Your Turn
11.1
11.2
11.3
11.4
11.6
11.7
11.8
3.
Different candidates won. It appears that the vote counts were so close that a small shift in either direction could change the results of either method.
11.9
11.10
11.11
11.12
11.13
11.14
11.16
11.17
11.18
11.19
11.20
4.
Increasing the ranking for a winner of a Borda count election on a ballot will increase that candidate’s Borda score while decreasing another candidate’s Borda score, but leaving the remaining candidates’ Borda score unchanged. So, a Borda count election will never violate the monotonicity criterion.
11.21
4.
It is possible that the monotonicity criterion would be met in other ranked-choice election scenarios, but overall, the ranked-choice voting method is said to fail the monotonicity criterion even if it failed in only one scenario.
11.24
11.26
4.
The ratio of State Population to Representative Seats seems to be either 700,000 or 800,000 to 1. There does appear to be a constant ratio of about 0.000001 to 1 of Representative Seats to State Population when rounding to six decimal places. This is the same as the top five states.
11.27
11.28
1.
The states are the Hernandez family and the Higgins family. The seats are the pieces of candy. The house size is 313. The state populations are three in the Hernandez family and four in the Higgins family. The total population is 7.
2.
The standard divisor is the ratio of the number of children to the number of pieces of candy. 0.0224 children per piece of candy.
11.29
11.30
1.
| Family | Family’s Standard Quota |
|---|---|
| Hernandez | 175.6667 candies |
| Higgins | 234.2222 candies |
| Ho | 117.1111 candies |
| Total | 527 |
The sum is very close to 527.
11.32
1.
It is not possible to give a fractional part of a gift card. Also, traditional rounding to the nearest integer results in 4 gift cards for each student, which leaves one extra gift card.
11.34
1.
The final Hamilton apportionment is Fictionville as follows: 1, Pretendstead 3, Illusionham 5, and Mythbury 26.
11.35
11.36
5.
The modified quota from part 3 was the smallest, because the divisor was the largest of the three. Dividing the same number by a larger value gives a smaller result.
11.37
1.
Each state would receive the following seats: Fictionville 1, Pretendstead 2, Illusionham 4, and Mythbury 28.
11.38
11.39
11.40
1.
The largest state is Mythbury. The citizens would likely favor the Jefferson method of apportionment most since they received the most seats by that method. They would likely favor the Adams and Webster methods of apportionment least because they received the least number of seats by those method.
2.
As a group the other three states received nine seats by either the Hamilton method, seven seats by the Jefferson Method, and ten seats by either the Adams method or the Webster method. They would likely favor the Adams method and Webster method most and favor the Jefferson methods least.
11.41
11.42
1.
The standard divisor is 214,079.1236 citizens per seat. The standard quota for Colorado is 2.5210 seats.
2.
The standard divisor is 213,479.4622 citizens per seat. The standard quota for Colorado is 2.5281 seats.
4.
It must have been the case that either the fractional part 0.5281 ranked lower amongst the other fractional parts of the state quotas than the fractional part 0.5210 did, or there were fewer remaining seats to be distributed, or both.
11.43
3.
Yes, this demonstrates the Alabama paradox because State C receives 14 seats if the house size is 323, but only 13 seats if the house size is 324.
11.44
1.
Hospital C lost a respirator while hospital A gained a seat, but hospital C has a higher growth rate than hospital A.
11.45
1.
The Hamilton reapportionment is: 19 for Mudston, 13 for WallaWalla, and 6 for Dilberta. This is an example of the population paradox because WallaWalla lost a seat to Dilberta, even though WallaWalla’s population grew by 2.24 percent while Dilberta’s only grew by 1.56 percent.
11.46
3.
The original state of Beaversdam lost a seat to the original state of Beruna when the new state of Chippingford was added.
11.47
1.
The reapportionment gives 38 seats to Neverwood, 46 seats to Mermaids Lagoon, and 9 seats to Marooners Rock. This is an example of the new-states paradox because the original state Mermaids Lagoon lost a seat to the original state Neverwood when the new state was added to the union.
11.48
1.
<{\text{Original Standard Divisor}} \approx \frac{{76,000,000}}{{391}} \approx 194,400>
<{\text{New Population}} \approx 76,000,000 + 300,000 = 76,300,000>
There are <392 - 391 = 1> new seats to be apportioned to New Mexico.
11.49
Check Your Understanding
7.
In two-round voting, only the top two candidates from Round 1 move on to Round 2, and there are only two rounds. In ranked-choice voting, all candidates except those in last place move on to the next round, and there can be many rounds of voting.
14.
False, because the ranked-choice method violates the Condorcet criterion, but it doesn’t violate the majority criterion.
15.
No. Arrow’s Impossibility Theorem does not apply to approval voting because it is not a ranked voting system
16.
False. Arrow’s Impossibility Theorem says that no ranked voting system is perfect and that voter profiles may arise that will lead to a violation of one or more fairness criteria, but it does not guarantee that those voter profiles will occur or are even likely to occur.
17.
True. Candidates are rated as approved or not approved, and voters can give multiple candidates the same rating.