| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Zero-sum game optimal mixed strategy |
| Difficulty | Standard +0.8 This D2 game theory question requires systematic analysis of a 3×3 pay-off matrix including identifying play-safe strategies, checking for saddle points, and solving for optimal mixed strategies using linear programming or algebraic methods. While the techniques are standard for D2, the multi-step nature and need to set up and solve equations for mixed strategies makes it moderately challenging, above average difficulty but within typical Further Maths scope. |
| 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 | 0 | - 2 | 6 |
| A plays 2 | 3 | 4 | 1 |
| A plays 3 | - 1 | 1 | - 3 |
| Answer | Marks |
|---|---|
| Row minima: -2, 1, -3 max is 1 | M1 A1 |
| Column maxima: 3, 4, 6 min is 3 | |
| Play safe is A plays 2 and B plays 1 | A1 (3) |
| Answer | Marks |
|---|---|
| Row maximum (1) ≠ Column minimax (3) so not stable | B1 (1) |
| Answer | Marks |
|---|---|
| Row 2 dominates row 3 so delete row 3 | B1 |
| Let A play 1 with probability \(p\) and 2 with probability \(1–p\) | |
| If B plays 1 A's expected winnings are \(3(1–p) = –3p + 3\) | |
| If B plays 2 A's expected winnings are \(–2p + 4(1–p) = –6p + 4\) | M1 A1 |
| If B plays 3 A's expected winnings are \(6p + (1–p) = 5p + 1\) | |
| Graph showing three lines with correct slant direction and relative intersection with 'axes' | M1 A1 |
| Answer | Marks |
|---|---|
| \(5p + 1 = 3 – 3p \Rightarrow p = \frac{1}{4}\) | DM1 A1 |
| A should play row 1 with probability \(\frac{1}{4}\), row 2 with probability \(\frac{3}{4}\) and row 3 never | A1 (8) |
| Answer | Marks |
|---|---|
| Value of the game to player B is \(–\frac{9}{4}\) | B1 (1) |
**Part (a):**
Row minima: -2, 1, -3 max is 1 | M1 A1
Column maxima: 3, 4, 6 min is 3 |
Play safe is A plays 2 and B plays 1 | A1 (3)
**Part (b):**
Row maximum (1) ≠ Column minimax (3) so not stable | B1 (1)
**Part (c):**
Row 2 dominates row 3 so delete row 3 | B1
Let A play 1 with probability $p$ and 2 with probability $1–p$ |
If B plays 1 A's expected winnings are $3(1–p) = –3p + 3$ |
If B plays 2 A's expected winnings are $–2p + 4(1–p) = –6p + 4$ | M1 A1
If B plays 3 A's expected winnings are $6p + (1–p) = 5p + 1$ |
Graph showing three lines with correct slant direction and relative intersection with 'axes' | M1 A1
**Finding the optimal point:**
$5p + 1 = 3 – 3p \Rightarrow p = \frac{1}{4}$ | DM1 A1
A should play row 1 with probability $\frac{1}{4}$, row 2 with probability $\frac{3}{4}$ and row 3 never | A1 (8)
**Part (d):**
Value of the game to player B is $–\frac{9}{4}$ | B1 (1)
---3. 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
& B plays 1 & B plays 2 & B plays 3 \\
\hline
A plays 1 & 0 & - 2 & 6 \\
\hline
A plays 2 & 3 & 4 & 1 \\
\hline
A plays 3 & - 1 & 1 & - 3 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Identify the play safe strategies for each player.
\item State, giving a reason, whether there is a stable solution to this game.
\item Find the best strategy for player A.
\item Find the value of the game to player B.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 2017 Q3 [13]}}