978-0393919684 Chapter 15 Solution Manual

Solutions to Chapter 15 Exercises

SOLVED EXERCISES

S1. Under truthful voting, A should match Geology and Sociology in the first vote, with the winner

S2. Allocate points in the following way: 5 for a first-place vote, 2 for a second-place vote, and 1 for

S3. (a) Under a plurality voting system (with truthful voting), Proposal 1 gets 20 votes and

Proposals 2 and 3 get 15 votes each.

(c) For both the second and third type of voter to gain, Proposal 3 must win rather than

Proposal 1. This can be achieved if types 2 and/or 3 change their votes in such a way that Proposal 1 gets

Note two things. First, other voting schemes will also result in Proposal 3’s winning. For example,

S4. (a) The preferences create a standard intransitive ordering, so none of the three alternatives

would win a majority vote with truthful voting.

(b) This example is virtually identical to the three councillors example in the text in Sections

2.A, 2.B, and 4.B. Let Soft here be equivalent to Generous (G) in the City Council example, let Medium

S5. (a) (i) Under the plurality method, Tsongas gets 18 votes, Clinton 12, Brown 10, Kerry 9,

and Harkin 6.

(b) Realizing that their preferred candidates can’t win (they are only playing the role of

(c) Here are some examples of strategic voting (others can also be constructed); assume that

all groups not described vote truthfully.

(i) Under the runoff system, members of Group I could vote for Kerry instead of Tsongas

(ii) Under the elimination method, members of Group I could vote as though Harkin were

(iii) Under the Borda count, if members of Group V drop Kerry into the fifth spot on their

ballots, his point total will fall below that of Harkin, so Group V gets its first choice rather than its third.

S6. (a) Since the IRV method has not been adopted many places yet, it is still unclear empirically

It may be the case that third-party candidates would garner more first-round votes under IRV if

voters are prompted to be more sincere in stating their true preferences. If so, they might be expected to

(b) The more fundamental criticisms of IRV are that the method can be sensitive to the

inclusion of irrelevant alternatives, it does not always select the Condorcet winner, and it doesn’t solve the

For a thorough summary of the criticisms of IRV, see Kathy Dopp, “Realities Mar Instant Runoff

S7. (a) First, note that if you locate just to the left of the leftmost candidate, your vote total

equals x. If you locate just to the right of the rightmost candidate, your vote total equals y. If you locate

(i) Note that x > y is given. We can rewrite 3x + y > 1 as x > (1 – y – x)/2. Thus, locating

just left of x gives the most votes.

(b)

0 1

0

1

x

y

1/3

(1/4, 1/4)

x + 3y = 1

3x + y = 1

x = y

(i)

(ii)

(iii)

(c) Given the goal of candidates (as stated in the question), a Nash equilibrium in locations

can exist only if each candidate is in the location that maximizes his share of the vote, given the locations

UNSOLVED EXERCISES

U1. Under truthful voting, B should match Philosophy and Geology in the first vote, with the winner

U2. Allocate points in the following way: 3 for a first-place vote, 2 for a second-place vote, and 1 for

(b) Under the restriction x ≥ y ≥ z, the scheme that minimizes his advantage in first- and

(c) White will have a higher Borda count than Peterson under a scheme with weights (x – 2 –

1) when

The lowest integer value of x that will give White a higher Borda count than Peterson is 6.

(d) If all of White’s first-place voters have Peterson as their second-place vote and they vote

(e) As seen in part (d), Leinhart still wins when Oklahoma fans vote strategically, so strategic

(f) Consider a winning candidate with n first-place votes and no second- or third-place votes.

U4. (a) Before Slutskaya skated, Judge 1 had the three other skaters ranked Hughes, Kwan,

Cohen; Judge 2 had Cohen, Kwan, Hughes; Judge 3 had Kwan, Hughes, Cohen; Judge 4 had Kwan,

(b) If Slutskaya were to have placed first across the board in the long program, she would

have been ranked first for the program and would have received 1.0 points. Hughes, Kwan, and Cohen

(c) On the final judges’ cards, Hughes received the most first rankings (5), so placed first in

the long program; Slutskaya finished second (4 firsts), Kwan finished third (no firsts, 5 seconds), and

Cohen finished fourth. Thus, Hughes got 1.0 for the long program and Slutskaya, Kwan, and Cohen got

(d) The scoring system violates the principle of independence of irrelevant alternatives

U5. (a) Under truthful voting, the pairwise votes would have been

Given these preferences, Romney would have been Condorcet winner with truthful voting.

(b) Under truthful voting, the results of the second round of the caucus—after the elimination

(c) Given that the 16% of McCain voters sincerely prefer Romney to Huckabee, it is initially

(d) The allegations of collusion between the McCain and Huckabee camps are likely

unfounded. No collusion or agreement between the supporters of McCain and Huckabee was needed to

U6. (a) Ana and Bernard have an incentive to vote strategically. They can prevent their

(b) This example shows that one of the necessary conditions for a guaranteed eventual

In this case if another voter, Harris, appears, it may be even more difficult for a majority to be

reached by either V or W, since five votes will be required instead of four.

U7. (a) The Center-type voter’s vote matters only if there is one vote for A and one vote for D

from the other voters. This Center voter assumes that both Right and Left types vote truthfully in round

(b) If this Center type votes truthfully for A in both rounds, then A will definitely win the first

round. Whether A will beat G in the second round, however, depends on whether the original vote for D

If this Center type votes strategically (for D) in the first round, then D wins the first round. In the