| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| 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 dominance reduction and mixed strategy calculation. While multi-step, each part follows a mechanical procedure taught directly in D2: checking for saddle points, identifying dominated strategies, and solving 2×2 games using the standard formula. No novel insight or complex problem-solving is required. |
| 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 |
| B plays 1 | B plays 2 | B plays 3 | |
| A plays 1 | - 5 | 6 | - 3 |
| A plays 2 | 1 | - 4 | 13 |
| A plays 3 | - 2 | 3 | - 1 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | Row minima \(\{-5, -4, -2\}\) row maximum \(= -2\); Column maxima \(\{1, 6, 13\}\) col minimax \(= 1\); \(-2 \neq 1\) therefore not stable. | M1 A1 A1 (3) |
| (b) | Column 1 dominates column 3, so column 3 can be deleted. | B1 (1) |
| (c) | \(\begin{array}{c | ccc} & \text{A plays 1} & \text{A plays 2} & \text{A plays 3} \\ \hline \text{B plays 1} & 5 & -1 & 2 \\ \text{B plays 2} & -6 & 4 & -3 \end{array}\) |
| (d) | Let B play row 1 with probability \(p\) and row 2 with probability \((1-p)\) | M1 A1 |
| If A plays 1, B's expected winnings are \(11p - 6\) | ||
| If A plays 2, B's expected winnings are \(4 - 5p\) | ||
| If A plays 3, B's expected winnings are \(5p - 3\) | ||
| Graph showing three lines with appropriate intercepts and intersections | M1 A1 | |
| \(5p - 3 = 4 - 5p\); \(10p = 7\); \(p = \frac{7}{10}\) | M1 | |
| B should play 1 with a probability of 0.7, 2 with a probability of 0.3 and never play 3 | A1 | |
| The value of the game is 0.5 to B | A1 (7) |
(a) | Row minima $\{-5, -4, -2\}$ row maximum $= -2$; Column maxima $\{1, 6, 13\}$ col minimax $= 1$; $-2 \neq 1$ therefore not stable. | M1 A1 A1 (3) | |
(b) | Column 1 dominates column 3, so column 3 can be deleted. | B1 (1) | |
(c) | $\begin{array}{c|ccc} & \text{A plays 1} & \text{A plays 2} & \text{A plays 3} \\ \hline \text{B plays 1} & 5 & -1 & 2 \\ \text{B plays 2} & -6 & 4 & -3 \end{array}$ | B1 B1 (2) | |
(d) | Let B play row 1 with probability $p$ and row 2 with probability $(1-p)$ | M1 A1 | |
| If A plays 1, B's expected winnings are $11p - 6$ | | |
| If A plays 2, B's expected winnings are $4 - 5p$ | | |
| If A plays 3, B's expected winnings are $5p - 3$ | | |
| Graph showing three lines with appropriate intercepts and intersections | M1 A1 | |
| $5p - 3 = 4 - 5p$; $10p = 7$; $p = \frac{7}{10}$ | M1 | |
| B should play 1 with a probability of 0.7, 2 with a probability of 0.3 and never play 3 | A1 | |
| The value of the game is 0.5 to B | A1 (7) | |
**Total [13]**
---3. 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 | }
\hline
& B plays 1 & B plays 2 & B plays 3 \\
\hline
A plays 1 & - 5 & 6 & - 3 \\
\hline
A plays 2 & 1 & - 4 & 13 \\
\hline
A plays 3 & - 2 & 3 & - 1 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Verify that there is no stable solution to this game.
\item Reduce the game so that player B has a choice of only two actions.
\item Write down the reduced pay-off matrix for player B.
\item Find the best strategy for player B and the value of the game to player B.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 2009 Q3 [13]}}