| Exam Board | OCR |
|---|---|
| Module | Further Discrete (Further Discrete) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| 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, dominance, Nash equilibria, and expected values with mixed strategies. While the individual calculations are straightforward, students must apply multiple game theory concepts systematically across four parts, which is more demanding than typical A-level questions but uses accessible mathematics (comparing numbers, basic probability). |
| Spec | 7.08c Pure strategies: play-safe strategies and stable solutions7.08d Nash equilibrium: identification and interpretation7.08e Mixed strategies: optimal strategy using equations or graphical method |
| \multirow{2}{*}{} | Player 2 | |||
| X | Y | Z | ||
| \multirow{3}{*}{Player 1} | A | \(( 6,0 )\) | \(( 1,7 )\) | \(( 5,6 )\) |
| B | \(( 9,4 )\) | \(( 2,6 )\) | \(( 8,1 )\) | |
| C | \(( 6,8 )\) | \(( 1,3 )\) | \(( 7,2 )\) | |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | Player 2 row min |
| Answer | Marks |
|---|---|
| Play-safe for player 2 (cols) is Y | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Calculating row minima for player 1 or column minima for player 2 |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (b) | Column Z is (strictly) dominated by column Y |
| 7 > 6, 6 > 1 and 3 > 2 | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 2.2a | Identifying Y (only) (as better / dominating) [NOT X and Y] |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (c) | Player 2 |
| Answer | Marks |
|---|---|
| Nash equilibrium at (B, Y) | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Identifying at least 4 of (A, Y), (B, Y), (C, X), (B, X), (B, Y), (B, Z) |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (d) | (i) |
| Answer | Marks |
|---|---|
| p = 0.5 | M1 * |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | Finding expressions for the expected number of points won by each |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (d) | (ii) |
| Answer | Marks |
|---|---|
| Hence player 1 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 2.2a | Calculating 9p + 2(1 – p) or 4p + 6(1 – p) for their p (seen) |
| Answer | Marks | Guidance |
|---|---|---|
| (6, 0) | (1, 7) | |
| (9, 4) | (2, 6) | |
| (6, 0) | (1, 7) | |
| (9, 4) | (2, 6) | (8, 1) |
Question 5:
5 | (a) | Player 2 row min
X Y Z
A (6, 0) (1, 7) (5, 6) 1
Player 1 B (9, 4) (2, 6) (8, 1) 2
C (6, 8) (1, 3) (7, 2) 1
col min 0 3 1
Play-safe for player 1 (rows) is B
Play-safe for player 2 (cols) is Y | M1
A1
A1
[3] | 1.1
1.1
1.1 | Calculating row minima for player 1 or column minima for player 2
B from sight of 1, 2, 1 as row minima
Y from sight of 0, 3, 1 as col minima
SC B1 Both play-safes correct but no (or insufficient) working
5 | (b) | Column Z is (strictly) dominated by column Y
7 > 6, 6 > 1 and 3 > 2 | B1
B1
[2] | 1.1
2.2a | Identifying Y (only) (as better / dominating) [NOT X and Y]
Three appropriate comparisons, or equivalent in words, e.g.
Y gives player 2 more points than Z, for each of player 1’s choices
5 | (c) | Player 2
X Y Z
A (6, 0) (1, 7) (5, 6) Y
Player 1 B (9, 4) (2, 6) (8, 1) Y
C (6, 8) (1, 3) (7, 2) X
B B B
Nash equilibrium at (B, Y) | B1
B1
[2] | 1.1
1.1 | Identifying at least 4 of (A, Y), (B, Y), (C, X), (B, X), (B, Y), (B, Z)
or at least 4 of B B B and Y Y X (with X for row C)
May be seen in table
May be convincingly argued in words
(B, Y)
5 | (d) | (i) | 6(p) + 1(1 – p) = 0(p) + 7(1 – p)
1 + 5p = 7 – 7p
p = 0.5 | M1 *
M1
dep*
A1
[3] | 3.1a
1.1
1.1 | Finding expressions for the expected number of points won by each
player using row A (not two or more rows for one player)
Equate these expressions or sketch graph o.e. and solve for p
(or implied from correct expressions seen and p = 0.5)
0.5, cao from valid working
5 | (d) | (ii) | Player 1 gets 9(0.5) + 2(0.5) = 5.5
Player 2 gets 4(0.5) + 6(0.5) = 5.0
Hence player 1 | M1
A1
[2] | 1.1
2.2a | Calculating 9p + 2(1 – p) or 4p + 6(1 – p) for their p (seen)
SC B1 only for both 2 + 7p and 6 – 2p o.e. without numerical p
‘1’ from valid correct working seen using p = 0.5
(6, 0) | (1, 7)
(9, 4) | (2, 6)
(6, 0) | (1, 7)
(9, 4) | (2, 6) | (8, 1)
(6, 8)5 In each turn of a game between two players they simultaneously each choose a strategy and then calculate the points won using the table below. They are each trying to maximise the number of points that they win.
In each cell the first value is the number of points won by player 1 and the second value is the number of points won by player 2 .
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{3}{|c|}{Player 2} \\
\hline
& & X & Y & Z \\
\hline
\multirow{3}{*}{Player 1} & A & $( 6,0 )$ & $( 1,7 )$ & $( 5,6 )$ \\
\hline
& B & $( 9,4 )$ & $( 2,6 )$ & $( 8,1 )$ \\
\hline
& C & $( 6,8 )$ & $( 1,3 )$ & $( 7,2 )$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the play-safe strategy for each player.
\item Explain why player 2 would never choose strategy Z .
\item Find the Nash equilibrium solution(s) or show that there is no Nash equilibrium solution.
Player 2 chooses strategy X with probability $p$ and strategy Y with probability $1 - p$.
You are given that when player 1 chooses strategy A, the expected number of points won by each player is the same.
\item \begin{enumerate}[label=(\roman*)]
\item Calculate the value of $p$.
\item Determine which player expects to win the greater number of points when player 1 chooses strategy B.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Discrete 2022 Q5 [12]}}