| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Moderate -0.5 This is a standard game theory question requiring dominance reduction and mixed strategy calculation using routine algorithms. While it involves multiple steps, the techniques are mechanical and well-practiced in D2, making it easier than average A-level maths questions that require more conceptual insight or proof. |
| Spec | 7.08b Dominance: reduce pay-off matrix7.08e Mixed strategies: optimal strategy using equations or graphical method |
| B plays 1 | B plays 2 | B plays 3 | |
| A plays 1 | 5 | 4 | - 6 |
| A plays 2 | - 1 | - 2 | 3 |
| A plays 3 | 1 | - 1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\begin{bmatrix} 4 & -6 \\ -2 & 3 \\ -1 & 2 \end{bmatrix}\) column 2 dominates column 1 | B1 | (1) |
| (b) \(\begin{bmatrix} -4 & 2 & 1 \\ 6 & -3 & -2 \end{bmatrix}\) | B1 B1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| If A plays 3: B's expected winnings = \(p - 2(1 - p) = -2 + 3p\) | 1M1A1 | (2) |
| [Graph showing three lines] | B2, 1ft, 0 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = \frac{8}{13}\) | 2M1 A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| The value of the game is \(-\frac{2}{13}\) to B | B1 B1 | (2) Total 11 |
**(a)** $\begin{bmatrix} 4 & -6 \\ -2 & 3 \\ -1 & 2 \end{bmatrix}$ column 2 dominates column 1 | B1 | (1)
**(b)** $\begin{bmatrix} -4 & 2 & 1 \\ 6 & -3 & -2 \end{bmatrix}$ | B1 B1 | (2)
**(c)** Let $p$ = probability that B plays new row 1
If A plays 1: B's expected winnings = $-4p + 6(1- p) = 6 - 10p$
If A plays 2: B's expected winnings = $2p - 3(1 - p) = -3 + 5p$
If A plays 3: B's expected winnings = $p - 2(1 - p) = -2 + 3p$ | 1M1A1 | (2)
[Graph showing three lines] | B2, 1ft, 0 (2) |
$6 - 10p = -2 + 3p$
$8 = 13p$
$p = \frac{8}{13}$ | 2M1 A1 | (2)
B should play 1: never, play 2 with probability $\frac{8}{13}$ and play 3 with probability $\frac{5}{13}$
The value of the game is $-\frac{2}{13}$ to B | B1 B1 | (2) Total 11
Supporting detail:
a1B1 CAO (accept reduced matrix or 'column 2 dominates column 1' or column crossed out). Allow recover in part (b)
b1B1 either $3 \times 2$ matrix with correct values (including signs) or $2 \times 3$ matrix with correct values (condone incorrect signs)
b2B1 CAO
c1M1 Setting up three probability expressions, implicit definition of 'p'.
c1A1 CAO (condone incorrect simplification)
c1B1ft Attempt at three lines (correct gradients and intersection with 'axes'), accept $p > 1$ or $p < 0$ here. Must be functions of p.
c2B1 CAO $0 \le p \le 1$, scale clear (or 1 line = 1), condone lack of labels. Rulers used.
c2M1 Finding their correct optimal point, must have three lines and set up an equation to find $0 \le p \le 1$. Dependent on first B mark in part (c). Must have three intersection points. Solving all three simultaneous equations only is M0.
c2A1 CSO
c3B1 All three options listed must fit from their p, check page 1 for B should never play 1. $0 \le$ probabilities $\le 1$.
c4B1 -2/13 CAO (accept awrt 0.154)
SC1: If column 2 deleted in (a) candidates can earn a **maximum of** (a) B0 (b) B1 B0 (c) M1 A0 B1 B0 M1 A0 B1 B1 B0 M1 A0 B1 B1 (max. of 6) – the final B mark is for the value of the game being -4/3
SC2: If column 3 is deleted in (a) candidates can earn a **maximum of** (a) B0 (b) B1 B0 (c) M1 A0 B1 B0 M1 A0 B1 B0 M0 A0 B0 B0
---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 | }
\hline
& B plays 1 & B plays 2 & B plays 3 \\
\hline
A plays 1 & 5 & 4 & - 6 \\
\hline
A plays 2 & - 1 & - 2 & 3 \\
\hline
A plays 3 & 1 & - 1 & 2 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Reduce the game so that player B has only two possible 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 him.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 2013 Q4 [11]}}