| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Standard +0.8 This is a game theory problem requiring verification of no saddle point, then solving a 2×3 game using linear programming or graphical methods to find optimal mixed strategies. While systematic, it demands understanding of minimax theorem, strategy dominance, and multi-step algebraic manipulation beyond routine A-level content. |
| 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 |
| I | II | III | |
| I | 5 | 2 | 3 |
| II | 3 | 5 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Payoff table with row minima: I gives 2, II gives 3; column maxima: I gives 5, II gives 5, III gives 4; maximin \(= 3\), minimax \(= 4\) | M1 A1 | |
| Since \(3 \neq 4\), not stable | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Let \(A\) play I with probability \(p\), play II with probability \((1-p)\) | ||
| If \(B\) plays I: \(A\)'s gain \(= 5p + 3(1-p) = 2p+3\) | M1 A1 | (2 marks) |
| If \(B\) plays II: \(A\)'s gain \(= 2p + 5(1-p) = 5-3p\) | ||
| If \(B\) plays III: \(A\)'s gain \(= 3p + 4(1-p) = 4-p\) | A2, 1, 0 | (2 marks) |
| Graph drawn showing lines; intersection of \(2p+3\) and \(4-p\) gives \(p = \dfrac{1}{3}\) | M1 A1 ft | |
| \(A\) should play I \(\frac{1}{3}\) of time and II \(\frac{2}{3}\) of time; value (to \(A\)) \(= 3\tfrac{2}{3}\) | A1 ft A1 ft | (2 marks) |
# Question 8:
## Part (a)
| Answer/Working | Marks | Notes |
|---|---|---|
| Payoff table with row minima: I gives 2, II gives 3; column maxima: I gives 5, II gives 5, III gives 4; maximin $= 3$, minimax $= 4$ | M1 A1 | |
| Since $3 \neq 4$, not stable | A1 | (3 marks) |
## Part (b)
| Answer/Working | Marks | Notes |
|---|---|---|
| Let $A$ play I with probability $p$, play II with probability $(1-p)$ | | |
| If $B$ plays I: $A$'s gain $= 5p + 3(1-p) = 2p+3$ | M1 A1 | (2 marks) |
| If $B$ plays II: $A$'s gain $= 2p + 5(1-p) = 5-3p$ | | |
| If $B$ plays III: $A$'s gain $= 3p + 4(1-p) = 4-p$ | A2, 1, 0 | (2 marks) |
| Graph drawn showing lines; intersection of $2p+3$ and $4-p$ gives $p = \dfrac{1}{3}$ | M1 A1 ft | |
| $A$ should play I $\frac{1}{3}$ of time and II $\frac{2}{3}$ of time; value (to $A$) $= 3\tfrac{2}{3}$ | A1 ft A1 ft | (2 marks) |
**Total: 15 marks**8. 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 | }
\hline
& I & II & III \\
\hline
I & 5 & 2 & 3 \\
\hline
II & 3 & 5 & 4 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Verify that there is no stable solution to this game.
\item Find the best strategy for player $A$ and the value of the game to her.\\
(Total 11 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 Q8 [11]}}