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ECN214 Games and Strategies – 2013 Questions and Answers £3.99
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  7. ECN214 Games and Strategies (ECN214)

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ECN214 Games and Strategies – 2013 Questions and Answers

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High-quality past paper questions and answers for the ECN214 Games and Strategies module for the Queen Mary University of London (QMUL) Economics Course. Each question is reproduced and high-quality full-mark scores are written up clearly for each one. Great for preparing for exams, studying and so...

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  • June 7, 2020
  • 9
  • 2012/2013
  • Exam (elaborations)
  • Questions & answers

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ECN214 Games and Strategies – 2013
Questions and Answers

Part A

Question 1




The Equilibrium Existence Theorem states that allowing for mixed strategies, any finite game
possesses at least one Nash equilibrium.

Question 2




a. They cannot get their most preferred alternative selected, as neither players 2 or 3 have that as
their most preferred alternative. Therefore, the best they can expect to achieve is their second-best,
which is policy b. Then they will choose player 2 who will choose b.

b. If alternative c is selected, then player 3 must have been made king. The player who would prefer
C to have been chosen the most, except for player 3, must be player 2. Therefore, player 2 must
have been the kingmaker.

c. They would choose someone who would choose them to be made king. This would allow them to
implement their most preferred option. They would then choose player 3 to be the kingmaker, who
would then choose player 1 to be made king.

, FOR MORE HIGH-QUALITY PAST PAPER MODEL ANSWERS, ONLINE TUTORING AND
ECONOMICS HELP, visit LondonEconomicsTutors.co.uk.
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Question 3




Underlining the best responses in pure strategies:

l r
U -1, -1 5, 1
D 1, 5 4, 4

Therefore, the pure strategy Nash equilibria are (d, l) and (u, r).

To find the mixed strategy Nash equilibria, we set the probabilities such that the expected value of
each action are equal for both players.

𝐸[𝑢] = (−1)𝜌 + (1 − 𝜌)5 = 5 − 6𝜌
𝐸[𝑑] = (1)𝜌 + (1 − 𝜌)4 = 4 − 3𝜌
5 − 6𝜌 = 4 − 3𝜌
1 = 3𝜌
𝜌 = 1/3
𝐸[𝑙] = (−1)𝜋 + (1 − 𝜋)5 = 5 − 6𝜋
𝐸[𝑟] = (1)𝜋 + (1 − 𝜌)𝜋 = 4 − 3𝜋
5 − 6𝜋 = 4 − 3𝜋
𝜋 = 1/3
b) The “Nash reversion” strategy profile is one where if both players have always cooperated in the
past, then they choose to cooperate again (the “friendly regime”). In this case, this would involve
playing the action profile <d, r>. If not (the “hostile regime”), then they chose to defect in this
period, which in this case would be the action profile <u, l>.

The discounted total payoff of the “friendly regime” is:

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