| Exam Board | OCR |
|---|---|
| Module | Further Discrete AS (Further Discrete AS) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Standard +0.8 This is a game theory problem requiring understanding of play-safe strategies, minimax concepts, and algebraic reasoning with parameters. Part (a) is straightforward, but part (b) requires setting up inequalities to ensure strategy A has the maximum row minimum (involving parameter k), and part (c) requires finding column maxima then their minimum. The multi-step reasoning with parameters and game theory concepts places this above average difficulty but not exceptionally hard for Further Maths students. |
| Spec | 7.08a Pay-off matrix: zero-sum games7.08c Pure strategies: play-safe strategies and stable solutions |
| Strategy E | Strategy F | Strategy G | Strategy H | |
| Strategy A | \(2 k\) | -2 | \(1 - k\) | 4 |
| Strategy B | -3 | 3 | 4 | -5 |
| Strategy C | 1 | 4 | -4 | 2 |
| Strategy D | 4 | -2 | -5 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | 5 |
| [1] | cao | |
| 5 | (b) | Row minima: |
| Answer | Marks |
|---|---|
| So A is a play-safe strategy when –2 ≤ k ≤ 5 | M1 |
| Answer | Marks |
|---|---|
| [6] | 2k, –2, 1 – k (may also include 4) |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | E: max{2k, 4}, F: 4 G: max{4, 1 – k}, H: 6 |
| Answer | Marks |
|---|---|
| Hence col minimax = 4 | M1 |
| Answer | Marks |
|---|---|
| [2] | Any three of these correct (not implied), ‘max’ may be |
Question 5:
5 | (a) | 5 | B1
[1] | cao
5 | (b) | Row minima:
A = min{2k, –2, 1 – k}
B = – 5
C = – 4
D = – 5
2k = – 4 when k = – 2
1 – k = – 4 when k = 5
So A is a play-safe strategy when –2 ≤ k ≤ 5 | M1
A1
B1
B1
M1
A1
[6] | 2k, –2, 1 – k (may also include 4)
– 4 seen or used appropriately
Critical value k = – 2 (or as an inequality)
Critical value k = 5 (or as an inequality)
Appropriate sketch graph showing y = 2k, y = 1 – k and at
least one of y = – 2, y = – 4
Or implied from answer
Accept < instead of (either or both ) ≤
5 | (c) | E: max{2k, 4}, F: 4 G: max{4, 1 – k}, H: 6
max{2k, 4} ≥ 4 and max{4, 1 – k} ≥ 4 and F = 4
Hence col minimax = 4 | M1
A1
[2] | Any three of these correct (not implied), ‘max’ may be
implied
4 from correct working seen5 The number of points won by player 1 in a zero-sum game is shown in the pay-off matrix below, where $k$ is a constant.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Player 2}
\begin{tabular}{|l|l|l|l|l|}
\hline
& Strategy E & Strategy F & Strategy G & Strategy H \\
\hline
Strategy A & $2 k$ & -2 & $1 - k$ & 4 \\
\hline
Strategy B & -3 & 3 & 4 & -5 \\
\hline
Strategy C & 1 & 4 & -4 & 2 \\
\hline
Strategy D & 4 & -2 & -5 & 6 \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item In one game, player 2 chooses strategy H.
Write down the greatest number of points that player 2 could win.
You are given that strategy A is a play-safe strategy for player 1.
\item Determine the range of possible values for $k$.
\item Determine the column minimax value.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Discrete AS 2020 Q5 [9]}}