University of Kansas, Fall 2006
Philosophy 666: Rational Choice Theory
Ben Eggleston—eggleston@ku.edu

test on game theory: answer key

(October 27, 2006)

Instructions:

  1. Answer all of the following questions on the answer sheets provided. You can write on this list of questions, but credit will be awarded only for answers written on answer sheets.
  2. Do not access any book, notebook, newspaper, calculator, computer, cell phone, or other possible source of inappropriate aid during the exam, do not leave the room before you are finished taking the exam, and be sure to finish the exam within this 50-minute class period—no credit will be given for any work done after you access any possible source of inappropriate aid, after you leave the room for any reason, or after the end of this class period.
  3. When you are finished, be sure your name is written on each of your answer sheets, and turn them in. You do not need to turn in this list of questions.

Questions:

Questions 1–3 involve zero-sum games, with each cell containing a number that represents the row player’s utility and the negation of the column player’s utility.
  1. Consider the following game:
  C1 C2 C3
R1 0 4 9
R2 9 6 8
R3 8 7 7

Using dominance considerations, identify the solution to this game. Write your answer as a strategy pair: (Rx, Cy), where x = 1, 2, or 3; and y is 1, 2, or 3.

The solution is (R3, C2).

  1. Explain why [(1/2 R1, 1/2 R2), (2/3 C1, 1/3 C2)] is not an equilibrium pair for the game represented by the following table. (Explain this in terms of the definition of an equilibrium pair, not just finding an equilibrium strategy pair and pointing out that it is different from the pair just mentioned.)
  C1 C2
R1 9 2
R2 5 8

answer:

It's not an equilibrium pair because at least one player can do better by unilaterally deviating from his or her stipulated strategy, on the assumption that the other player is, in fact, playing his or her stipulated strategy.

First, consider the row player.
EU(R1) = (2/3)(9) + (1/3)(2) = 18/3 + 2/3 = 20/3
EU(R2) = (2/3)(5) + (1/3)(8) = 10/3 + 8/3 = 18/3
Since EU(R1) > EU(R2), the row player can do better by playing R1 than by playing any mixed strategy that gives him a nonzero probability of playing R2—such as (1/2 R1, 1/2 R2).

That’s enough to show that at least one player can do better by unilaterally deviating from his or her stipulated strategy, on the assumption that the other player is, in fact, playing his or her stipulated strategy. But let’s look at the column player’s situation, too.
EU(C1) = (1/2)(9) + (1/2)(5) = 9/2 + 5/2 = 14/2
EU(C2) = (1/2)(2) + (1/2)(8) = 2/2 + 8/2 = 10/2
Since EU(C2) > EU(C1), the column player can do better by playing C2 than by playing any mixed strategy that gives her a nonzero probability of playing C1—such as (2/3 C1, 1/3 C2).

So, just looking at things from the column player’s point of view also turns out to be enough to show that at least one player can do better by unilaterally deviating from his or her stipulated strategy, on the assumption that the other player is, in fact, playing his or her stipulated strategy.

  1. Find the values for p and q that make [(p R1, 1 – p R2), (q C1, 1 – q C2)] an equilibrium strategy pair for the game given in question 2. (Show your calculations.)

solving for p:
EU(C1) = EU(C2)
p(9) + (1 – p)(5) = p(2) + (1 – p)(8)
9p + 5 – 5p = 2p + 8 – 8p
4p + 5 = –6p + 8
10p = 3
p = 3/10

solving for q:
EU(R1) = EU(R2)
q(9) + (1 – q)(2) = q(5) + (1 – q)(8)
9q + 2 – 2q = 5q + 8 – 8q
7q + 2 = –3q + 8
10q = 6
q = 6/10
q = 3/5

  1. Write any equilibrium strategy pair(s) for the following game. Is it a battle of wills, a prisoner’s dilemma, both, or neither?
  C1 C2
R1 2, 8 8, 7
R2 3, 4 9, 1

There is one equilibrium strategy pair—(R2, C1)—and the game is a prisoner’s dilemma.

  1. Write any equilibrium strategy pair(s) for the following game. Is it a battle of wills, a prisoner’s dilemma, both, or neither?
  C1 C2
R1 3, 8 8, 3
R2 1, 4 9, 5

There are two equilibrium strategy pairs—(R1, C1) and (R2, C2)—and the game is a battle of wills.

  1. This is a four-part problem.
    1. True or false? In a prisoner’s dilemma, any equilibrium outcome is also a Pareto-optimal outcome.

False.

  1. True of false? In a prisoner’s dilemma, any Pareto-optimal outcome is also an equilibrium outcome.

False.

  1. True or false? In a prisoner’s dilemma, there is always at least one outcome that is both an equilibrium outcome and an Pareto-optimal outcome.

False.

  1. True or false? In a prisoner’s dilemma, no equilibrium outcome is a Pareto-optimal outcome.

True.

  1. Why is it that, in a particular round of an iterated prisoner’s dilemma, it might be rational for a player to choose a different strategy than the one that would be dominant if he or she were in a non-iterated, one-shot, standard prisoner’s dilemma?

This is because the other player might be playing a strategy that is sensitive to his or her opponent’s prior moves; in particular, that strategy might lead the other player to play cooperatively if his or her opponent has done so in a previous round. (The tit-for-tat strategy investigated by Axelrod is like this.) Getting one’s opponents to play cooperatively, rather than uncooperatively, can often be a desirable prospect for a player, and rational to pursue.

  1. Suppose two people are in a bargaining situation and that, after being told about the Nash approach to such situations, they report that (A) the Nash approach does not identify a unique solution to their problem, because they have found that (B) there are two points with the same Nash score, with that score being as high as the Nash score of any other achievable point. How would you go about proving that they are mistaken about B? (You do not have to supply any such proof; you just have to explain how such a proof would proceed.)

You would show that, whenever two points have the same Nash score, a third point—such as the average of the first two, or indeed any point on the line segment connecting them—has a higher Nash score.

  1. Suppose that, in some bargaining situation, player 1’s maximum possible payoff is 10, player 2’s maximum possible payoff is 8, and the no-agreement point is (4, 3).
    1. What is the Nash score of the outcome (7, 5)?

It is (7 – 4)(5 – 3) = (3)(2) = 6.

  1. What is relative benefit of player 1 under the outcome (7, 5)?

It is (7 – 4)/(10 – 4) = 3/6 = 1/2

  1. In an n-person game (with a characteristic function, coalitions, imputations, and so on), what conditions must be satisfied in order for one imputation to dominate another with respect to some coalition? As you state each condition, also give the motivation, or meaning, behind it.

There are two conditions that are individually necessary and jointly sufficient.

The first condition is that each member of the coalition must do better under the first imputation than under the second one. The point of this is that if not every member of the coalition preferred the first imputation to the second, then it doesn't make sense to think of the first one being preferable to the second, from the point of view of that particular coalition.

The second condition is that sum of the payoffs that the members of the coalition would receive under the first imputation must not exceed the value of the coalition, as given by the characteristic function. This is because there is little point in talking about one imputation being preferable to another, for some coalition, if that coalition could not obtain the payoffs of the preferred imputation without the assistance of other players (via, for example, forming a larger coalition with them).

Instructions, revisited:

As stated in item 3 of the instructions, you do not need to turn in this list of questions.