| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Discrete (Further AS Paper 2 Discrete) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Non-group structures |
| Difficulty | Moderate -0.5 This is a standard game theory question requiring routine application of maximin/minimax principles to find a saddle point, then basic arithmetic reasoning about repeated games. The concepts are A-level appropriate but mechanical, requiring less problem-solving than a typical pure maths question. |
| 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 interpretation |
| Strategy | \(\mathbf { Y } _ { \mathbf { 1 } }\) | \(\mathbf { Y } _ { \mathbf { 2 } }\) | \(\mathbf { Y } _ { \mathbf { 3 } }\) |
| \(\mathbf { X } _ { \mathbf { 1 } }\) | - 4 | 1 | - 3 |
| \(\mathbf { X } _ { \mathbf { 2 } }\) | 4 | - 3 | - 3 |
| \(\mathbf { X } _ { \mathbf { 3 } }\) | - 1 | 1 | - 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| row minima: \((-4, -3, -2)\); col maxima: \((4, 1, -2)\); OR identifies at least one dominated strategy | M1 | Identifies correctly the row minima or column maxima |
| \(\max(\text{row minima}) = -2\), \(\min(\text{col maxima}) = -2\); stable solution exists | A1 | Finds correctly \(\max(\text{row minima})\) and \(\min(\text{col maxima})\) OR finds a \(1 \times 1\) pay-off matrix by removing all dominated strategies |
| As \(\max(\text{row minima}) = -2 = \min(\text{col maxima})\), a stable solution exists | R1 | Completes a reasoned argument to show that the game has a stable solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Play-safe strategy for Xander is \(X_3\); Play-safe strategy for Yvonne is \(Y_3\) | B1 | States the correct play-safe strategy for each player |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The maximum amount of marbles that can be won or lost by either player in a game is \(4\). Because the game is zero-sum, when one player wins \(4\), the other player loses \(4\), so the difference is \(8\) | E1 | Explains that the maximum outcome for a game for either player from the pay-off matrix |
| \(24 \div 8 = 3\); if the players repeated the same strategies for each game, it would take \(3\) games to reach a difference of \(24\) marbles | E1 | Explains clearly how a value from the pay-off matrix relates to the minimum number of games needed to complete the challenge |
# Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| row minima: $(-4, -3, -2)$; col maxima: $(4, 1, -2)$; OR identifies at least one dominated strategy | M1 | Identifies correctly the row minima or column maxima |
| $\max(\text{row minima}) = -2$, $\min(\text{col maxima}) = -2$; stable solution exists | A1 | Finds correctly $\max(\text{row minima})$ and $\min(\text{col maxima})$ OR finds a $1 \times 1$ pay-off matrix by removing all dominated strategies |
| As $\max(\text{row minima}) = -2 = \min(\text{col maxima})$, a stable solution exists | R1 | Completes a reasoned argument to show that the game has a stable solution |
---
# Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Play-safe strategy for Xander is $X_3$; Play-safe strategy for Yvonne is $Y_3$ | B1 | States the correct play-safe strategy for each player |
---
# Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The maximum amount of marbles that can be won or lost by either player in a game is $4$. Because the game is zero-sum, when one player wins $4$, the other player loses $4$, so the difference is $8$ | E1 | Explains that the maximum outcome for a game for either player from the pay-off matrix |
| $24 \div 8 = 3$; if the players repeated the same strategies for each game, it would take $3$ games to reach a difference of $24$ marbles | E1 | Explains clearly how a value from the pay-off matrix relates to the minimum number of games needed to complete the challenge |
---6 Xander and Yvonne are playing a zero-sum game.
The game is represented by the pay-off matrix for Xander.
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Yvonne}
Xander \begin{tabular}{ | c | c | c | c | }
\hline
Strategy & $\mathbf { Y } _ { \mathbf { 1 } }$ & $\mathbf { Y } _ { \mathbf { 2 } }$ & $\mathbf { Y } _ { \mathbf { 3 } }$ \\
\hline
$\mathbf { X } _ { \mathbf { 1 } }$ & - 4 & 1 & - 3 \\
\hline
$\mathbf { X } _ { \mathbf { 2 } }$ & 4 & - 3 & - 3 \\
\hline
$\mathbf { X } _ { \mathbf { 3 } }$ & - 1 & 1 & - 2 \\
\hline
\end{tabular}
\end{center}
\end{table}
6
\begin{enumerate}[label=(\alph*)]
\item Show that the game has a stable solution.\\
6
\item State the play-safe strategy for each player.
Play-safe strategy for Xander is $\_\_\_\_$\\
Play-safe strategy for Yvonne is $\_\_\_\_$
6
\item The game that Xander and Yvonne are playing is part of a marbles challenge.
The pay-off matrix values represent the number of marbles gained by Xander in each game.
In the challenge, the game is repeated until one player has 24 marbles more than the other player.
Explain why Xander and Yvonne must play at least 3 games to complete the challenge.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Discrete 2023 Q6 [6]}}