| Exam Board | Edexcel |
|---|---|
| Module | FD2 AS (Further Decision 2 AS) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Zero-sum game optimal mixed strategy |
| Difficulty | Standard +0.3 This is a standard textbook exercise in game theory requiring routine application of well-defined algorithms: checking for saddle points, finding optimal mixed strategies using linear equations, and calculating game values. While it involves multiple parts and some algebraic manipulation, each step follows a mechanical procedure taught directly in the FD2 syllabus with no novel insight required. It's slightly easier than average because the methods are algorithmic and the arithmetic is straightforward. |
| 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 interpretation7.08e Mixed strategies: optimal strategy using equations or graphical method7.08f Mixed strategies via LP: reformulate as linear programming problem |
| \cline { 2 - 4 } \multicolumn{2}{c|}{} | June | ||
| \cline { 3 - 4 } \multicolumn{2}{c|}{} | Option X | Option Y | |
| \multirow{4}{*}{Terry} | Option A | 1 | 4 |
| \cline { 2 - 4 } | Option B | - 2 | 6 |
| \cline { 2 - 4 } | Option C | - 1 | 5 |
| \cline { 2 - 4 } | Option D | 8 | - 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| In a zero-sum game each participant's gain/loss is exactly balanced by the loss/gain of the other participant | B1 | Must convey that one person's losses equal the other's gains |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Row minima: \(1, -2, -1, -4\); max is \(1\). Column maxima: \(8, 6\); min is \(6\) | M1 | Finding row minimums and column maximums; condone one error |
| Row(maximin) \(\neq\) Col(minimax) therefore game is not stable | A1 | Dependent on all correct 6 values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix} -1 & 2 & 1 & -8 \\ -4 & -6 & -5 & 4 \end{pmatrix}\) | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Let June play X with probability \(p\) and Y with probability \(1-p\) | B1 | Must use word 'probability'; minimum: 'X with probability \(p\), Y with probability \(1-p\)' |
| If Terry plays A: \(-p + (-4)(1-p) = 3p - 4\); If B: \(2p + (-6)(1-p) = 8p - 6\); If C: \(p + (-5)(1-p) = 6p - 5\); If D: \(-8p + 4(1-p) = -12p + 4\) | M1, A1 | M1: setting up four expressions in \(p\); A1: all four correctly simplified |
| At least three lines correctly drawn for their expressions | M1 | Lines must be in correct position relative to each other |
| \(3p - 4 = -12p + 4 \Rightarrow p = \dfrac{8}{15}\) | M1 | Using graph to obtain probability equation leading to value of \(p\) |
| June should play X with probability \(\dfrac{8}{15}\) and Y with probability \(\dfrac{7}{15}\) | A1 | Must refer to 'play' and associated probabilities; dependent on completely correct graph |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Value of game to Terry is \(2.4\) | A1 | cao; dependent on all previous M marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(-t + (-8)(1-t) = -4t + 4(1-t)\) or equivalent forms | M1 | Setting up linear equation in \(t\) using two valid options from graph in (d)(i); sign slips only allowed |
| \(t = \dfrac{4}{5}\) | A1 | cao |
| Terry should play option A with probability \(0.8\), never play B and C, and play option D with probability \(0.2\) | A1ft | Interpret value of \(t\) in context; must include the two options never played |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| In a zero-sum game each participant's gain/loss is exactly balanced by the loss/gain of the other participant | B1 | Must convey that one person's losses equal the other's gains |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Row minima: $1, -2, -1, -4$; max is $1$. Column maxima: $8, 6$; min is $6$ | M1 | Finding row minimums and column maximums; condone one error |
| Row(maximin) $\neq$ Col(minimax) therefore game is not stable | A1 | Dependent on all correct 6 values |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix} -1 & 2 & 1 & -8 \\ -4 & -6 & -5 & 4 \end{pmatrix}$ | B1 | cao |
## Part (d)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Let June play X with probability $p$ and Y with probability $1-p$ | B1 | Must use word 'probability'; minimum: 'X with probability $p$, Y with probability $1-p$' |
| If Terry plays A: $-p + (-4)(1-p) = 3p - 4$; If B: $2p + (-6)(1-p) = 8p - 6$; If C: $p + (-5)(1-p) = 6p - 5$; If D: $-8p + 4(1-p) = -12p + 4$ | M1, A1 | M1: setting up four expressions in $p$; A1: all four correctly simplified |
| At least three lines correctly drawn for their expressions | M1 | Lines must be in correct position relative to each other |
| $3p - 4 = -12p + 4 \Rightarrow p = \dfrac{8}{15}$ | M1 | Using graph to obtain probability equation leading to value of $p$ |
| June should play X with probability $\dfrac{8}{15}$ and Y with probability $\dfrac{7}{15}$ | A1 | Must refer to 'play' and associated probabilities; dependent on completely correct graph |
## Part (d)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Value of game to Terry is $2.4$ | A1 | cao; dependent on all previous M marks |
## Part (e)
| Answer | Mark | Guidance |
|--------|------|----------|
| $-t + (-8)(1-t) = -4t + 4(1-t)$ or equivalent forms | M1 | Setting up linear equation in $t$ using two valid options from graph in (d)(i); sign slips only allowed |
| $t = \dfrac{4}{5}$ | A1 | cao |
| Terry should play option A with probability $0.8$, never play B and C, and play option D with probability $0.2$ | A1ft | Interpret value of $t$ in context; must include the two options never played |
---3. Terry and June play a zero-sum game. The pay-off matrix shows the number of points that Terry scores for each combination of strategies.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\cline { 2 - 4 }
\multicolumn{2}{c|}{} & \multicolumn{2}{c|}{June} \\
\cline { 3 - 4 }
\multicolumn{2}{c|}{} & Option X & Option Y \\
\hline
\multirow{4}{*}{Terry} & Option A & 1 & 4 \\
\cline { 2 - 4 }
& Option B & - 2 & 6 \\
\cline { 2 - 4 }
& Option C & - 1 & 5 \\
\cline { 2 - 4 }
& Option D & 8 & - 4 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Explain the meaning of 'zero-sum' game.
\item Verify that there is no stable solution to the game.
\item Write down the pay-off matrix for June.
\item \begin{enumerate}[label=(\roman*)]
\item Find the best strategy for June, defining any variables you use.
\item State the value of the game to Terry.
Let Terry play option A with probability $t$.
\end{enumerate}\item By writing down a linear equation in $t$, find the best strategy for Terry.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD2 AS 2022 Q3 [14]}}