| Exam Board | OCR |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Easy -2.0 This is a straightforward game theory question requiring only basic definitions and routine checking. Part (i) is trivial arithmetic, (ii) involves simple comparison of payoffs, (iii) requires finding maximin/minimax values using standard algorithms, and (iv) needs only conceptual understanding of play-safe strategies. No proof, novel insight, or complex multi-step reasoning required—purely mechanical application of definitions from Decision Mathematics. |
| Spec | 7.08a Pay-off matrix: zero-sum games7.08c Pure strategies: play-safe strategies and stable solutions |
| \(X\) | \(Y\) | \(Z\) | ||
| \cline { 2 - 5 } | \(X\) | 5 | - 3 | 1 |
| \cline { 2 - 5 } Rebecca | \(Y\) | 3 | 2 | - 2 |
| \cline { 2 - 5 } | \(Z\) | - 1 | 1 | 3 |
| \cline { 2 - 5 } | ||||
| \cline { 2 - 5 } |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(-5\) | B1 |
| [1] | ||
| (ii) | Because \(-3 < 2\) in column \(Y\) and \(2 > -2\) in row \(Y\) | M1 |
| A1 | Both of these comparisons and no others | |
| (iii) | Play-safe for Rebecca is \(Z\) | B1 |
| Play-safe for Claire is \(Y\) | B1 | Indicating column \(Y\) |
| Best choice is \(X\) | B1 ft | The correct choice with their play-safe |
| [3] | ||
| (iv) | For Rebecca: \(-1 >\) smaller of \(\{-3\), value that 5 becomes\(\}\) | B1 |
| For Claire: \(2 <\) larger of \(\{3\), value that 5 becomes\(\}\) | B1 | This, or equivalent (but NOT '5 is not in the play-safe column') |
| [2] |
(i) | $-5$ | B1 | $-5$ |
| | [1] | |
(ii) | Because $-3 < 2$ in column $Y$ and $2 > -2$ in row $Y$ | M1 | Either of these, possibly with others |
| | | A1 | Both of these comparisons and no others |
(iii) | Play-safe for Rebecca is $Z$ | B1 | Indicating row $Z$ |
| | Play-safe for Claire is $Y$ | B1 | Indicating column $Y$ |
| | Best choice is $X$ | B1 ft | The correct choice with their play-safe |
| | | [3] | |
(iv) | For Rebecca: $-1 >$ smaller of $\{-3$, value that 5 becomes$\}$ | B1 | This, or equivalent, or 5 is not in the play-safe row |
| | For Claire: $2 <$ larger of $\{3$, value that 5 becomes$\}$ | B1 | This, or equivalent (but NOT '5 is not in the play-safe column') |
| | | [2] | |
---3 Rebecca and Claire repeatedly play a zero-sum game in which they each have a choice of three strategies, $X , Y$ and $Z$.
The table shows the number of points Rebecca scores for each pair of strategies.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Claire}
\begin{tabular}{ | c | c | r | r | r | }
\hline
& \multicolumn{1}{|c|}{$X$} & \multicolumn{1}{c|}{$Y$} & \multicolumn{1}{c|}{$Z$} & \\
\cline { 2 - 5 }
& $X$ & 5 & - 3 & 1 \\
\cline { 2 - 5 }
Rebecca & $Y$ & 3 & 2 & - 2 \\
\cline { 2 - 5 }
& $Z$ & - 1 & 1 & 3 \\
\cline { 2 - 5 }
& & & & \\
\cline { 2 - 5 }
\end{tabular}
\end{center}
\end{table}
(i) If both players choose strategy $X$, how many points will Claire score?\\
(ii) Show that row $X$ does not dominate row $Y$ and that column $Y$ does not dominate column $Z$.\\
(iii) Find the play-safe strategies. State which strategy is best for Claire if she knows that Rebecca will play safe.\\
(iv) Explain why decreasing the value ' 5 ' when both players choose strategy $X$ cannot alter the playsafe strategies.
\hfill \mbox{\textit{OCR D2 2007 Q3 [8]}}