AS June 2019 Q4

EdexcelCurrent spec15 marksGame Theory

4. The table below gives the pay-off matrix for a zero-sum game between two players, Aljaz and Brendan. The values in the table show the pay-offs for Aljaz.

		Brendan
		Option X	Option Y	Option Z
Aljaz	Option P	\(-6\)	\(-1\)	\(2\)
	Option Q	\(5\)	\(4\)	\(-7\)
	Option R	\(5\)	\(6\)	\(3\)

(a)

(i) Show that this game is stable.

(ii) State the value of this game to Brendan. (3)

Option R is removed from Aljaz’s choices and the reduced game, with option R removed, is no longer stable.

(b) Find the best strategy for Aljaz in this reduced game, defining any variable you use. (7)

Let Brendan play option X with probability \(q\)

(d)

(i) Explain why \(q\) satisfies the equation \(6q - 2(1 - q) = 1.6\)

(ii) Hence find the best strategy for Brendan in this reduced game. (4)

Mark scheme (a)

Scheme	Marks	AO
(i) Row minima: \(-6, -7, 3\) max is \(3\) Column maxima: \(5, 6, 3\) min is \(3\)	M1	1.1b
Row(maximin) = Col(minimax) therefore game is stable	A1	2.4
(ii) value of the game to B is \(-3\)	B1	2.2a
	(3)

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 (\(-3\))

graph of A’s expected pay-off against p from p = 0 to p = 1, with lines for option X (5 to −6), option Y (4 to −1) and option Z (−7 to 2) — Mark scheme (b)

Scheme	Marks	AO
Let A play option P with probability \(p\) and option Q with probability \(1 - p\)	B1	3.3
If \(B\) plays option X, \(A\)’s gains are \(-6p + 5(1 - p) = 5 - 11p\) If \(B\) plays option Y, \(A\)’s gains are \(-p + 4(1 - p) = 4 - 5p\) If \(B\) plays option Z, \(A\)’s gains are \(2p + (-7)(1 - p) = -7 + 9p\)	M1 A1	1.1b 1.1b
	M1dep A1	1.1b 1.1b
\(5 - 11p = -7 + 9p \Rightarrow p = 3/5\)	A1	1.1b
A should play option P with probability 0.6 and option Q with probability 0.4	A1ft	3.2a
	(7)

Mark scheme (c)

Scheme	Marks	AO
As indicated by the graph Brendan can, for all values of \(p\), gain more by playing either options X or Z e.g. for \(0 \leqslant p \leqslant \frac{3}{5}\) Brendan would be better off playing Z and for \(\frac{3}{5} \lt p \leqslant 1\) Brendan would be better off playing X than playing Y	B1	3.2a
	(1)

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

Mark scheme (d)

Scheme	Marks	AO
(i) If A plays option P then B can expect to gain \(-\left(-6q + 2(1 - q)\right)\) as the values in the table are the pay-offs for A	B1	2.2a
The value of the game to B is \(-\left(5 - 11(0.6)\right) = 1.6\) (or equivalent calculation e.g. \(-\left(-7 + 9(0.6)\right)\))	B1	2.1
(ii) \(6q - 2(1 - q) = 1.6 \Rightarrow q = 0.45\)	B1	1.1b
B should play option X with probability 0.45 and option Z with probability 0.55	B1	3.2a
	(4)
	(15 marks)

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’