D2 June 2005 Q7

EdexcelOld spec17 marksGame Theory

(a) Explain briefly what is meant by a zero-sum game. (1)

A two person zero-sum game is represented by the following pay-off matrix for player \(A\).

	I	II	III
I	5	2	3
II	3	5	4

(b) Verify that there is no stable solution to this game. (3)

(d) Formulate the game as a linear programming problem for player \(B\). Write the constraints as inequalities and define your variables clearly. (5)

Scheme	Marks
A zero-sum game is one in which the sum of the gains for all players is zero. (o.e.)	B1
	(1)

Scheme	Marks
Let \(A\) play I with probability \(p\) Let \(A\) play II with probability \((1 - p)\)
If \(B\) plays I \(A\)’s gains are \(5p + 3(1 - p) = 2p + 3\) If \(B\) plays II \(A\)’s gains are \(2p + 5(1 - p) = 5 - 3p\) If \(B\) plays III \(A\)’s gains are \(3p + 4(1 - p) = 4 - p\)	M1 A1
	A2, 1, 0
Intersection of \(2p + 3\) and \(4 - p \Rightarrow p = \dfrac{1}{3}\)	M1 A1ft
\(\therefore A\) should play I \(\dfrac{1}{3}\) of time and II \(\dfrac{2}{3}\) of time; value (to \(A\)) \(= 3\dfrac{2}{3}\)	A1ft A1ft
	(8)

Mark scheme (d)

Scheme	Marks
Let \(B\) play I with probability \(q_1\), II with probability \(q_2\) and III with probability \(q_3\)	B1
e.g. Let \(x_1 = \dfrac{q_1}{v}\quad x_2 = \dfrac{q_2}{v}\quad x_3 = \dfrac{q_3}{v}\)	M1
Maximise \(P = x_1 + x_2 + x_3\)	A1
subject to \(5x_1 + 2x_2 + 3x_3 \leqslant 1\) \(3x_1 + 5x_2 + 4x_3 \leqslant 1\) \(x_1, x_2, x_3 \geqslant 0\)	A2, 1, 0
	(5)
	(17 marks)

(Corrected from the printed mark scheme: the objective is printed as \(P = x_1 + x_2 = x_3\).)

e.g. \(\begin{bmatrix}-5&-3\\-2&-5\\-3&-4\end{bmatrix} \to \begin{bmatrix}1&3\\4&1\\3&2\end{bmatrix}\)
maximise \(P = V\)
subject to \(v - q_1 - 4q_2 - 3q_3 \leqslant 0\) \(v - 3q_1 - q_2 - 2q_3 \leqslant 0\) \(q_1 + q_2 + q_3 \leqslant 1\) or \(= 1\) \(v, q_1, q_2, q_3 \geqslant 0\)