solving a 2×2 reduced game. While it involves multiple steps (checking for saddle point, reducing the matrix, setting up equations for mixed strategy), the procedures are algorithmic and well-practiced, making it slightly easier than average for Further Maths students who have studied this topic. |
| Spec | 7.08a Pay-off matrix: zero-sum games7.08c Pure strategies: play-safe strategies and stable solutions7.08e Mixed strategies: optimal strategy using equations or graphical method |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | B plays 1 | B plays 2 | B plays 3 |
| A plays 1 | 4 | 1 | 2 |
| A plays 2 | 2 | 4 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Row minima: 1, 2; max is 2. Column maxima: 4, 4, 3; min is 3 | M1, A1 | Finding row minimums and column maximums – condone one error; row minima and column maxima correct |
| Row maximin \((2) \neq\) Column minimax \((3)\) so not stable | A1 | Explanation involving \(2 \neq 3\) and a conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Let A play strategy 1 with probability \(p\) and strategy 2 with probability \(1-p\), using this to get at least one equation in \(p\) | M1 | Translating situation into model by defining variables and constructing at least one equation |
| If B plays strategy 1, A's gains are \(4p + 2(1-p) = 2p + 2\); If B plays strategy 2, A's gains are \(p + 4(1-p) = 4 - 3p\); If B plays strategy 3, A's gains are \(2p + 3(1-p) = 3 - p\) | A1, A1 | One row correct; all three rows correct |
| Graph with axes correct, at least one line correctly drawn | M1 | Axes correct, at least one line correctly drawn for their expression |
| Correct graph | A1 | Correct graph |
| Intersection of \(2p + 2\) and \(3 - p\) occurs where \(p = \frac{1}{3}\) | dM1, A1ft | Using probability expectation graph to find probability by equating two correct expressions and attempting to solve as far as \(p =\) |
| Therefore player A should play strategy 1 \(\frac{1}{3}\) of the time and play strategy 2 \(\frac{2}{3}\) of the time | A1ft | Interpret their value of \(p\) in context – must refer to play, player A |
| The value of the game to player A is \(2\frac{2}{3}\) | A1 | cao |
## Question 4:
### Part (a):
| Row minima: 1, 2; max is 2. Column maxima: 4, 4, 3; min is 3 | M1, A1 | Finding row minimums and column maximums – condone one error; row minima and column maxima correct |
| Row maximin $(2) \neq$ Column minimax $(3)$ so not stable | A1 | Explanation involving $2 \neq 3$ and a conclusion |
### Part (b):
| Let A play strategy 1 with probability $p$ and strategy 2 with probability $1-p$, using this to get at least one equation in $p$ | M1 | Translating situation into model by defining variables and constructing at least one equation |
| If B plays strategy 1, A's gains are $4p + 2(1-p) = 2p + 2$; If B plays strategy 2, A's gains are $p + 4(1-p) = 4 - 3p$; If B plays strategy 3, A's gains are $2p + 3(1-p) = 3 - p$ | A1, A1 | One row correct; all three rows correct |
| Graph with axes correct, at least one line correctly drawn | M1 | Axes correct, at least one line correctly drawn for their expression |
| Correct graph | A1 | Correct graph |
| Intersection of $2p + 2$ and $3 - p$ occurs where $p = \frac{1}{3}$ | dM1, A1ft | Using probability expectation graph to find probability by equating two correct expressions and attempting to solve as far as $p =$ |
| Therefore player A should play strategy 1 $\frac{1}{3}$ of the time and play strategy 2 $\frac{2}{3}$ of the time | A1ft | Interpret their value of $p$ in context – must refer to play, player A |
| The value of the game to player A is $2\frac{2}{3}$ | A1 | cao |4. A two person zero-sum game is represented by the following pay-off matrix for player A.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & B plays 1 & B plays 2 & B plays 3 \\
\hline
A plays 1 & 4 & 1 & 2 \\
\hline
A plays 2 & 2 & 4 & 3 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Verify that there is no stable solution.
\item \begin{enumerate}[label=(\roman*)]
\item Find the best strategy for player A.
\item Find the value of the game to her.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel FD2 AS Q4 [12]}}