AS June 2025 Q3

EdexcelCurrent spec11 marksGame Theory

3. Layla and Mohsin play a zero-sum game represented by the following pay-off matrix for Layla.

(a)

(i) Find the play-safe strategies for each player.

(ii) State, giving a reason, whether there is a stable solution to this game. (3)

(b) Option R is now removed from Layla’s choices.

(i) For each of Mohsin’s three options, find the expected pay-off to Layla when she plays options P and Q equally often. (1)

(ii) Use a graphical method to determine Layla’s optimal mixed strategy. You should define any variables you use. (7)

Mark scheme (a)

Scheme	Marks	AO
(i) Row minima: \(-2, -5, -3\) maximum is \(-2\) Column maxima: \(1, 3, 2\) minimum is \(1\) Layla’s play-safe is play P and Mohsin’s play-safe is play X.	M1 A1	1.1b 1.1b
(ii) The game is not stable because Row(maximin) \((-2) \neq\) Column(minimax) \((1)\)	B1	2.4
	(3)

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)

Scheme	Marks	AO
(i) If Mohsin plays X, Layla’s expected pay-off \(= 0.5\times 1 + 0.5\times(-4) = -1.5\) If Mohsin plays Y, Layla’s expected pay-off \(= 0.5\times(-2) + 0.5\times 3 = 0.5\) If Mohsin plays Z, Layla’s expected pay-off \(= 0.5\times 2 + 0.5\times(-5) = -1.5\)	B1	1.1b
	(1)
(ii) Let Layla play P with probability \(p\) and play Q with probability \((1-p)\)	B1	3.3
If Mohsin plays X, Layla’s expected win \(= p - 4(1-p) = 5p - 4\) If Mohsin plays Y, Layla’s expected win \(= -2p + 3(1-p) = -5p + 3\) If Mohsin plays Z, Layla’s expected win \(= 2p - 5(1-p) = 7p - 5\)	M1 A1	1.1b 1.1b
	M1 A1	1.1b 1.1b
\(-5p + 3 = 5p - 4 \Rightarrow p = \dfrac{7}{10}\)	A1	1.1b
Layla should play P with probability \(7/10\) and play Q with probability \(3/10\)	A1ft	3.2a
	(7)
	(11 marks)