| Exam Board | Edexcel |
|---|---|
| Module | FD2 AS (Further Decision 2 AS) |
| Year | 2018 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Zero-sum game stable solution |
| Difficulty | Standard +0.3 This is a standard textbook zero-sum game problem with straightforward play-safe strategy identification, dominance checking, and a routine graphical method for finding mixed strategy equilibrium. While it requires multiple steps, each component follows algorithmic procedures taught directly in FD2 with no novel insight required. The graphical method is particularly mechanical. Slightly easier than average due to its procedural nature. |
| 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 method |
| \cline { 3 - 5 } \multicolumn{2}{c|}{} | Team B | |||
| \cline { 3 - 5 } \multicolumn{2}{c|}{} | Paul | Qaasim | Rashid | |
| \multirow{3}{*}{Team A} | Mischa | 5 | 6 | 3 |
| \cline { 2 - 5 } | Noel | 4 | 1 | 7 |
| \cline { 2 - 5 } | Olive | 4 | 5 | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The gains (or losses) made by one player are exactly balanced by the losses (or gains) made by the other player. | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. if a member of team \(A\) gains \(x\) points then a member of team \(B\) gains \(10-x\) points. Subtracting 5 from both gives \(A\): \(x-5\) and \((10-x)-5=5-x\). The sum is \((x-5)+(5-x)=0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) Row minima: \(-2, -4, -1\); max is \(-1\). Column maxima: \(0, 1, 3\); min is \(0\). Play safe for Team \(A\) is Olive and for Team \(B\) is Paul. | M1, A1, A1 | |
| (ii) Row maximin \((-1) \neq\) Col minimax \((0)\) so not stable | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| If \(B\) plays strategy 1, \(A\)'s gains are \(-1(1-p) = p-1\) | M1 | |
| If \(B\) plays strategy 2, \(A\)'s gains are \(p + -4(1-p) = 5p-4\) | A1 | |
| If \(B\) plays strategy 3, \(A\)'s gains are \(-2p + 2(1-p) = 2-4p\) | ||
| Correct graph of three lines plotted | M1, A1 | |
| \(2-4p = 5p-4 \Rightarrow p = \frac{2}{3}\) | A1 | |
| Team \(A\) should play Mischa with probability \(\frac{2}{3}\) and Noel with probability \(\frac{1}{3}\) | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) \(\frac{13}{3}\) | B1 | |
| (ii) \(\frac{17}{3}\) | B1ft |
# Question 2:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The gains (or losses) made by one player are exactly balanced by the losses (or gains) made by the other player. | B1 | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3$ | B1 | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. if a member of team $A$ gains $x$ points then a member of team $B$ gains $10-x$ points. Subtracting 5 from both gives $A$: $x-5$ and $(10-x)-5=5-x$. The sum is $(x-5)+(5-x)=0$ | B1 | |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** Row minima: $-2, -4, -1$; max is $-1$. Column maxima: $0, 1, 3$; min is $0$. Play safe for Team $A$ is Olive and for Team $B$ is Paul. | M1, A1, A1 | |
| **(ii)** Row maximin $(-1) \neq$ Col minimax $(0)$ so not stable | B1 | |
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| If $B$ plays strategy 1, $A$'s gains are $-1(1-p) = p-1$ | M1 | |
| If $B$ plays strategy 2, $A$'s gains are $p + -4(1-p) = 5p-4$ | A1 | |
| If $B$ plays strategy 3, $A$'s gains are $-2p + 2(1-p) = 2-4p$ | | |
| Correct graph of three lines plotted | M1, A1 | |
| $2-4p = 5p-4 \Rightarrow p = \frac{2}{3}$ | A1 | |
| Team $A$ should play Mischa with probability $\frac{2}{3}$ and Noel with probability $\frac{1}{3}$ | A1ft | |
## Part (f):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** $\frac{13}{3}$ | B1 | |
| **(ii)** $\frac{17}{3}$ | B1ft | |2. (a) Explain what the term 'zero-sum game' means.
Two teams, A and B , are to face each other as part of a quiz.\\
There will be several rounds to the quiz with 10 points available in each round.\\
For each round, the two teams will each choose a team member and these two people will compete against each other until all 10 points have been awarded. The number of points that Team A can expect to gain in each round is shown in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\cline { 3 - 5 }
\multicolumn{2}{c|}{} & \multicolumn{3}{|c|}{Team B} \\
\cline { 3 - 5 }
\multicolumn{2}{c|}{} & Paul & Qaasim & Rashid \\
\hline
\multirow{3}{*}{Team A} & Mischa & 5 & 6 & 3 \\
\cline { 2 - 5 }
& Noel & 4 & 1 & 7 \\
\cline { 2 - 5 }
& Olive & 4 & 5 & 8 \\
\hline
\end{tabular}
\end{center}
The teams are each trying to maximise their number of points.\\
(b) State the number of points that Team B will expect to gain each round if Team A chooses Noel and Team B chooses Rashid.\\
(c) Explain why subtracting 5 from each value in the table will model this situation as a zero-sum game.\\
(d) (i) Find the play-safe strategies for the zero-sum game.\\
(ii) Explain how you know that the game is not stable.
At the last minute, Olive becomes unavailable for selection by Team A.\\
Team A decides to choose its player for each round so that the probability of choosing Mischa is $p$ and the probability of choosing Noel is $1 - p$.\\
(e) Use a graphical method to find the optimal value of $p$ for Team A and hence find the best strategy for Team A.
For this value of $p$,\\
(f) (i) find the expected number of points awarded, per round, to Team A,\\
(ii) find the expected number of points awarded, per round, to Team B.\\
\hfill \mbox{\textit{Edexcel FD2 AS 2018 Q2 [15]}}