| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Game theory LP formulation |
| Difficulty | Standard +0.3 This is a standard D2 game theory question following a well-rehearsed algorithm: apply dominance rules to reduce the matrix, then solve the 2×2 game using the standard formulas for mixed strategies. While it requires multiple steps, each follows a mechanical procedure taught explicitly in the syllabus with no novel problem-solving required. |
| Spec | 7.08b 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 { 3 - 5 } \multicolumn{2}{c|}{} | \(B\) | |||
| \cline { 2 - 5 } \multicolumn{2}{c|}{} | I | II | III | |
| \multirow{3}{*}{\(A\)} | I | 3 | 5 | - 2 |
| \cline { 2 - 5 } | II | 7 | - 4 | - 1 |
| \cline { 2 - 5 } | III | 9 | - 4 | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Player \(A\) can ignore row II since row III dominates row II (each entry \(\geq\) corresponding entry in row II) | B1 | Correct row with reason |
| Player \(B\) can ignore column I since column II dominates column I (\(B\) minimises; col II entries \(\leq\) col I in reduced table) | B1B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Reduced \(2\times 2\) table used; let \(p\) = prob \(A\) plays I | M1 | |
| \(3p + 9(1-p) = -2p + 8(1-p)\) solved for player \(A\) | M1A1 | |
| \(p = \frac{1}{4}\), so \(A\) plays I with prob \(\frac{1}{4}\), III with prob \(\frac{3}{4}\) | A1 | |
| Player \(B\): \(q\) = prob plays II; \(5q-2(1-q)=8q+8(1-q)\)... correct equation | M1A1 | |
| \(q\) found correctly | A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Value substituted and calculated | M1 | |
| Value of game stated correctly | A1 |
# Question 6:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Player $A$ can ignore row II since row III dominates row II (each entry $\geq$ corresponding entry in row II) | B1 | Correct row with reason |
| Player $B$ can ignore column I since column II dominates column I ($B$ minimises; col II entries $\leq$ col I in reduced table) | B1B1 | |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Reduced $2\times 2$ table used; let $p$ = prob $A$ plays I | M1 | |
| $3p + 9(1-p) = -2p + 8(1-p)$ solved for player $A$ | M1A1 | |
| $p = \frac{1}{4}$, so $A$ plays I with prob $\frac{1}{4}$, III with prob $\frac{3}{4}$ | A1 | |
| Player $B$: $q$ = prob plays II; $5q-2(1-q)=8q+8(1-q)$... correct equation | M1A1 | |
| $q$ found correctly | A1A1 | |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Value substituted and calculated | M1 | |
| Value of game stated correctly | A1 | |
---6. The payoff matrix for player $A$ in a two-person zero-sum game is shown below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\cline { 3 - 5 }
\multicolumn{2}{c|}{} & \multicolumn{3}{|c|}{$B$} \\
\cline { 2 - 5 }
\multicolumn{2}{c|}{} & I & II & III \\
\hline
\multirow{3}{*}{$A$} & I & 3 & 5 & - 2 \\
\cline { 2 - 5 }
& II & 7 & - 4 & - 1 \\
\cline { 2 - 5 }
& III & 9 & - 4 & 8 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Applying the dominance rule, explain, with justification, which strategy can be ignored by
\begin{enumerate}[label=(\roman*)]
\item player $A$,
\item player $B$.
\end{enumerate}\item For the reduced table, find the optimal strategy for
\begin{enumerate}[label=(\roman*)]
\item player $A$,
\item player $B$.
\end{enumerate}\item Find the value of the game.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 Q6 [13]}}