| Exam Board | OCR |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Marks | 12 |
| 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 understanding of saddle points, graphical solution methods for mixed strategies, and calculation of game values. While the techniques are standard for D2, the question demands careful execution across multiple steps (checking for saddle point, constructing and interpreting a graph, solving simultaneous equations for probabilities, and computing expected value). This is more challenging than routine A-level questions but remains within the D2 syllabus 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 |
| \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 | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (a) Row minimums: \(-8, -4, -8\); Column maximums: 4, 2 | M1 | |
| \(\max(\text{row min}) = -4\), \(\min(\text{col max}) = 2\); since these are not equal, no saddle point | A1 | |
| (b) Expected loss for \(B\): \(AI: 4q-8(1-q)=12q-8\); \(AII: 2q-4(1-q)=6q-4\); \(AIII: -8q+2(1-q)=2-10q\) | M1 A1 | |
| Graph of three lines plotted correctly | M1 | |
| Strategy I not worth \(A\) considering (dominated) | A1 | |
| For optimal strategy: \(6q-4=2-10q \Rightarrow 16q=6, \ q=\frac{3}{8}\) | M1 | |
| \(B\) should play I \(\frac{3}{8}\) of time and II \(\frac{5}{8}\) of time | A1 | |
| (c) \(BI: 2p-8(1-p)=10p-8\); \(BII: -4p+2(1-p)=2-6p\) | M1 A1 | |
| For optimal: \(10p-8=2-6p \Rightarrow 16p=10, \ p=\frac{5}{8}\) | ||
| \(A\) should play I never, II \(\frac{5}{8}\) of time, III \(\frac{3}{8}\) of time | A1 | |
| (d) Value of game \(= (6\times\frac{5}{8})-4 = -1\frac{3}{4}\) | A1 | (12) |
# Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(a)** Row minimums: $-8, -4, -8$; Column maximums: 4, 2 | M1 | |
| $\max(\text{row min}) = -4$, $\min(\text{col max}) = 2$; since these are not equal, no saddle point | A1 | |
| **(b)** Expected loss for $B$: $AI: 4q-8(1-q)=12q-8$; $AII: 2q-4(1-q)=6q-4$; $AIII: -8q+2(1-q)=2-10q$ | M1 A1 | |
| Graph of three lines plotted correctly | M1 | |
| Strategy I not worth $A$ considering (dominated) | A1 | |
| For optimal strategy: $6q-4=2-10q \Rightarrow 16q=6, \ q=\frac{3}{8}$ | M1 | |
| $B$ should play I $\frac{3}{8}$ of time and II $\frac{5}{8}$ of time | A1 | |
| **(c)** $BI: 2p-8(1-p)=10p-8$; $BII: -4p+2(1-p)=2-6p$ | M1 A1 | |
| For optimal: $10p-8=2-6p \Rightarrow 16p=10, \ p=\frac{5}{8}$ | | |
| $A$ should play I never, II $\frac{5}{8}$ of time, III $\frac{3}{8}$ of time | A1 | |
| **(d)** Value of game $= (6\times\frac{5}{8})-4 = -1\frac{3}{4}$ | A1 | **(12)** |6. 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{OCR D2 Q6 [12]}}