C_{1} | C_{2} | C_{3} | |
R_{1} | 0 | 4 | 9 |
R_{2} | 9 | 6 | 8 |
R_{3} | 8 | 7 | 7 |
Using dominance considerations, identify the solution to this game. Write your answer as a strategy pair: (R_{x}, C_{y}), where x = 1, 2, or 3; and y is 1, 2, or 3.
The solution is (R_{3}, C_{2}).
C_{1} | C_{2} | |
R_{1} | 9 | 2 |
R_{2} | 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(R_{1}) = (2/3)(9) + (1/3)(2) = 18/3 + 2/3 = 20/3
EU(R_{2}) = (2/3)(5) + (1/3)(8) = 10/3 + 8/3 = 18/3
Since EU(R_{1}) > EU(R_{2}), the row player can do better by playing R_{1} than by playing any mixed strategy that gives him a nonzero probability of playing R_{2}—such as (1/2 R_{1}, 1/2 R_{2}).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(C_{1}) = (1/2)(9) + (1/2)(5) = 9/2 + 5/2 = 14/2
EU(C_{2}) = (1/2)(2) + (1/2)(8) = 2/2 + 8/2 = 10/2
Since EU(C_{2}) > EU(C_{1}), the column player can do better by playing C_{2} than by playing any mixed strategy that gives her a nonzero probability of playing C_{1}—such as (2/3 C_{1}, 1/3 C_{2}).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.
solving for p:
EU(C_{1}) = EU(C_{2})
p(9) + (1 – p)(5) = p(2) + (1 – p)(8)
9p + 5 – 5p = 2p + 8 – 8p
4p + 5 = –6p + 8
10p = 3
p = 3/10solving for q:
EU(R_{1}) = EU(R_{2})
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
C_{1} | C_{2} | |
R_{1} | 2, 8 | 8, 7 |
R_{2} | 3, 4 | 9, 1 |
There is one equilibrium strategy pair—(R_{2}, C_{1})—and the game is a prisoner’s dilemma.
C_{1} | C_{2} | |
R_{1} | 3, 8 | 8, 3 |
R_{2} | 1, 4 | 9, 5 |
There are two equilibrium strategy pairs—(R_{1}, C_{1}) and (R_{2}, C_{2})—and the game is a battle of wills.
False.
- True of false? In a prisoner’s dilemma, any Pareto-optimal outcome is also an equilibrium outcome.
False.
- 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.
- True or false? In a prisoner’s dilemma, no equilibrium outcome is a Pareto-optimal outcome.
True.
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.
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.
It is (7 – 4)(5 – 3) = (3)(2) = 6.
- What is relative benefit of player 1 under the outcome (7, 5)?
It is (7 – 4)/(10 – 4) = 3/6 = 1/2
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).
As stated in item 3 of the instructions, you do not need to turn in this list of questions.