| Exam Board | OCR |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Basic committee/group selection |
| Difficulty | Standard +0.3 This is a standard D2 game theory question requiring routine application of expected value formulas and graphical methods. While it involves multiple parts, each step follows a textbook procedure: calculating expected payoffs (linear expressions in p), plotting three lines, finding their intersection, and interpreting the maximin value. No novel insight or complex problem-solving is required, making it slightly easier than average for A-level. |
| 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 { 2 - 5 } | Strategy \(X\) | Strategy \(Y\) | Strategy \(Z\) | |
| \cline { 2 - 5 } Rowan | Strategy \(P\) | 5 | - 3 | - 2 |
| \cline { 2 - 5 } | Strategy \(Q\) | - 4 | 3 | 1 |
| \cline { 2 - 5 } | ||||
| \cline { 2 - 5 } |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(5p - 4(1-p) = 9p - 4\) | M1 |
| A1 | \(9p - 4\) or \(-4 + 4 + 9p\) | |
| [2] | ||
| (ii) | Correct structure to graph with lines \(E = 9p - 4\), \(E = 3 - 6p\), \(E = 1 - 3p\) | M1 |
| Line \(E = 9p - 4\) plotted from \((0,4)\) to \((1, 5)\) | A1 | Line \(E = 9p - 4\) plotted from (0,4) to (1, 5) |
| Line \(E = 3 - 6p\) plotted from \((0, 3)\) to \((1,-3)\) | A1 | Line \(E = 3 - 6p\) plotted from (0, 3) to (1,-3) |
| Line \(E = 1 - 3p\) plotted from \((0, 1)\) to \((1,-2)\) | A1 | Line \(E = 1 - 3p\) plotted from (0, 1) to (1,-2) |
| Withhold an A1 for horizontal scale beyond 0 to 1 | ||
| [4] | ||
| (iii) | \(9p - 4 = 1 - 3p \Rightarrow p = 5/12\) or \(0.41\) to \(0.42\) (or better) | M1 |
| A1 ft | Correct value for their lines | |
| [2] | ||
| (iv) | If Colin plays \(X\) or \(Z\), Rowan's expected winnings are \(-0.25\) so Colin's expected winnings are \(+0.25\) | B1 |
| Even if Colin plays optimally he cannot expect, in the long run, to do better on average than to win what Rowan loses | B1 | Realising that Colin need to play his optimal strategy as well as Rowan |
| [2] |
(i) | $5p - 4(1-p) = 9p - 4$ | M1 | This, or implied |
| | | A1 | $9p - 4$ or $-4 + 4 + 9p$ |
| | | [2] | |
(ii) | Correct structure to graph with lines $E = 9p - 4$, $E = 3 - 6p$, $E = 1 - 3p$ | M1 | Correct structure to graph |
| | Line $E = 9p - 4$ plotted from $(0,4)$ to $(1, 5)$ | A1 | Line $E = 9p - 4$ plotted from (0,4) to (1, 5) |
| | Line $E = 3 - 6p$ plotted from $(0, 3)$ to $(1,-3)$ | A1 | Line $E = 3 - 6p$ plotted from (0, 3) to (1,-3) |
| | Line $E = 1 - 3p$ plotted from $(0, 1)$ to $(1,-2)$ | A1 | Line $E = 1 - 3p$ plotted from (0, 1) to (1,-2) |
| | | | Withhold an A1 for horizontal scale beyond 0 to 1 |
| | | [4] | |
(iii) | $9p - 4 = 1 - 3p \Rightarrow p = 5/12$ or $0.41$ to $0.42$ (or better) | M1 | Solving the correct pair of lines for their graph |
| | | A1 ft | Correct value for their lines |
| | | [2] | |
(iv) | If Colin plays $X$ or $Z$, Rowan's expected winnings are $-0.25$ so Colin's expected winnings are $+0.25$ | B1 | Showing why it is $+0.25$ for Colin |
| | Even if Colin plays optimally he cannot expect, in the long run, to do better on average than to win what Rowan loses | B1 | Realising that Colin need to play his optimal strategy as well as Rowan |
| | | [2] | |
---4 The table gives the pay-off matrix for a zero-sum game between two players, Rowan and Colin.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Colin}
\begin{tabular}{ | l | l | c | c | c | }
\cline { 2 - 5 }
& & Strategy $X$ & Strategy $Y$ & Strategy $Z$ \\
\cline { 2 - 5 }
Rowan & Strategy $P$ & 5 & - 3 & - 2 \\
\cline { 2 - 5 }
& Strategy $Q$ & - 4 & 3 & 1 \\
\cline { 2 - 5 }
& & & & \\
\cline { 2 - 5 }
\end{tabular}
\end{center}
\end{table}
Rowan makes a random choice between strategies $P$ and $Q$, choosing strategy $P$ with probability $p$ and strategy $Q$ with probability $1 - p$.\\
(i) Write down and simplify an expression for the expected pay-off for Rowan when Colin chooses strategy $X$.\\
(ii) Using graph paper, draw a graph to show Rowan's expected pay-off against $p$ for each of Colin's choices of strategy.\\
(iii) Using your graph, find the optimal value of $p$ for Rowan.\\
(iv) Rowan plays using the optimal value of $p$. Explain why, in the long run, Colin cannot expect to win more than 0.25 per game.
\hfill \mbox{\textit{OCR D2 2007 Q4 [10]}}