OCR Further Discrete 2023 June — Question 7 12 marks

Exam BoardOCR
ModuleFurther Discrete (Further Discrete)
Year2023
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
DifficultyChallenging +1.8 This is a game theory problem requiring understanding of play-safe strategies, stability, dominance, and LP formulation for mixed strategies. While the individual components (finding row/column minima/maxima, comparing strategies, solving a simplified LP) are mechanical, the problem requires synthesis across multiple game theory concepts and careful interpretation of the LP setup. The simplex reduction and final optimization require multi-step reasoning beyond standard textbook exercises, placing it well above average difficulty.
Spec7.08a Pay-off matrix: zero-sum games7.08e Mixed strategies: optimal strategy using equations or graphical method

7 Player 1 and player 2 are playing a two-person zero-sum game.
In each round of the game the players each choose a strategy and simultaneously reveal their choice. The number of points won in each round by player 1 for each combination of strategies is shown in the table below. Each player is trying to maximise the number of points that they win.
Player 2 Player 1
ABC
X2- 3- 4
Y013
Z- 224
    1. Determine play-safe strategies for each player.
    2. Show that the game is not stable.
  1. Show that the number of strategies available to player 1 cannot be reduced by dominance. You must make it clear which values are being compared. Player 1 intends to make a random choice between strategies \(\mathrm { X } , \mathrm { Y } , \mathrm { Z }\), choosing strategy X with probability \(x\), strategy Y with probability \(y\) and strategy Z with probability \(z\).
    Player 1 formulates the following LP problem so they can find the optimal values of \(x , y\) and \(z\) using the simplex algorithm. Maximise \(M = m - 4\) subject to \(m \leqslant 6 x + 4 y + 2 z\) $$\begin{aligned} & m \leqslant x + 5 y + 6 z \\ & m \leqslant 7 y + 8 z \\ & x + y + z \leqslant 1 \end{aligned}$$ and \(m \geqslant 0 , x \geqslant 0 , y \geqslant 0 , z \geqslant 0\)
  2. Explain how the inequality \(m \leqslant 6 x + 4 y + 2 z\) was formed. The problem is solved by running the simplex algorithm on a computer.
    The printout gives a solution in which \(\mathrm { x } + \mathrm { y } = 1\).
    This means that the LP problem can be reduced to the following formulation.
    Maximise \(M = m - 4\) subject to \(m \leqslant 4 + 2 x\) \(\mathrm { m } \leqslant 5 - 4 \mathrm { x }\) \(m \leqslant 7 - 7 x\) and \(m \geqslant 0 , x \geqslant 0\)
  3. Solve this problem to find the optimal values of \(x , y\) and \(z\) and the corresponding value of the game to player 1.

Question 7:
AnswerMarks Guidance
7(a) (i)
X 2 –3 –4 –4
Y 0 1 3 0
Z –2 2 4 –2
max 2 2 4
Play-safe strategy for player 1 is Y
Play-safe strategy for player 2 is either A or
AnswerMarks
BM1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
2.2a–4, 0, –2 as row minima for player 1
2, 2, 4 as column maxima for player 2
(or –2, –2, –4)
1 = Y, and 2 = A and/or B
AnswerMarks Guidance
(ii)0  2
so not stableB1
[1]2.1 row maximin  col minimax (o.e.)
(0, -2) is not a possible cell in a zero-sum game
Or an appropriate description of ‘chasing cells’
AnswerMarks Guidance
7(b) X and Y: 2 > 0 but -3 < 1 (or -4 < 3)
X and Z: 2 > -2 but -3 < 2 (or -4 < 4)
AnswerMarks
Y and Z: 0 > -2 but 1 < 2 (or 3 < 4)M1
A11.1
2.4Showing no dominance between two of the rows (e.g. X and Y)
Allow XA > YA, XB < YB etc.
6 appropriate comparisons (or equivalent)
Alternative method 1
AnswerMarks Guidance
A: X > Y > Z but B,C: Z > Y > XM1 Either of these, or ‘first, second, third’
A1A1 Both together with evidence of reasoning
Alternative method 2
For player 1
AnswerMarks Guidance
Best in A is X and best in B (or C) is ZM1 Use col max to show that none of X, Y, Z is always best (o.e.)
Worst in A is Z and worst in B (or C) is XWorst in A is Z and worst in B (or C) is X A1
(or compare Y with both X and Z)
[2]
AnswerMarks Guidance
7(c) Each entry is increased by 4 to make them
all non-negative.
Using the augmented values,
when player 2 chooses A
player 1 expects to win 6x + 4y + 2z
For each (x, y, z) m is the minimum
expected win so m  6x + 4y + 2z
The optimum is the maximum value of m
AnswerMarks
over x, y and zB1
B1
AnswerMarks
[2]3.3
3.4Add 4 throughout
Add 4 to all values
Appropriate reference to strategy A or first column
AnswerMarks Guidance
7(d) 4 + 2x = 5 – 4x
1 5
x = , y = , z = 0
6 6
1 1
m = 4  M = so the value of the game to
3 3
1
player 1 is
AnswerMarks
3B1
M1
A1
A1
AnswerMarks
[4]3.1a
3.4
1.1
AnswerMarks
3.2aSketch graph showing
m = 4 + 2x, m = 5 – 4x, m = 7 – 7x for 0  x  1
Solve as simultaneous equations to find x  [0, 1] or implied from
1 1 2
a correct x value [x = , x = or x = ] (o.e. solving for m)
6 3 3
cao
cao
PMT
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Though we make every effort to check our resources, there may be contradictions between published support and the
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Question 7:
7 | (a) | (i) | A B C min
X 2 –3 –4 –4
Y 0 1 3 0
Z –2 2 4 –2
max 2 2 4
Play-safe strategy for player 1 is Y
Play-safe strategy for player 2 is either A or
B | M1
M1
A1
[3] | 1.1
1.1
2.2a | –4, 0, –2 as row minima for player 1
2, 2, 4 as column maxima for player 2
(or –2, –2, –4)
1 = Y, and 2 = A and/or B
(ii) | 0  2
so not stable | B1
[1] | 2.1 | row maximin  col minimax (o.e.)
(0, -2) is not a possible cell in a zero-sum game
Or an appropriate description of ‘chasing cells’
7 | (b) | X and Y: 2 > 0 but -3 < 1 (or -4 < 3)
X and Z: 2 > -2 but -3 < 2 (or -4 < 4)
Y and Z: 0 > -2 but 1 < 2 (or 3 < 4) | M1
A1 | 1.1
2.4 | Showing no dominance between two of the rows (e.g. X and Y)
Allow XA > YA, XB < YB etc.
6 appropriate comparisons (or equivalent)
Alternative method 1
A: X > Y > Z but B,C: Z > Y > X | M1 | Either of these, or ‘first, second, third’
A1 | A1 | Both together with evidence of reasoning | Both together with evidence of reasoning
Alternative method 2
For player 1
Best in A is X and best in B (or C) is Z | M1 | Use col max to show that none of X, Y, Z is always best (o.e.)
Worst in A is Z and worst in B (or C) is X | Worst in A is Z and worst in B (or C) is X | A1 | A1 | Use col min to show that none of X, Y, Z is always worst (o.e.)
(or compare Y with both X and Z)
[2]
7 | (c) | Each entry is increased by 4 to make them
all non-negative.
Using the augmented values,
when player 2 chooses A
player 1 expects to win 6x + 4y + 2z
For each (x, y, z) m is the minimum
expected win so m  6x + 4y + 2z
The optimum is the maximum value of m
over x, y and z | B1
B1
[2] | 3.3
3.4 | Add 4 throughout
Add 4 to all values
Appropriate reference to strategy A or first column
7 | (d) | 4 + 2x = 5 – 4x
1 5
x = , y = , z = 0
6 6
1 1
m = 4  M = so the value of the game to
3 3
1
player 1 is
3 | B1
M1
A1
A1
[4] | 3.1a
3.4
1.1
3.2a | Sketch graph showing
m = 4 + 2x, m = 5 – 4x, m = 7 – 7x for 0  x  1
Solve as simultaneous equations to find x  [0, 1] or implied from
1 1 2
a correct x value [x = , x = or x = ] (o.e. solving for m)
6 3 3
cao
cao
PMT
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If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in
touch with our customer support centre.
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/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.
7 Player 1 and player 2 are playing a two-person zero-sum game.\\
In each round of the game the players each choose a strategy and simultaneously reveal their choice.

The number of points won in each round by player 1 for each combination of strategies is shown in the table below.

Each player is trying to maximise the number of points that they win.\\
Player 2

Player 1

\begin{center}
\begin{tabular}{ c | r | r | r }
 & A & B & C \\
\hline
X & 2 & - 3 & - 4 \\
\hline
Y & 0 & 1 & 3 \\
\hline
Z & - 2 & 2 & 4 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Determine play-safe strategies for each player.
\item Show that the game is not stable.
\end{enumerate}\item Show that the number of strategies available to player 1 cannot be reduced by dominance. You must make it clear which values are being compared.

Player 1 intends to make a random choice between strategies $\mathrm { X } , \mathrm { Y } , \mathrm { Z }$, choosing strategy X with probability $x$, strategy Y with probability $y$ and strategy Z with probability $z$.\\
Player 1 formulates the following LP problem so they can find the optimal values of $x , y$ and $z$ using the simplex algorithm.

Maximise $M = m - 4$\\
subject to $m \leqslant 6 x + 4 y + 2 z$

$$\begin{aligned}
& m \leqslant x + 5 y + 6 z \\
& m \leqslant 7 y + 8 z \\
& x + y + z \leqslant 1
\end{aligned}$$

and $m \geqslant 0 , x \geqslant 0 , y \geqslant 0 , z \geqslant 0$
\item Explain how the inequality $m \leqslant 6 x + 4 y + 2 z$ was formed.

The problem is solved by running the simplex algorithm on a computer.\\
The printout gives a solution in which $\mathrm { x } + \mathrm { y } = 1$.\\
This means that the LP problem can be reduced to the following formulation.\\
Maximise $M = m - 4$\\
subject to $m \leqslant 4 + 2 x$\\
$\mathrm { m } \leqslant 5 - 4 \mathrm { x }$\\
$m \leqslant 7 - 7 x$\\
and $m \geqslant 0 , x \geqslant 0$
\item Solve this problem to find the optimal values of $x , y$ and $z$ and the corresponding value of the game to player 1.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Discrete 2023 Q7 [12]}}