Notes

M1: Attempt to calculate row minima and column maxima (all 5 values) – condone one error

A1: Correct reasoning that the game is not stable (accept \(-1 \neq 1\) + statement) – dependent on correct row maximin and column minimax which must be clearly identified either around the table or in their statement (must have all 5 max and min values correct)

Notes

M1: setting up three expressions in terms of \(p\)

A1: all three expressions correct

M1: axes correct, at least two lines correctly drawn for their expressions, horizontal scale from 0 to 1

A1: correct graph

A1: using the graph to obtain the correct probability expressions leading to the correct value of \(p\)

A1ft: interpret their value of \(p\) in the context of the question – must refer to play and name options

Notes

(c) Condone use of \(p_1\ p_2\ p_3\) instead of \(x\ y\ z\)

B1: Correct augmentation (by at least 5) – possibly implied by later working in tableau

B1: Any one (numerical in nature) constraint row (\(r\), \(s\), \(t\) or \(u\)) correct (ignore labelling of b.v. column) or one correct constraint equation stated (so must be using columns)

B1: Any two (numerical in nature) constraint row (\(r\), \(s\), \(t\) or \(u\)) correct (ignore labelling of b.v. column) or two correct constraint equations stated

B1: CAO (including b.v. column)

Notes

B1: correct value for \(z\) (may be implied by subsequent working)

M1: Attempts to calculate the value of the \(3 \times 3\) game – must substitute in to at least two of their equations which may either be augmented or unaugmented

(if using the original game then expect to see (if correct) \(V \leqslant \dfrac{3}{7}, \dfrac{3}{7}, \dfrac{17}{7} \Rightarrow V = \dfrac{3}{7}\) so the value of the \(3 \times 3\) game is \(\dfrac{3}{7}\))

A1: Correct values for both games seen together with some indication of how much better the new game is (could be given as a % increase, for example ‘71.4% better’, or equivalent or as a multiplier e.g, \(\dfrac{12}{7}\) or 1.714)

Notes

aiM1: Finding row minimums and column maximums - condone one error.

aiA1: Correct play-safes stated for both players (P and X not values)

aiiB1: Row maximin \((-2) \neq\) column minimax \((1)\) (must be clearly identified)

Notes

bi1B1: Correct calculations, allow one slip.

bii1B1: Defining variable \(p\)

bii1M1: Setting up three expressions in terms of \(p\)

bii 1A1: All three expressions correct. (must be fully simplified)

bii 2M1: Axes correct, at least one line correctly drawn for their expressions. Horizontally the graph must be drawn between 0 and 1 only

bii 2A1: Correct graph.

bii 3A1: Using the graph to obtain the correct probability expressions leading to the correct value of \(p\)

bii4A1ft: Interpret their value of \(p\) in the context of the question.

Notes

M1: Attempt to calculate row minima and column maxima – condone one error (note row max are 3, 4, 6 and column min are -1, -2, -3 so we must see where values come from)

A1: Correct reasoning that the game is not stable (accept \(-1 \neq 3\)) – dependent on correct row maximin and column minimax

Notes

(b) Note – a fully correct tableau implies all marks in (b)

B1: Correct augmentation – possibly implied by later working in tableau

M1: At least three equations in \(V, p_1, p_2, p_3\) and at least one dummy variable seen (must be using columns)

A1: CAO for all four equations (possibly implied by later working in tableau)

M1: Any two (numerical in nature) row correct (ignore labelling of b.v. column)

A1: CAO

Notes

B1: Correct optimal strategy in context (dependent on both M marks in (b))

M1: For \(\pm\left(\dfrac{73}{16} \pm 3\right)\)

A1: CAO

Notes

M1: Attempt to set up at least three equations in \(q_1, q_2, q_3\) using the value of the game from (c)

A1: CAO (for any three of the four correct equations)

A1: CAO in context (must have at least three correct equations)

Notes

M1: finding row minimums and column maximums – condone one error

A1: row maximin \(\neq\) col minimax so not stable (dependent on correct row mins and col maxs)

Notes

B1: Correct reduced table for Meera (2 rows only)

Notes

M1: setting up four expressions in terms of \(p\)

A1: all four expressions correct and fully simplified

M1: axes correct, at least one line correctly drawn for their expressions

A1: correct graph (all lines drawn with a ruler, scale clear with correct relative position of lines)

A1: using the graph to obtain the correct probability expressions leading to the correct value of \(p\) and interpret their value of \(p\) in the context of the question (dependent on previous A mark)

B1ft: correct value of game to Haruki (ft their \(p\))

B1: CAO – B and D

Notes

B1: States value = 0  (may be implied by forming equation in \(p\) or \(q = 0\))

M1: considers Option A and obtains value for \(p\) (alt considers \(H\) and obtains value for \(q\))

M1: considers Option C and sets up equation in \(k\) (dependent on previous M mark)

A1: CAO (\(-4/5\) o.e.)

Alternative

Considers \(H\) playing A with probability \(q\) and C with probability \((1-q)\)
When \(M\) plays Z \(H\)’s gain  \(1 - 6q = 0 \quad q = 1/6\)
When \(M\) plays X \(H\)’s gain  \(4/6 + 5k/6 = 0\)
\(k = -4/5\)

Notes

M1: Clear attempt to find the row maximin and column minimax (either the row minimums or column maximums correct or at least five (of the seven) values stated correctly)

A1: CAO (dependent on all rowmins and colmaxs correct) states \(-1 \neq 2\) (or row (maximin) \(\neq\) col (minimax) as long as \(-1\) is clearly identified as the row maximin and 2 as the column minimax)

Notes

B1: Correct statement – must include the word ‘dominate’ (or exact equivalent)

B1: Correct inequalities – must be clear that all inequalities must hold

Notes

(c)(i)

B1: Correct augmentation – possibly implied by later working

B1: Correct objective defined

M1: At least three equations in \(V, p_1, p_2, p_3\) and at least one dummy variable seen

A1: CAO

B1: Correct probability equation

(ii)

M1: Any two (numerical in nature) row correct

A1: CAO including correct row and column labels for Simplex tableau

Alternative if augmenting by +4

Notes

B1: \(p_3\) correctly stated

M1: Attempt to find \(V\) by substituting probabilities into all three equations or corresponding inequalities involving \(V, p_1, p_2, p_3\) (or using the original value of the game) (Note candidates may reject B plays Z here as \(V\) is worse than playing X or Y)

B1: Defining probabilities for Player B

dM1: Dependent on previous M mark - setting up at least three equations in their \(q_1, q_2, q_3\) using their \(V\) (or original value) (may be just \(q\) and \((1 - q)\) if Option Z rejected)

A1ft: Correct three equations (two if Option Z rejected) following through their augmented matrix in (c) – A0 if option R for player A is used (and not rejected later)

A1: Correct options for Player B in context (must say that Player B should never play option Z but this may appear earlier in their working) dependent on previous A mark

Alternative if augmenting by +4

Alternative for (d)

(d) B1: \(p_3\) correctly stated

M1: Uses dominance argument (may be implied) to reduce either the original Matrix or their augmented Matrix to either 3 x 2 or 2 x 2 (if A plays R deleted). The Matrix may be transposed and signs changed.

B1: Defining probabilities for Player B (condone use of \(p\) instead of \(q\))

dM1: Dependent on previous M mark - setting up at least two equations (using A plays Q and S) or expressions in terms of their \(q\)

A1ft: Solves their expressions or equations to obtain \(q\) (may use graphical approach). If solving multiple pairs of expressions, they must clearly choose their solution – A0 if option R for player A is used (and not rejected later)

A1: Correct options for Player B in context (must say that Player B should never play option Z but this may appear earlier in their working) dependent on previous A mark

Notes

M1: finding row minimums and column maximums – condone one error

A1: row maximin 3 = col minimax 3 (so stable) – dependent on all 6 correct values – as a minimum must see the two 3’s explicitly being considered

A1: cao (\(-3\)) – ‘lose 3’ is A0

Notes

B1: cao (must be the reduced game so with option S removed)

Notes

In (c) if the candidate includes option S (and so has four expressions and lines) then this can score the first 3 marks only in this part

M1: setting up three expressions in terms of \(p\)

A1: all three expressions correctly simplified

M1: at least two lines correctly drawn for their expressions – if values on at least one vertical axis not given then lines must be in the right position relative to each other

A1: completely correct graph with clear indication of the correct intersection points at the ends where \(p = 0\) and \(p = 1\) (so if no scaling on the vertical axis assume that 1 line = 1 unit (unless other suitable scaling is clear), and lines must not extend past \(p \lt 0\) and/or \(p \gt 1\))

A1: using the graph (with 3 lines) to obtain the correct probability expressions leading to the correct value of \(p\)

A1ft: interpret their value of \(p\) in the context of the question – must refer to ‘play’ and the associated probabilities (but do not need to explicitly use the word ‘probability’)

Notes

(d)(i) B1: cao (\(\frac{4}{5}\) or 0.8)

(d)(ii) B1: cao (option T)

(d)(iii) M1: Setting up a linear equation with their V(\(A\)) from (d)(i) wither either of the two correct expressions \(2q + (-1)(1-q)\) or \(-2q + 5(1-q)\) or setting up the correct equation \(-2q + 5(1-q) = 2q + (-1)(1-q)\) (or \(2q - 5(1-q) = -2q + (1-q)\))

A1: interpret the correct value of \(q\) in the context of the question – must refer to ‘play’ and the associated probabilities (but do not need to explicitly use the word ‘probability’)

(corrected from the printed mark scheme: the printed line in (d)(i) reads “6 7/10 −5 = −4/5”, with the bracket/multiplication between 6 and 7/10 missing)

Notes

M1: Attempt to calculate row minima. Condone ‘\(k\) or \(-4\)’ for \(\min(k, -4)\).

A1: Correct argument/conditions for why the play-safe for player A is always their option Q

Notes

B1: Attempt to calculate column maxima. Condone ‘\(k\) or \(-2\)’ for \(\max(k, -2)\)

B1: CAO ignore lower limit figure, if given

Notes

B1: Correct statement – must include the word ‘dominate’. Also e.g. option Z dominates option X

B1: Correct inequalities – must be clear that all three inequalities must hold

Notes

(d)(i) B1: Correct augmentation – possibly implied by later working, X column may be included

M1: At least three equations in \(V, p_1, p_2, p_3\) and at least one dummy variable seen

A1: CAO (ignore extra probability equation, if seen, see note below)

B1: Correct probability equation

(ii) B1: Correct row and column labels for Simplex tableau

M1: Any one (numerical in nature) row correct

A1: CAO

Special case

(d) If augmentation with +5 maximum marks possible B0 M1 A1 B1 B1 M1 A0

Note: (d)(i) A1 The extra probability equation you may see is \(V - 7p_1 - 9p_2 - 10p_3 + \text{slack} = 0\)

Notes

B1: \(p_3\) correctly stated

M1: Attempts to calculate \(V\) using either equation from (d) not involving \(k\), or attempts to eliminate \(V\) from two equations, one involving \(k\)

dM1: Dependent on previous M mark – either uses equation in \(k\) and their \(V\) to calculate \(k\), or eliminates \(V\) from two equations to calculate \(k\)

A1: CAO \(k = 1\)

Notes

B1: cao (give bod but must get the idea across that one person’s losses are equal to the other’s gains)

Notes

M1: finding row minimums and column maximums – condone one error

A1: row maximin (1) \(\neq\) col minimax (6) (so not stable) – dependent on all correct 6 values

Notes

B1: cao

Notes

(d)(i)

B1: defining variable \(p\) (or any other letter e.g., \(q\)) – must use the word ‘probability’ (do not have to mention ‘June’ so as a minimum accept ‘X with probability \(p\) and Y with probability \(1 - p\)’)

M1: setting up four expressions in terms of their \(p\)

A1: all four expressions correctly simplified

M1: at least three lines correctly drawn for their expressions – if values on at least one vertical axis not given then lines must be in the right position relative to each other

M1: using their graph to obtain their correct probability equation leading to a value of \(p\) (dependent on both previous M marks)

A1: interpret the correct value of \(p\) in the context of the question – must refer to ‘play’ and the associated probabilities (need not say ‘probability’ again). This mark is dependent on a completely correct graph (so if no scaling on the vertical axis assume that 1 line = 1 unit, and lines must not extend past \(p \lt 0\) and/or \(p \gt 1\))

(d)(ii)

A1: cao (2.4) oe (dependent on all previous M marks)

Notes

M1: Setting up a linear equation in \(t\) (with possibly their value from (d)(ii)), using only the two valid options from their graph in (d)(i) (so if the graph was correct in (d) they must be using Terry’s options A and D) – allow sign slips only (allow any other choice of letter)

A1: cao (0.8)

A1ft: interpret their value of \(t\) in the context of the question – do not penalise a lack of ‘play’ twice. Must also include the two options that are never played

Notes

M1: Either one correct row or column

A1: cao

SC M1 A0 for \(\begin{pmatrix} 6 & 0 & 4 \\ 1 & 8 & 2 \end{pmatrix}\)

Notes

M1: Setting up three expressions in terms of \(p_1\) (either in terms of the original or modified game) or \(p_2\)

A1: All three expressions correct – or equivalent e.g., \(5p_1 - 2, -8p_1 + 5, 2p_1 - 1\)

M1: Axes correct, at least one line correctly drawn for their expressions

A1: Correct graph

A1: Using a correct graph to obtain the correct probability expressions leading to the correct value of \(p_1\) or \(p_2\)

A1ft: Interpret their values in the context of the question – must refer to play and the associated probabilities

Notes

B1: cao

Notes

M1: Setting up two equations in \(q_2\) and \(q_3\) with their value of either the original or modified game

A1: Correct two equations

A1: Interpret their values in the context of the question – must refer to play and the associated probabilities

For part (d) candidates might set up three equations in three unknowns e.g.,

Notes

B1: correct statement – must include the word ‘dominate’ (note that T dominates S too)

B1: correct inequalities – must be clear that all inequalities must hold

Notes

M1: attempt at row minima and column maxima – condone one error

A1: correct max(row min) and min(col max)

A1: correct reasoning that the game is not stable (accept \(1 \neq 3\) + statement)

Notes

B1: an understanding that for each value of \(p\) we are seeking the minimum possible output

Notes

B1: as a minimum accept an answer that implies that an inequality is required so that we can apply the Simplex algorithm

Notes

B1: cao

M1: substitute their \(p\) values into all three expressions for the upper bound of \(V\)

A1: cao for the value of the game to player A

Notes

M1: Attempt to set up at least three equations in \(q_1, q_2, q_3\) using the value of the game from (e)

A1ft: Two correct ft “their” \(V\)

A1: cao (for exactly three equations correct)

A1: cao in context

Notes

(i) B1: cao

(ii) B1: cao

Notes

(i) M1: finding row minimums and column maximums – condone one error

A1: row minima and column maxima correct

A1: correct play safes for both teams

(ii) B1: row maximin (\(-2\)) \(\neq\) col minimax (2) so not stable

Notes

B1: cao (or equivalent – e.g. Qaasim because \(-2\) is the lowest value in Noel’s row) – explanation must involve consideration of values and not just (for example) a general statement that Qaasim will gain the most

Notes

B1: defining variable \(q\)

M1: setting up three expressions in terms of \(q\)

A1: all three expressions correct – allow correct un-simplified expressions for this mark

M1: axes correct, at least one line correctly drawn for their expressions

A1: correct graph

A1: using the graph to obtain the correct probability expressions leading to the correct value of \(q\)

A1ft: interpret their value of \(q\) in the context of the question – must refer to play/choose and the two players

Note that in (d) candidates may use \(p\) (or another letter) instead of \(q\) which is fine for full marks. Also, the three expressions may be the negative of what is giving in the main scheme (e.g. \(10q - 6\), \(2q - 2\) and \(-8q + 2\)) and this is fine for the first 5 marks in (d). For the final two marks though they would need to consider the optimal point reading from the top (rather than the bottom) of their graph. No follow through for the final mark if they do not read off their graph correctly.

Notes

1M1: attempt at row minima and column maxima – condone one error

1A1: correct reasoning that the game is not stable (accept “\(-1 \neq 0\)” + statement) – dependent on correct row minima and column maxima

Notes

1B1: Indicates that coefficients are incorrect because \(V\) must be non-negative. Must convey both underlined aspects. Condone ‘positive’ for ‘non-negative’

2B1: Indicates that inequality signs are the wrong way because \(V\) is the minimum (so expected winnings are \(\geqslant V\)). Must convey both aspects but give bod.

3B1: CAO (all three inequalities). Give this mark for correct equations with slack variables provided 2B1 has been awarded.

Notes

1B1: All row and column labels correct for Simplex tableau

1M1: Setting up the Simplex model - any two of my ‘\(r\)’, ‘\(s\)’ or ‘\(t\)’ rows correct. Or a completely correct answer with either one column or one row missing – condone lack of basic variable column. Should follow from changed constraints of the correct form from b). So, constraints must have been of the form \(V \leqslant ap_1 + bp_2 + cp_3\ (a, b, c \geqslant 0)\) o.e.

1A1: CAO on numerical values.

Note: The B mark is for labelling the simplex tableau correctly; the M and A marks are for values only.

Notes

1M1: substitutes their \(p\) values into all three expressions for the upper bound of \(V\). Condone use of an equals sign here. But not ‘\(V \geqslant \ldots\)’
May see one of:

• \(V \leqslant 3(0.6) - 4(0) + 0.4 = 2.2\ \{= \tfrac{11}{5}\};\ V \leqslant -2(0.6) - 4(0) + 2(0.4) = -0.4\ \{= -\tfrac{2}{5}\};\ V \leqslant -2(0) - (0.4) = -0.4\ \{= -\tfrac{2}{5}\}\)
• \(V \leqslant 8(0.6) + (0) + 6(0.4) = 7.2\ \{= \tfrac{36}{5}\};\ V \leqslant 3(0.6) + 9(0) + 7(0.4) = 4.6\ \{= \tfrac{23}{5}\};\ V \leqslant 5(0.6) + 3(0) + 4(0.4) = 4.6\ \{= \tfrac{23}{5}\}\)
• \(V \leqslant 7(0.6) + 5(0.4) = 6.2\ \{= \tfrac{31}{5}\};\ V \leqslant 2(0.6) + 8(0) + 6(0.4) = 3.6\ \{= \tfrac{18}{5}\};\ V \leqslant 4(0.6) + 2(0) + 3(0.4) = 3.6\ \{= \tfrac{18}{5}\}\)

1A1: CAO for the value of the game to player A

Notes

1B1: CAO – uses the model to determine that Player B only plays Y and Z

1M1: A correct equation for B (where value of game to B = – 1 x their value of game to A)

1A1: CAO for \(q\) (the probability that B plays Y)

2A1ft: Correct optimal strategy in context (not just in terms of \(q\)) following through their \(q\)

Alternative for (e)

1B1: CAO – uses the model to determine that Player B only plays Y and Z

1M1: Formulates two correct expressions for the expected value of the game to B and finds the intersection: \(2q = -2q + 1(1 - q)\ (= 1 - 3q)\)

1A1: As above

2A1ft: As above

Notes

M1: finding row minimums and column maximums – condone one error

A1: row maximin (3) = col minimax (3) so stable (dependent on correct row minimums and col maximums) - as a minimum accept ‘3 = 3 so stable’

B1: CAO (\(-3\))

Notes

B1: defining variable \(p\) (must mention ‘probability’ – as a minimum accept ‘P with probability \(p\) and Q with probability \(1 - p\)’)

M1: setting up three expressions in terms of \(p\) (need not be simplified)

A1: all three expressions correctly simplified

M1dep: axes correct, at least one line correctly drawn from their expressions – dependent on previous M mark in (b)

A1: correct graph – if no scaling on vertical axis assume 1 line = 1 unit (A0 if graph extends for \(p \lt 0\) and/or \(p \gt 1\))

A1: using the graph to obtain the correct probability expressions leading to the correct value of \(p\)

A1ft: interpret their value of \(p\) in the context of the question – must refer to ‘play’ and the associated probabilities (need not say ‘probability’ again) – this mark is dependent on both previous M marks in this part

Notes

B1: correct explanation in context (of playing only X and Z for all \(p\)) with specific reference to the modelling of the problem by the graph

Notes

(i) B1: correctly deducing the lhs of the given equation (with clear reasoning for the change in sign) – only allow stating \(6q - 2(1 - q)\) (without seeing the change of sign) if the game is restated for player B

B1: correctly deriving the rhs of the given equation (must indicate that this is the value of the game to B although as a minimum accept V(B)) – stating \(\pm 1.6\) without any working is B0 - note that for either mark in (d)(i) candidates must explain where the two parts of the given equation came from

Note that candidates may explain the formulation of the given equation by considering \(-6q + 2(1 - q) = -1.6\) (which is what player B can expect to lose if player A plays option P which is equal to the value of the game to player A)

(ii) B1: CAO for the value of \(q\) (must come from solving \(6q - 2(1 - q) = 1.6\))

B1: CAO in context and must refer to ‘play’

Notes

B1: cao - indication that either the losses of one (player) are balanced by the gains of the other (player) or that the total points scored by both (players) is zero

Notes

B1: cao (3)

Notes

B1: correct explanation – could explain their reasoning using a numerical example

Notes

(i) M1: finding row minimums and column maximums – condone one error

A1: max (row minima) and min (column maxima) correct – dependent on all correct values of row minimums and column maximums

A1: correct play safes for both teams (Olive (O) and Paul (P))

(ii) B1: row maximin (\(-1\)) \(\neq\) col minimax (0) (so not stable)

SC for (d) and (e) for those candidates who do not consider the zero-sum game

For (d) M1 only for finding row minimums (3, 1, 4) and column maximums (5, 6, 8) – condone one error. Then allow possibility of full marks in (e) – expressions should be \(p + 4\), \(5p + 1\) and \(-4p + 7\) and the graph should lead to \(5p + 1 = -4p + 7\)

Notes

M1: setting up three expressions in terms of \(p\) with at least one correct

A1: all three expressions correct

M1: axes correct, at least one line correctly drawn for their expressions

A1: correct graph with consistent scaling (lines must not extend past \(p \lt 0\) and \(p \gt 1\))

A1: correct probability expressions leading to correct value of \(p\)

A1ft: interpret their value of \(p\) in the context of the question – must refer to play and the team members

SC for (d) and (e) for those candidates who do not consider the zero-sum game

For (d) M1 only for finding row minimums (3, 1, 4) and column maximums (5, 6, 8) – condone one error. Then allow possibility of full marks in (e) – expressions should be \(p + 4\), \(5p + 1\) and \(-4p + 7\) and the graph should lead to \(5p + 1 = -4p + 7\)

Notes

(i) B1: cao

(ii) B1ft: 10 – their answer to (f)(i)