| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Zero-sum game optimal mixed strategy |
| Difficulty | Standard +0.8 This is a multi-part game theory question requiring graphical solution of a 3×2 zero-sum game. While the individual techniques (checking for saddle points, graphical method for mixed strategies) are standard D2 content, the question requires careful execution across multiple steps including graph construction, finding intersection points, and interpreting results. The 3×2 structure adds complexity compared to simpler 2×2 games, making this moderately challenging for A-level Decision Maths. |
| 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 |
| \cline { 3 - 4 } \multicolumn{2}{c|}{} | \(B\) | ||
| \cline { 3 - 4 } | I | II | |
| \multirow{2}{*}{\(A\)} | I | 4 | \({ } ^ { - } 8\) |
| \cline { 2 - 4 } | II | 2 | \({ } ^ { - } 4\) |
| \cline { 2 - 4 } | III | \({ } ^ { - } 8\) | 2 |
| Answer | Marks |
|---|---|
| Row maximin: \(\max(-8,-4,-8) = -4\) (row II) | M1 |
| Column minimax: \(\min(4,2) = 2\) (col II) | M1 |
| Maximin \(\neq\) minimax (\(-4 \neq 2\)), so no saddle point | A1 |
| Answer | Marks |
|---|---|
| Let \(p\) = probability B plays I. Lines for A's strategies: \(4p - 8(1-p)\), \(2p - 4(1-p)\), \(-8p + 2(1-p)\) | M1 A1 |
| Graph lines, find highest intersection (optimal for B is to maximise minimum) | M1 |
| Intersection of \(12p - 8 = -10p + 2 \Rightarrow 22p = 10 \Rightarrow p = \frac{5}{11}\) | M1 A1 |
| B plays I with probability \(\frac{5}{11}\), II with probability \(\frac{6}{11}\) | A1 |
| Answer | Marks |
|---|---|
| At optimum, A mixes rows I and III (domination removes II) | M1 |
| \(4p - 8(1-p) = -8p + 2(1-p)\); solve for A's probability \(q\) of playing I | M1 A1 |
| A plays I with probability \(\frac{10}{22} = \frac{5}{11}\), III with probability \(\frac{6}{11}\) | A1 |
| Answer | Marks |
|---|---|
| Value \(= 4\left(\frac{5}{11}\right) - 8\left(\frac{6}{11}\right) = \frac{20-48}{11} = -\frac{28}{11}\) | M1 A1 |
# Question 4:
## Part (a)
| Row maximin: $\max(-8,-4,-8) = -4$ (row II) | M1 | |
| Column minimax: $\min(4,2) = 2$ (col II) | M1 | |
| Maximin $\neq$ minimax ($-4 \neq 2$), so no saddle point | A1 | |
## Part (b)
| Let $p$ = probability B plays I. Lines for A's strategies: $4p - 8(1-p)$, $2p - 4(1-p)$, $-8p + 2(1-p)$ | M1 A1 | |
| Graph lines, find highest intersection (optimal for B is to maximise minimum) | M1 | |
| Intersection of $12p - 8 = -10p + 2 \Rightarrow 22p = 10 \Rightarrow p = \frac{5}{11}$ | M1 A1 | |
| B plays I with probability $\frac{5}{11}$, II with probability $\frac{6}{11}$ | A1 | |
## Part (c)
| At optimum, A mixes rows I and III (domination removes II) | M1 | |
| $4p - 8(1-p) = -8p + 2(1-p)$; solve for A's probability $q$ of playing I | M1 A1 | |
| A plays I with probability $\frac{10}{22} = \frac{5}{11}$, III with probability $\frac{6}{11}$ | A1 | |
## Part (d)
| Value $= 4\left(\frac{5}{11}\right) - 8\left(\frac{6}{11}\right) = \frac{20-48}{11} = -\frac{28}{11}$ | M1 A1 | |
---4. The payoff matrix for player $A$ in a two-person zero-sum game is shown below.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\cline { 3 - 4 }
\multicolumn{2}{c|}{} & \multicolumn{2}{c|}{$B$} \\
\cline { 3 - 4 }
& I & II & \\
\hline
\multirow{2}{*}{$A$} & I & 4 & ${ } ^ { - } 8$ \\
\cline { 2 - 4 }
& II & 2 & ${ } ^ { - } 4$ \\
\cline { 2 - 4 }
& III & ${ } ^ { - } 8$ & 2 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Explain why the game does not have a saddle point.
\item Using a graphical method, find the optimal strategy for player $B$.
\item Find the optimal strategy for player $A$.
\item Find the value of the game.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 Q4 [15]}}