| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2003 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Moderate -0.5 This is a standard game theory question from Decision Mathematics requiring systematic application of well-defined algorithms (maximin/minimax, dominance, and solving 2×2 games). While it has multiple parts, each step follows textbook procedures with no novel insight required, making it slightly easier than average A-level difficulty. |
| Spec | 7.08a Pay-off matrix: zero-sum games7.08b 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 { 2 - 4 } \multicolumn{1}{c|}{} | \(B\) plays I | \(B\) plays II | \(B\) plays III |
| \(A\) plays I | 2 | - 1 | 3 |
| \(A\) plays II | 1 | 3 | 0 |
| \(A\) plays III | 0 | 1 | - 3 |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | \(B\) plays I | \(B\) plays II | \(B\) plays III |
| \(A\) plays I | 2 | - 1 | 3 |
| \(A\) plays II | 1 | 3 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | Player A: row minimums are −1, 0, −3 so maximin choice is play II. Player B: column maximums are 2, 3, 3 so minimax choice is play I | M1 A1 M1 A1 |
| (b) | Since A's maximum (0) ≠ B's minimax (2) there is no stable solution | B1 |
| (c) | For player A row II dominates row III, so A will now play III | B2, 1, 0 |
| (d) | Let A play I with probability \(p\) and II with probability \((1-p)\). If B plays I, A's expected winnings are \(2p + (1-p) = 1 + p\). If B plays II, A's expected winnings are \(-p + 3(1-p) = 3 - 4p\). If B plays III, A's expected winnings are \(3p\) | M1, A2, 1, 0 |
| Graph showing: \(3 - 4p = 3p \Rightarrow p = \frac{3}{7}\) | M1 A1 | - |
| A should play I with probability \(\frac{3}{7}\) and A should play II with probability \(\frac{4}{7}\) and never play III. The value of the game is \(\frac{9}{7}\) to A | A1 A1 ft | 4 |
(a) | Player A: row minimums are −1, 0, −3 so maximin choice is play II. Player B: column maximums are 2, 3, 3 so minimax choice is play I | M1 A1 M1 A1 | 4 |
(b) | Since A's maximum (0) ≠ B's minimax (2) there is no stable solution | B1 | 1 |
(c) | For player A row II dominates row III, so A will now play III | B2, 1, 0 | 2 |
(d) | Let A play I with probability $p$ and II with probability $(1-p)$. If B plays I, A's expected winnings are $2p + (1-p) = 1 + p$. If B plays II, A's expected winnings are $-p + 3(1-p) = 3 - 4p$. If B plays III, A's expected winnings are $3p$ | M1, A2, 1, 0 | 3 |
Graph showing: $3 - 4p = 3p \Rightarrow p = \frac{3}{7}$ | M1 A1 | - |
A should play I with probability $\frac{3}{7}$ and A should play II with probability $\frac{4}{7}$ and never play III. The value of the game is $\frac{9}{7}$ to A | A1 A1 ft | 4 |4. A two person zero-sum game is represented by the following pay-off matrix for player $A$.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & $B$ plays I & $B$ plays II & $B$ plays III \\
\hline
$A$ plays I & 2 & - 1 & 3 \\
\hline
$A$ plays II & 1 & 3 & 0 \\
\hline
$A$ plays III & 0 & 1 & - 3 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Identify the play safe strategies for each player.
\item Verify that there is no stable solution to this game.
\item Explain why the pay-off matrix above may be reduced to
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & $B$ plays I & $B$ plays II & $B$ plays III \\
\hline
$A$ plays I & 2 & - 1 & 3 \\
\hline
$A$ plays II & 1 & 3 & 0 \\
\hline
\end{tabular}
\end{center}
\item Find the best strategy for player $A$, and the value of the game.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 2003 Q4 [14]}}