Gametheory Homework

Solutions to Problem Set 2

The following note was very important for the solutions:

In all problems below a rational preference relation is understood as one that satisfies the axioms of

von Neumann and Morgenstern’s utility theory. When solving these problems involving the

expected utility theory use the von Neumann-Morgenstern theorem. In other words, you prove that

a preference relation is rational by showing utility values that satisfy corresponding conditions and

you prove that a preference relation is not rational by showing that no utility values can possibly

satisfy these conditions. SOLUTIONS THAT DON’T USE THIS METHOD WILL NOT BE

ACCEPTED !!!

Problem 1 (3p) Suppose you have asked your friend Peter if he prefers a sure payment of

$20 or a lottery in which he gets $15 with probability 0.5 and $10 with probability 0.5. Is it

rational for Peter to prefer the sure payment over the lottery? Is it rational to prefer the

lottery over the sure payment? Is it rational to be indifferent between the lottery and the sure

payment? Would your answer be any different had I asked you the same question but with

A substituted for $20, B for $15 and C for $10? What is the general lesson to learn from

this exercise?

SOLUTION: You can assign numbers to u($20), u($15) and u($10) in such a way that

u($20) will be larger than, or equal to, or smaller than 0.5u($15)+0.5u($10). This shows that

all three preferences are rational. If instead of $20, $15 and $10 you write A, B and C the

solution to this problem, which does not depend in any way on the specifics of the three

alternatives, should be obvious. A few general lessons here: (1) Expected utility theory,

just like preference theory, does not “impose any values” on your preferences. (2) Be

careful never to use assumptions that are not clearly stated. (3) If you are given a single

piece of information about decision maker’s preferences then no matter what this

information is it cannot be possibly irrational. Rationality is, in essence, a requirement of

consistency of preferences. If there is only one condition, what would it be possibly

inconsistent with?

Problem 2 (3p) George tells you that he prefers more money over less. George also tells

you about his preference between a lottery in which he gets $30 with probability 0.9 and 0

with probability 0.1 and a sure payment of $20. Assume that George is rational. Is it

possible for him to prefer the lottery over the sure payment? Is it possible to prefer the sure

payment over the lottery? Is it possible for him to be indifferent between the sure payment

and the lottery? What is the general lesson to learn from this exercise?

SOLUTION: Suppose you have assigned numbers to u($30), u($20) and u($0) in such a

way that u($30)>u($20)>u($0):

Can such numbers satisfy u($20) < 0.9u($30)+0.1u($0)? Yes, they can. For instance,

u($30)=1, u($20)=0.5 and u($0)=0.

Can such numbers satisfy u($20) > 0.9u($30)+0.1u($0)? Yes, they can. For instance,

u($30)=1, u($20)=0.95 and u($0)=0.

Can such numbers satisfy u($20) = 0.9u($30)+0.1u($0)? Yes, they can. For instance,

u($30)=1, u($20)=0.9 and u($0)=0.

Hence, it is possible for George to have all three preferences. In general, then, the expected

utility theory does not assume anything about decision maker’s attitude towards risk.

2

Problem 3 (3p)

Paul told you that he is indifferent between a lottery in which he gets A with probability 0.8

and C with probability 0.2 and a lottery in which he gets A with probability 0.5 and B with

probability 0.5. Paul told you also that he prefers a lottery in which he gets A with

probability 0.3 and C with probability 0.7 over a lottery in which he gets B with probability

0.5 and C with probability 0.5. Is Paul’s preference relation rational?

SOLUTION: The first condition gives us 0.8u($20)+0.2u($10)= 0.5u($20)+0.5u($15)

which simplifies to 0.3u($20)+0.2u($10)= 0.5u($15). The second condition gives us

0.3u($20)+0.7u($10)> 0.5u($15)+0.5u($10) which simplifies to 0.3u($20)+0.2u($10)>

0.5u($15). But the two conditions are inconsistent, hence Paul’s preference relation is not

rational.

Problem 4 (3p)

Tom prefers A over B and B over C. Also, Tom is indifferent between a lottery in which he

gets C with probability p and A with probability 1-p and a lottery in which he gets B with

probability p and C with probability 1-p. The value of p in both lotteries is the same. For

what values of p would Tom’s preferences be rational in the sense of von Neumann-

Morgenstern’s expected utility theory?

SOLUTION: Since A B C and since the utility function (assuming that it exists, i.e., the decision maker is rational) constitutes interval scale measurement (we can pick our own

zero and our own unit) we can assume that u(A) = 1and u(C) = 0. From these assumptions it

follows that 0 < u(B) < 1, since A B C .

Now from Tom’s preference on the two lotteries we gather that:

pu(C) + (1-p)u(A) = pu(B) + (1-p)u(C)

which for u(A) = 1and u(C) = 0 becomes 1-p = pu(B). Solving this inequality for u(B)

gives:

1-p

u(B) = p

But 0 < u(B) < 1 which means that 0 < (1-p)/p < 1 which, in turn, means that 0.5 < p < 1. In

conclusion, Tom’s preferences are rational for 0.5 < p < 1.

Problem 5 (Dixit and Skeath p.117) (3p)

An old lady is looking for help crossing the street. Only one person is needed to help her;

more are okay but no better than one. You and I are the two people in the vicinity who can

help; we have to choose simultaneously whether to do so. Each of us will gain (get

pleasure) 3 “utiles” from her success, no matter who helps her. But each one who goes to

help will bear a cost of 1 utile, this being the utility of our time taken up in helping. With no

cost incurred and no pleasure derived our payoff is 0. Set this up as a normal form game.

Can you solve the game through iterated dominance?

3

SOLUTION: The game looks as follows:

Help

Not Help

Help

2

2

3

2

Not Help

2

3

0

0

This game cannot be solved by iterated dominance.

Problem 6 (Dixit and Skeath p117) (3p)

The game known as the battle of the Bismarck Sea is a model of an actual naval engagement

between the US and Japan in World War II. In 1943, a Japanese admiral was ordered to

move a convoy of ships to New Guinea; he had to choose between a rainy northern route

and a sunnier southern route, both of which required 3 days sailing time. The Americans

knew that the convoy would sail and wanted to send bombers after it, but they didn’t know

which route it would take. The Americans had to send reconnaissance planes to scout for

the convoy, but they had only enough reconnaissance planes to explore one route at a time.

Both the Japanese and the Americans had to make their decisions with no knowledge of the

plans being made by the other side.

If the convoy was on route explored by the Americans first, they could send

bombers right away; if not, they lost a day of bombing. Poor weather on the northern route

would also hamper bombing. If the Americans explored the northern route and found the

Japanese right away, they could expect only 2 (out of 3) bombing days; if they explored the

northern route and found that the Japanese had gone south, they could also expect 2 days of

bombing. If the Americans chose to explore the southern route first, they could expect 3 full

days of bombing if they found the Japanese right away but only one day of bombing if they

found that the Japanese had gone north. For payoffs use the days of bombing, positive

number for Americans and negative one for Japanese.

(i) Construct a game with (ordinal) payoffs that corresponds to this situation.

(ii) Can you solve the game through iterated dominance? Why, why not?

SOLUTION: The game looks as follows:

North

North South

South

US

Japan

2 2

-2 -2

1

-1 -3

3

4

This game cannot be solved by iterated dominance. In game theory “dominance” means

strict dominance—payoffs have to be strictly larger. For this reason strategy North of Japan

does not dominate its strategy South. Hence the game cannot be solved by iterated

dominance.

TWO EXTRA CREDIT PROBLEMS

Problem 7 (1 extra credit point)

Using only 0 and 1 as payoffs construct a 44 game which can be solved through iterated

dominance in the maximal possible number of steps.

There are many games that satisfy the two conditions. Here is one that works:

Problem 8 (2 extra credit points)

(after Kreps 1988) Assume that the President has the following preferences over any two

strategies S and S* on how to conduct a war: When choosing between S and S* prefer S

if and only if (1) it gives a lower probability of losing or (2) in case they both give the

same probability of losing, when S gives a higher probability of winning. Suppose that

we have three possible outcomes of a war: win, lose and draw. A strategy is understood

as a probability distribution on the three possible outcomes.

(i) Is this preference relation rational in the sense defined by the preference

theory?

(ii) (ii) Is this preference relation rational in the sense defined by the expected

utility theory?

X

Y

Z

W

A

1

1

0

1

0

1

0

1

B

1

0

1

0

0

1

0

1

C

1

0

1

0

1

0

0

1 D

1

0

1

0

1

0

1

0

5

Prove your conclusions.

PS. For part (ii) assume that if you have two strategies defined by the vectors of

probabilities (p 1 , p

2 , 1-p

1 -p

2 ) and (p

1 *, p

2 *, 1-p

1 *-p

2 *) which give you the probabilities of

(lose, win, draw) respectively then a lottery that gives you the first strategy with probability

q and the second with probability 1-q is equivalent to the following strategy (q p 1 + (1-q) p

1 *,

q p 2 + (1-q) p

2 *, q (1-p

1 -p

2 ) + (1-q) 1-p

1 *-p

2 *)).

SOLUTION:

(i) To prove that preference relation as defined in the problem is asymmetric and

negatively transitive, hence it is rational in the sense defined by the preference theory,

note the following. First, note that asymmetry is obvious from the definition of .

Another obvious conclusion from the definition of is that indifference between (p1, p2,

1-p1-p2) and (p1*, p2*, 1-p1*-p2*) holds only if all three probabilities are equal. But this

means that indifference relation is symmetric and transitive. Thus the only property that

remains to be proved is transitivity of . (Recall that asymmetry and transitivity of

taken together with symmetry and transitivity of ~ is equivalent to being asymmetric

and negatively transitive.)

For clarity I will write (p1, p2) instead of (p1, p2, 1-p1-p2): once p1 and p2 are known the

third probability follows from their values. In other words, the only two independent

parameters are the first two probabilities.

Note that if S = (p1, p2), S' = (p1', p2'), S'' = (p1'', p2'') are such that S S' and

S' S'' then it is either

(1) p1' < p1 and p1'' < p1' in which case p1'' < p1 and thus S S'', or

(2) p1' = p1 and p2 < p2' and p1'' < p1' in which case p1'' < p1 and thus S S'', or

(3) p1' < p1 and p1'' = p1' and p2' < p2'' in which case p1'' < p1 and thus S S'', or

(4) p1' = p1 and p2 < p2' and p1'' = p1' and p2' < p2'' in which case p1'' = p1 and p2 < p2''

and thus S S'' which completes the proof that is transitive: (S S' & S' S'')  S

S''.

Conclusion: is rational in the sense defined by the preference theory.

(ii) This preference relation is not rational in the sense defined by the expected utility

theory. To see why take the following three strategies S = (0.4, 0.4), S' = (0.3, 0.3), S'' =

(0.3, 0.4). Note first that S S' S''. From the axiom of continuity there must be a p in

(0,1) such that a lottery, call it S*, in which S is obtained with probability p and S'' with

probability 1-p must fall in between S' and S'', i.e., S' S* S''. But note that

S* = (0.3 + 0.1p, 0.4), since p > 0, and this means that 0.3 + 0.1p > 0.3 which, in turn,

means that (0.3 + 0.1p, 0.4) = S* S' = (0.3, 0.3) which contradicts S' S* S''.