Solution s 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. .

1. Solution
s 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

2. 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

3. 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

4. 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

5. 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.

6. 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

7. 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

8. 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:

9. 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)

10. 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

11. SOLUTION: The game looks as follows:
Help
Not Help
Help
2
2
3

12. 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

13. 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.

14. 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.

15. (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

16. 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

17. 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)
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:

18. Problem 8 (2 extra credit points)
(after Kreps 1988) Assume that the President has the following
preferences over any two

19. 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?

20. X
Y
Z
W
A
1
1
0

21. 1
0
1
0
1
B
1
0

22. 1
0
0
1
0
1
C
1

23. 0
1
0
1
0
0
1
D
1

24. 0
1
0
1
0
1
0

25. 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

26. 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

27. *, 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 .

28. 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

29. 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''.

30. 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''.

31. Problem Set 3: Due in class on Tuesday July 21.