Abstract
This paper compares the vulnerability of Borda Elimination Rule (BER) and of Nanson Elimination Rule (NER) to monotonicity paradoxes under both fixed and variable electorates. It is shown that while NER is totally immune and BER is vulnerable to monotonicity failure in 3-candidate elections, neither of these two rules dominates the other in n-candidate elections (n > 3) when no Condorcet Winner exists. When the number of competing alternatives is larger than three and no Condorcet Winner exists, we find profiles where NER violates monotonicity while BER does not, profiles where BER violates monotonicity while NER does not, as well as profiles where both NER and BER violate monotonicity. These findings extend to both fixed and variable electorates, as well as to situations where the initial winners under both rules are the same, as well as to situations where the initial winners under both rules are different. So, which of the two rules should be preferred in terms of monotonicity in n-candidate elections (n > 3) where no Condorcet Winner exists, depends on the kind of profiles one can expect to encounter in practice most often. Nevertheless, in view of the results of 3-candidate elections under other scoring elimination rules, we conjecture that inasmuch as BER and NER exhibit monotonicity failures, it is more likely to occur in closely contested elections.
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Notes
Smith (1973) has shown that a whole class of SERs is subject to monotonicity failure in fixed electorates. (It should be noted that despite its title, the focus of Smith’s article is on fixed electorates, as defined by us above). However, many voting procedures which are not SERs are also susceptible to monotonicity failures. See Felsenthal (2012), Felsenthal and Tideman (2013, 2014), as well as Felsenthal and Nurmi (2016, 2017).
Impartial Anonymous Culture (IAC) assumes that all possible voting situations are equally likely to be observed. Although this assumption is unrealistic, it is useful when one wishes to get an impression regarding the relative frequency in which various voting procedures may exhibit some phenomenon or regarding how similar to each other are procedures when it comes to exhibiting the phenomenon.
Assuming that voters do not change their preferences between rounds, PER can be described also as an elimination procedure based on voters’ rank order of the candidates where the voters go to the polls only once.
Borda’s rule (or Borda’s count) (1784 [1995]) is named after Jean Charles de Borda who proposed it in a paper he delivered before the French Royal Academy of Sciences in 1770 entitled ‘Memorandum on election by ballot’ (‘Mémoire sur les élections au scrutin’). This paper was included in the 1781 annals of the Academy which were published in 1784. See Black (1958), McLean and Urken (1995, pp. 83–89).
It should be noted that BER is distinct from ‘IRV (or more generally STV) with Borda elimination’, as proposed by Geller (2005). This rule (like other IRV variants) entails vote transfers, is not Condorcet-consistent, and is subject to monotonicity violations. Fishburn (1977, p. 472) called BER ‘Nanson’s function’.
NPER is distinct from Coombs’s (1964) method in that, inter alia, under NPER a candidate who is ranked last by the largest number of voters will be eliminated even if s/he also constitutes the top preference of an absolute majority of the voters, whereas under Coombs’s method this candidate would be declared the winner. Thus, for example, if we have 100 voters 26 of whom with preference ordering a > b > c, 25 with preference ordering a > c > b, 25 with preference ordering b > c > a and 24 with preference ordering c > b > a, then according to Coombs’s method a would be elected because s/he is ranked first by an absolute majority of the voters, whereas under NPER a would be eliminated first because s/he is ranked last by the largest number of voters and thereafter b would beat c and become the winner.
Given voters’ preference orderings among the n competing candidates (n ≥ 2), a Condorcet Winner is the candidate who is preferred by the majority of voters to any other candidate. This candidate is named after the Marquis de Condorcet (1785) who suggested that if this candidate exists then s/he ought to be elected.
Of course if all the remaining candidates have equal Borda scores then one must use some tie-breaking rule. We italicize the words ‘equal to or lower than…’ because many authors describing NER err in their description by stating that in every round only the candidates whose Borda score is less than the average Borda score are eliminated. When no Condorcet Winner exists the erroneous description of NER may result, ceteris paribus, in electing a different candidate than the one elected under the correct description of NER.
Although NER and BER satisfy the strong Condorcet condition, i.e., these rules always elect a candidate who is preferred by the majority of voters over every other candidate, these rules may not satisfy the weak Condorcet condition which requires that if there exist(s) candidate(s) who is (are) unbeaten by any other candidate but is (are) tied with at least one other candidate, then this (these) candidate(s)—and only this (these) candidate(s)—ought to be elected. Such candidates constitute the core of the voting game. For an example of violation of the weak Condorcet condition by NER and BER see Niou (1987). A stronger case of violating the weak Condorcet condition by NER has been discussed by Nurmi (1989, p. 202). This case is one with six voters and five candidates (a, b, c, d, e) where the core and the NER winner are distinct candidates. The profile under which this happens is as follows: 1 voter: a > b > c > d > e; 1 voter: a > d > b > c > e; 1 voter: a > d > e > b > c; 1 voter: b > c > e > d > a; 2 voters: c > e > d > b > a. Here only candidate a is unbeaten (and is, therefore, the weak Condorcet Winner and the only candidate who belongs to the core) but the NER winner is candidate c. A similar violation of the weak Condorcet condition by BER is not possible.
We wish to thank referee #1 of this paper for suggesting the following proof for case (1) under variable electorates—which was originally missing.
In Examples 5–10 in the Appendix we deal with situations where one of the investigated procedures violates monotonicity while the other does not when the initial winners are different under these procedures. Hence one must explain, as we do, why the procedure which does not violate monotonicity cannot modify the initial profile in any way such that its original winner will become a loser. However, in Examples 11–15 we deal with situations where one of the investigated procedures violates monotonicity while the other does not when the initial winners are the same under these procedures. Hence one needs only to show, as we do, that under the same modification in the initial profile one of the procedures violates monotonicity while the other does not.
All the examples we devised to demonstrate violation of monotonicity by BER and/or NER when there are more than three competing alternatives contain either four or six alternatives.
It should also be noted that an algebraic analysis of the underlying structure of the paradox of non-monotonicity in 3-candidate run-off elections is provided by Lepelley et al. (1996) and by Miller (2017). Recent examples of empirical evaluations of 3-candidate voting rules that include run-offs are Plassmann and Tideman (2014),Green-Armytage et al. (2016), and Miller (2017). Plassmann and Tideman (2014) estimated that of the 14 voting procedures they investigated the Black and Nanson procedures—both of which are Condorcet-consistent—are less vulnerable to voting paradoxes in 3-candidate elections than the other investigated voting procedures, especially in elections with few voters.
The sum of all voters’ Borda scores (240) can be computed independently of the matrix. It is always equal to vn(n − 1)/2 where v is the number of voters and n is the number of candidates.
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Acknowledgements
The authors are very grateful to Nicholas R. Miller for his thoughtful and constructive suggestions on earlier versions of this article, as well as to two anonymous referees of this article for their helpful comments.
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Appendix
Appendix
Following are 16 examples. Examples 1–4 prove Proposition 2(b) and the remaining 12 examples prove Proposition 3.
Examples Proving Proposition 2 (b)
BER may demonstrate both upward and downward monotonicity failures in 3-candidate elections with fixed and with variable electorates when a Condorcet Winner does not exist. To see an upward violation of monotonicity by BER in 3-candidate elections under fixed electorate consider the following example:
Example 1
This example is due to Fishburn (1977: 478, Example E3). Suppose there are 20 voters whose preference orderings among three candidates, a, b, and c, are as follows:
No. of voters | Preference orderinga |
|---|---|
8 | a > b > c |
5 | b > c > a |
5 | c > a > b |
2 | c > b > a |
According to BER the initial Borda scores of a, b, and c are 21, 20, and 19, respectively. So according to BER c is eliminated at the end of the first counting round and in the contest between a and b in the second counting round a beats b (13:7) and is elected (note that a is elected here under NER too).
Now suppose that, ceteris paribus, the two c > b > a voters change their preference ordering to c > a > b thereby increasing the support of a. As a result we get the following distribution:
No. of voters | Preference orderings |
|---|---|
8 | a > b > c |
5 | b > c > a |
7 | c > a > b |
Here the Borda scores in the first counting round of a, b, and c according to BER are 23, 18, and 19, respectively. So according to BER one eliminates b at the end of the first counting round and in the contest between a and c in the second counting round c beats a (12:8) and is elected—thereby displaying monotonicity failure. Note that here NER maintains monotonicity by continuing to elect a.
To demonstrate upward monotonicity failure by BER in 3-candidate elections under variable electorates when a Condorcet Winner does not exist, consider the following example.
Example 2
Suppose there are 13 voters whose preference orderings among three candidates, a, b, and c, are as follows:
No. of voters | Preference orderings |
|---|---|
5 | a > b > c |
4 | b > c > a |
4 | c > a > b |
According to BER the Borda scores of a, b, and c in the first counting round is 14, 13, and 12, respectively. So according to BER c is eliminated at the end of the first counting round and in the contest between a and b in the second counting round a beats b (9:4) and is elected.
Suppose now that, ceteris paribus, two additional voters with preference ordering a > c > b join the electorate thereby increasing a’s support. As a result according to BER the Borda scores of a, b, and c in the first counting round is 18, 13, and 14, respectively. So according to BER b is eliminated at the end of the first counting round and in the contest between a and c in the second counting round c beats a (8:7) and is elected—in violation of monotonicity. Note that in this example the initial winner under both BER and NER is the same and, as proven in Proposition 2a, NER does not violate monotonicity in the revised electorate.
To demonstrate downward monotonicity failure by BER in 3-candidate elections under fixed electorate consider the following example.
Example 3
Suppose there are 29 voters whose preference orderings among three candidates, a, b, and c, are as follows:
No. of voters | Preference orderings |
|---|---|
6 | a > b > c |
3 | a > c > b |
4 | b > a > c |
8 | b > c > a |
6 | c > a > b |
2 | c > b > a |
According to BER the Borda scores of a, b, and c in the first counting round is 28, 32, and 27, respectively. So according to BER c is eliminated at the end of the first counting round and in the contest between a and b in the second counting round a beats b (15:14) and is elected.
As b has not been elected, suppose that, ceteris paribus, two of the eight voters with preference ordering b > c > a change their preference ordering to c > b > a thereby lowering b’s support. As a result, according to BER the Borda scores of a, b, and c in the first counting round change to 28, 30, and 29, respectively. So according to BER a is now eliminated at the end of the first counting round and in the contest between b and c in the second counting round b beats c (16:13) and is elected—thus demonstrating downward monotonicity failure. Note that in this example the initial winner under NER (b) is different than that under BER and, as proven in Proposition 2a, NER does not demonstrate downward monotonicity failure in the revised 3-candidate electorate.
To demonstrate downward monotonicity failure by BER in 3-candidate elections under variable electorate consider the following example.
Example 4
Suppose that the initial situation is the same as the initial situation in Example 3, i.e., according to BER a beats b (15:14) and is elected.
As b has not been elected, suppose now that, ceteris paribus, two additional voters with preference ordering c > a > b join the electorate thereby decreasing further b’s support. As a result, according to BER the Borda scores of a, b, and c in the first counting round change to 30, 32, and 31, respectively. So according to BER a is now eliminated at the end of the first counting round and in the contest between b and c in the second counting round b beats c (18:13) and is elected—thus demonstrating that BER is susceptible to downward monotonicity failure under variable electorate. Note that in this example the initial winner under NER (b) is different than that under BER and, as proven in Proposition 2a, NER does not demonstrate downward monotonicity failure in the revised 3-candidate electorate.
Examples proving Proposition 3 : BER vs. NER in n-candidate elections (n > 3) when the initial winner is different
In the following six examples we show that in n-candidate elections (n > 3)—with either fixed or variable electorates—it is possible that, ceteris paribus, when the initial winners under NER and BER are different then one of these rules may violate monotonicity while the other does not, as well as cases where both rules violate monotonicity.
Example 5
The following example of 4-candidate election under fixed electorate shows that, ceteris paribus, monotonicity is violated by NER but not by BER when no Condorcet Winner exists and the initial winners under these two rules are different.
Suppose there are 43 voters whose preference orderings among four candidates, a, b, c, d, are as follows:
No. of voters | Preference ordering |
|---|---|
9 | a > b > d > c |
5 | a > c > b > d |
2 | a > c > d > b |
5 | b > a > c > d |
9 | b > d > c > a |
13 | c > a > d > b |
According to both BER and NER the Borda scores of a, b, c, and d in the first counting round is 84, 65, 67, and 42, respectively. As the average Borda score here is 64.5, d is eliminated at the end of the first counting round according to both NER and BER. After the elimination of d the revised Borda score of a, b, and c is 50, 37, and 42, respectively. As the average Borda score now is 43, both b and c are eliminated at the end of the second counting round according to NER and hence a is declared the winner. However, according to BER only b is eliminated at the end of the second counting round and in the contest between a and c in the third counting round c beats a (22:21) and is elected.
Now suppose that, ceteris paribus, the five voters with preference ordering b > a > c > d change their preference ordering to a > b > c > d thereby increasing the support of a, the original winner according to NER. As a result the initial Borda scores of a and b change to 89 and 60, respectively, while the initial Borda scores of c and d remain the same (67 and 42). As the average Borda score also remains the same (64.5), it follows that now b and d are eliminated according to NER at the end of the first counting round and in the contest between a and c in the second counting round c beats a (22:21) and becomes the winner—thereby demonstrating monotonicity failure under NER.
Note that in this example BER does not violate monotonicity. As we have seen, c is elected according to BER in the original situation. Now, no matter which voters, ceteris paribus, will increase their support of c, c remains the winner according to BER. This is so because:
-
(1)
As in the original situation, d remains the first alternative to be eliminated. This is so because in the original situation d had the lowest Borda score. So if any of the voters who ranked originally d higher than c will reverse their preference ordering in order to increase their support of c then, ceteris paribus, d’s Borda score will decrease further and otherwise it will not change.
-
(2)
No matter how one increases the support of c, its Borda score does not decrease, while that of some other alternative(s) does (do) decrease.
-
(3)
It follows that after d is eliminated at the end of the first counting round, c has at least the second largest (revised) Borda score among a, b and c.
-
(4)
If b is eliminated next, then c remains the winner (since it defeats a already in the original profile and a fortiori in the modified one).
-
(5)
If a is eliminated at the end of the second counting round then its Borda score in the modified profile has to be less than 37, i.e., at least 13 Borda points have to be switched from a to b. However, the increase in c’s support cannot, ceteris paribus, affect the original preference orderings between a and b, hence it is b, not a, that must be eliminated at the end of the second counting round.
-
(6)
As c beats a in the third counting round, c remains the winner—thereby BER does not violate monotonicity in this example.
Example 6
The following example of 4-candidate election under fixed electorate shows that, ceteris paribus, monotonicity is violated by BER but not by NER when no Condorcet Winner exists and the initial winners under these two rules are different.
Suppose there are 28 voters whose preference orderings among four candidates, a, b, c, d, are as follows:
No. of voters | Preference orderings |
|---|---|
6 | a > b > c > d |
3 | a > c > b > d |
4 | b > a > c > d |
6 | b > c > a > d |
6 | c > a > b > d |
2 | c > b > a > d |
1 | d > b > c > a |
According to both BER and NER the Borda scores of a, b, c, and d in the first counting round is 55, 57, 53, and 3, respectively. As the average Borda score here is 42, d is eliminated at the end of the first counting round according to both NER and BER. After the elimination of d the revised Borda score of a, b, and c is 28, 30, and 26, respectively. As the average Borda score now is 28, both a and c are eliminated at the end of the second counting round according to NER and hence b is declared the winner. However, according to BER only c is eliminated at the end of the second counting round and in the contest between a and b in the third counting round a beats b (15:13) and is elected.
As candidate a has been elected under BER, suppose now that, ceteris paribus, three of the four voters with preference ordering b > a > c > d change their preference ordering to a > b > c > d and the two voters with preference ordering c > b > a > d change their preference ordering to c > a > b > d—thereby increasing the support of a, the original winner according to BER. As a result the initial Borda scores of a and b change to 60 and 52, respectively, while the Borda scores of c and d remain the same (53 and 3, respectively). So now, as before, d is eliminated in the first counting round according to BER. After the elimination of d the revised Borda score of a, b, and c is 33, 25, and 26, respectively. So at the end of the second counting round b is eliminated according to BER and in the contest between a and c in the third counting round c beats a (15:13) and is elected—thereby demonstrating monotonicity failure under BER.
Note that in this example NER does not violate monotonicity. As we have seen, b is elected according to NER in the original situation. Now, no matter which voters, ceteris paribus, will increase their support of b, b remains the winner according to NER. This is so because:
-
(1)
As in the original situation, d remains the first alternative to be eliminated. This is so because in the original situation d had the lowest Borda score. As all voters except one ranked d at the bottom of their preference orderings, only the single voter who ranked d at the top of his/her preference ordering may increase b’s support by switching his/her preference ordering between b and d. But if this voter switches his/her preference ordering in this way then, ceteris paribus, d’s Borda score will decrease further and otherwise it will not change.
-
(2)
After eliminating d in the first counting round, then regardless of which voter(s) change(s)his/her/their preference ordering in b’s favor, ceteris paribus, the Borda score of b must increase while the Borda scores of at least one of the remaining candidates (a or c) must decrease. As the average Borda score remains the same (in the fixed electorate), it follows that as both a and c have been eliminated according to NER in the original situation at the end of the second counting round, these candidates must be eliminated according to NER a fortiori at the end of the second counting round in the revised profile. So b is elected according to NER in this example in the revised profile too without violating monotonicity.
Example 7
The following example of 4-candidate election under variable electorate shows that monotonicity is violated by BER but not by NER when no Condorcet Winner exists and the initial winners under these two rules are different.
Suppose there are 64 voters whose preference orderings among four candidates, a, b, c, d, are as follows:
No. of voters | Preference ordering |
|---|---|
22 | a > b > c > d |
2 | b > a > c > d |
16 | b > c > a > d |
18 | c > a > b > d |
5 | c > b > a > d |
1 | d > b > c > a |
The initial Borda scores of a, b, c, d, are 127, 128, 126, and 3, respectively, with an average of 96. So according to both BER and NER candidate d is eliminated at the end of the first counting round and the revised Borda scores of a, b, c are 64, 65, and 63, respectively, with an average of 64. So according to NER both a and c are eliminated at the end of the second counting round and hence b is declared the NER winner. However, after d is eliminated at the end of the first counting round, c is eliminated according to BER at the end of the second counting round, and thereafter a beats b (40:24) and becomes the BER winner.
Now suppose that, ceteris paribus, three additional voters with preference ordering a > c > b > d join the electorate thus increasing a’s support. As a result, the Borda scores of a, b, c, d, are now 136, 131, 132, and 3, respectively. So according to BER candidate d is eliminated at the end of the first counting round and the revised Borda scores of a, b, c are 70, 65, and 66, respectively. So now candidate b is eliminated according to BER at the end of the second counting round, and thereafter c beats a (40:27) and becomes the BER winner—thereby violating monotonicity.
However, given the first part of this example (where the initial winner is b according to NER), it is impossible to demonstrate, ceteris paribus, violation of monotonicity by NER by adding to the electorate even one voter whose top preference is b—regardless of this voter’s preference orderings among the three remaining candidates.
Example 8
The following example of 4-candidate election under variable electorate shows that monotonicity is violated by NER but not by BER when no Condorcet Winner exists and the initial winners under these two rules are different.
Suppose there are 18 voters whose preference orderings among four candidates, a, b, c, d, are as follows:
No. of voters | Preference ordering |
|---|---|
5 | a > b > d > c |
5 | b > c > d > a |
6 | c > a > d > b |
1 | c > b > a > d |
1 | c > b > d > a |
The Borda scores of a, b, c, and d in the first counting round are 28, 29, 34, and 17, respectively. As the average Borda score here is 27, d is eliminated at the end of the first counting round according to both NER and BER. After the elimination of d the revised Borda scores of a, b, and c are 16, 17, and 21, respectively. As the average Borda score now is 18, both a and b are eliminated at the end of the second counting round according to NER and hence c is declared the winner. However, according to BER only a is eliminated at the end of the second counting round and in the contest between b and c in the third counting round b beats c (10:8) and is elected.
Now suppose that, ceteris paribus, one additional voter with preference ordering c > b > d > a joins the electorate—thereby increasing c’s support. As a result the initial Borda scores of a, b, c, and d are now 28, 31, 37, and 18, respectively, with an average of 28.5. So according to NER both a and d are eliminated in the first counting round and in the second counting round b beats c (10:9) and is declared the winner—thereby demonstrating the violation of monotonicity under NER.
However, monotonicity is not violated by BER in this example. This is so because the only candidate who originally beats b in pairwise comparisons is a, so if, ceteris paribus, at least five additional voters whose top preference is b join the electorate, then b becomes a Condorcet Winner and is elected under BER even if a constitutes the second preference of these additional voters. Moreover, even if only one additional voter whose top preference is b and second preference is a joins the electorate then, ceteris paribus, b will continue to be elected according to BER. This is so because regardless of whether d and c constitute the third and last preference, respectively, of this voter, or vice versa, d will be eliminated in the first counting round, a will be eliminated in the second counting round, and in the contest between b and c in the third counting round b will beat c, and hence will be declared the winner—hence BER does not violate monotonicity in this example.
Example 9
The following example (adapted from Felsenthal and Tideman 2014: 64) shows violation of monotonicity by both BER and NER in variable electorate when no Condorcet Winner exists, n > 3 and the initial winners under both rules are different.
Suppose there are 16 voters whose preference orderings among six candidates, a–f are as follows:
Voter no. | Preference ordering |
|---|---|
1 | a > b > d > c > e > f |
2 | a > d > c > f > b > e |
3 | b > d > c > e > a > f |
4 | b > d > e > c > a > f |
5 | b > e > a > f > d > c |
6 | c > b > e > a > f > d |
7 | c > d > a > f > b > e |
8 | d > a > c > b > e > f |
9 | d > c > b > e > a > f |
10 | e > a > b > d > c > f |
11 | e > f > b > d > c > a |
12 | f > a > c > e > b > d |
13 | f > c > a > b > d > e |
14 | f > e > a > d > c > b |
15 | f > e > b > c > a > d |
16 | f > e > d > c > a > b |
This preference list can be transformed into the following paired comparison matrix. Each number in the body of the matrix, below, is the number of voters who rank the candidate listed at the left of the row ahead of the candidate listed at the top of the column. Thus, for example, the number 7 in row d and column a signifies that seven voters rank d ahead of a in their ballots, and therefore the complementary number, 9, appears in row a and column d. The sum of each row is the Borda score of the candidate listed at the left of the row.Footnote 19
a | b | c | d | e | f | Sum | |
|---|---|---|---|---|---|---|---|
a | – | 9 | 7 | 9 | 6 | 10 | 41 |
b | 7 | – | 7 | 10 | 10 | 8 | 42 |
c | 9 | 9 | – | 5 | 9 | 9 | 41 |
d | 7 | 6 | 11 | – | 8 | 8 | 40 |
e | 10 | 6 | 7 | 8 | – | 9 | 40 |
f | 6 | 8 | 7 | 8 | 7 | – | 36 |
240 |
As can be seen from the matrix above, the Borda scores of candidates a, b, c, d, e, f are 41, 42, 41, 40, 40, and 36, respectively, with an average of 40 (240/6). So according to NER candidates d, e, f are eliminated at the end of the first counting round and the revised Borda scores of candidates a, b, c are 16, 14, and 18, respectively, with an average of 16. So according to NER both a and b are now eliminated and c becomes the NER winner.
According to BER candidate f is eliminated at the end of the first counting round and the revised Borda scores of a, b, c, d, e are 31, 34, 32, 32, and 31, respectively; so a and e are eliminated at the end of the second counting round and the revised Borda scores of b, c, d, are 17, 14, and 17, respectively; so c is eliminated at the end of the third counting round and thereafter b beats d (10:6) and becomes the BER winner.
Now suppose that, ceteris paribus, an additional voter joins the electorate whose preference ordering is c > d > a > b > e > f thus increasing c’s support. As a result the revised Borda scores of a, b, c, d, e, f are now 44, 44, 46, 44, 41, and 36, respectively, with an average of 42.5. So according to NER candidates e, f are eliminated at the end of the first counting round and the revised Borda scores of a, b, c, d are 26, 24, 26, and 26, respectively with an average of 25.5. So b is eliminated at the end of the second counting round and the revised Borda scores of a, c, d are 16, 16, and 19, respectively, so d is declared the NER winner—thus demonstrating monotonicity failure.
Suppose instead that, ceteris paribus, an additional voter joins the electorate whose preference ordering is b > f > a > c > e > d—thus increasing b’s support. As a result, the revised Borda scores of a, b, c, d, e, f are now 44, 47, 43, 40, 41, and 40, respectively. So according to BER f is eliminated at the end of the first counting round and the revised Borda scores of a, b, c, d, e are 34, 38, 34, 32, and 32, respectively. So according to BER both d and e are eliminated at the end of the second counting round and the revised Borda scores of a, b, and c are 17, 16, and 18, respectively. So b is eliminated at the end of the third counting round, and thereafter c beats a (9:8) and becomes the BER winner—thus demonstrating monotonicity failure of BER.
Example 10
The following example (due to Felsenthal and Tideman 2014: 64) shows violation of monotonicity by both BER and NER in fixed electorate when no Condorcet Winner exists, n > 3 and the initial winners under both rules are different.
Suppose there are 16 voters whose preference orderings among six candidates, a–f are as follows:
Voter no. | Preference ordering |
|---|---|
1 | a > b > d > c > e > f |
2 | a > d > c > f > b > e |
3 | b > d > c > e > a > f |
4 | b > d > e > c > a > f |
5 | b > e > a > f > d > c |
6 | c > b > e > a > d > f |
7 | c > d > a > f > b > e |
8 | d > a > c > b > e > f |
9 | d > c > b > e > a > f |
10 | e > a > b > d > c > f |
11 | e > f > b > d > c > a |
12 | f > a > c > e > b > d |
13 | f > c > a > b > d > e |
14 | f > e > a > d > c > b |
15 | f > e > b > c > a > d |
16 | f > e > d > c > a > b |
This preference list can be transformed into the following paired comparison matrix.
a | b | c | d | e | f | Sum | |
|---|---|---|---|---|---|---|---|
a | – | 9 | 7 | 9 | 6 | 10 | 41 |
b | 7 | – | 7 | 10 | 10 | 8 | 42 |
c | 9 | 9 | – | 5 | 9 | 9 | 41 |
d | 7 | 6 | 11 | – | 8 | 9 | 41 |
e | 10 | 6 | 7 | 8 | – | 9 | 40 |
f | 6 | 8 | 7 | 7 | 7 | – | 35 |
240 |
As can be seen from this matrix, the Borda scores of candidates a, b, c, d, e, f are 41, 42, 41, 41, 40, and 35, respectively. Since the average Borda score is 40 (240/6), candidates e and f are eliminated according to NER. The revised Borda scores of candidates a, b, c, and d are 25, 24, 23, and 24, respectively, with an average of 24. So according to NER all candidates except a are eliminated, and a is declared the NER winner.
According to BER candidate f is eliminated at the end of the first counting round, and the revised Borda scores of a, b, c, d, and e are 31, 34, 32, 32, and 31, respectively. So now candidates a and e are eliminated and the revised Borda scores of b, c and d are 17, 14, and 17, respectively. So now c is eliminated and then b beats d (10:6) and becomes the BER winner.
Now suppose that, ceteris paribus, voter #8 changes his/her ranking from d > a > c > b > e > f to a > d > c > b > e > f—thereby increasing a’s support. As a result the Borda scores of a, b, c, d, e, and f change to 42, 42, 41, 40, 40, and 35, respectively. As the average score remains 40, candidates d, e, f are eliminated at the end of the first counting round according to NER and the revised Borda scores of a, b, c are 16, 14, and 18, respectively, with an average of 16. So both a and b are eliminated and c is declared the NER winner—in violation of monotonicity!
Suppose, alternatively, that voter #14 changes his/her preference ordering from f > e > a > d > c > b to f > e > a > b > d > c and that voter #15 changes his/her preference ordering from f > e > b > c > a > d to f > b > e > c > a > d—thereby increasing b’s support. As a result, the initial Borda scores are now 41, 45, 40, 40, 39, and 35, for a, b, c, d, e, and f, respectively. So according to BER candidate f is removed after the first counting round and the revised Borda scores of a, b, c, d, and e, are 31, 37, 31, 31, and 30, respectively. So candidate e is removed at the end of the second counting round and the revised scores of a, b, c, and d are 25, 26, 22, and 23, respectively. So candidate c is removed at the end of the third counting round and the revised Borda scores of a, b, and d are 18, 18, and 12, respectively. So candidate d is removed at the end of the third counting round, and then candidate a beats b (9:7) and is declared the BER winner—in violation of monotonicity!
Examples proving Proposition 3 : BER vs. NER in n-candidate elections (n > 3) when the initial winner is the same
In the following six examples we show that in n-candidate elections (n > 3)—with either fixed or variable electorates—it is possible that, ceteris paribus, when the initial winners under NER and BER are the same then one of these rules may violate monotonicity while the other does not, as well as cases where both rules violate monotonicity.
Example 11
The following example of 4-candidate election under fixed electorate shows that, ceteris paribus, monotonicity is violated by BER but not by NER when the initial winner under these two rules is the same (and is not a Condorcet Winner).
Suppose there are 27 voters whose preference orderings among four candidates, a, b, c, d, are as follows:
No. of voters | Preference orderings |
|---|---|
6 | a > d > b > c |
3 | a > d > c > b |
4 | b > a > d > c |
6 | b > c > d > a |
6 | c > a > b > d |
2 | c > b > a > d |
The Borda scores of a, b, c, and d in the first counting round are 49, 46, 39, and 28, respectively. As the average Borda score here is 40.5, c and d are eliminated at the end of the first counting round according to NER and in the contest between a and b in the second counting round a beats b (15:12) and is declared the NER winner. After the elimination of d at the end of the first counting round according to BER, the revised Borda scores of a, b, and c are 28, 28, and 25, respectively, so c is eliminated according to BER at the end of the second counting round, and in the contest between a and b in the third counting round a beats b (15:12) and is also declared as the BER winner.
Now suppose that, ceteris paribus, in order to increase a’s support three of the four voters with preference ordering b > a > d > c change their preference ordering to a > b > d > c and the two voters with preference ordering c > b > a > d change their preference ordering to c > a > b > d.
As a result we get the following distribution of voters and preference orderings:
No. of voters | Preference orderings |
|---|---|
3 | a > b > d > c |
6 | a > d > b > c |
3 | a > d > c > b |
1 | b > a > d > c |
6 | b > c > d > a |
8 | c > a > b > d |
The revised Borda scores of a, b, c, and d in the first counting round are 54, 41, 39, and 28, respectively. As the average Borda score here is also 40.5, c and d are eliminated at the end of the first counting round according to NER, and in the contest between a and b in the second counting round a beats b (20:7) and is declared the NER winner as in the original situation (without violation of monotonicity). However, according to BER after the elimination of d at the end of the first counting round the revised Borda scores of a, b, and c are 33, 23, and 25, respectively, so b is eliminated at the end of the second counting round, and in the contest between a and c in the third counting round c beats a (14:13) and is declared as the BER winner—thereby violating monotonicity.
Example 12
The following example of 6-candidate election under fixed electorate (adapted from Felsenthal and Tideman 2014: 64) shows that, ceteris paribus, monotonicity is violated by NER but not by BER when the initial winner under these two rules is the same (and is not a Condorcet Winner).
Suppose there are 16 voters having the following preference ordering among six candidates a–f:
Voter No. | Preference ordering |
|---|---|
1 | a > b > d > c > e > f |
2 | a > d > c > f > b > e |
3 | b > d > c > a > e > f |
4 | b > d > e > c > a > f |
5 | b > e > a > f > d > c |
6 | c > b > e > a > d > f |
7 | c > d > a > f > b > e |
8 | d > a > c > b > e > f |
9 | d > c > b > e > a > f |
10 | e > a > b > d > c > f |
11 | e > f > b > d > c > a |
12 | f > a > c > e > b > d |
13 | f > c > a > b > d > e |
14 | f > e > a > d > c > b |
15 | f > e > b > c > a > d |
16 | f > e > d > c > a > b |
This preference list can be transformed into the following pairwise comparison matrix:
a | b | c | d | e | f | Sum | |
|---|---|---|---|---|---|---|---|
a | – | 9 | 7 | 9 | 7 | 10 | 42 |
b | 7 | – | 7 | 10 | 10 | 8 | 42 |
c | 9 | 9 | – | 5 | 9 | 9 | 41 |
d | 7 | 6 | 11 | – | 8 | 9 | 41 |
e | 9 | 6 | 7 | 8 | – | 9 | 39 |
f | 6 | 8 | 7 | 7 | 7 | – | 35 |
240 |
As can be seen from this matrix, the initial Borda scores of a, b, c, d, e, and f, are 42, 42, 41, 41, 39, and 35, respectively, with an average of 40. So according to NER candidates e and f are eliminated in the first counting round and the revised Borda scores of a, b, c, and d, are 25, 24, 23, and 24, respectively. As the average Borda score of these four candidates is 24 (96/4), all candidates except a are eliminated thereby a becomes the NER winner.
According to BER candidate f is eliminated in the first counting round and the revised Borda scores of a, b, c, d, and e, are 32, 34, 32, 32, and 30, respectively. So at the end of the second counting round candidate e is eliminated and the revised Borda scores of candidates a, b, c, and d, are 25, 24, 23, and 24, respectively. So at the end of the third counting round candidate c is eliminated and the revised Borda scores of candidates a, b, and d, are 18, 17, and 13, respectively. So now d is eliminated and thereafter a beats b (9:7) and thus becomes also the BER winner.
Now suppose that, ceteris paribus, voter #7 changes his/her preference ordering from c > d > a > f > b > e to c > a > d > f > b > e—thus increasing a’s support. As a result the initial Borda scores of a, b, c, d, e, f are 43, 42, 41, 40, 39, 35, respectively. Since the average score remains the same (40), candidates d, e, f are eliminated at the end of the first counting round according to NER and the revised Borda scores of a, b, c are now 16, 14, 18, respectively—with an average of 16. So now both a and b are eliminated and c is declared the NER winner—thus demonstrating monotonicity failure.
However, ceteris paribus, according to BER candidate f is eliminated after the first counting round and the revised Borda scores of a, b, c, d, e are 33, 34, 32, 31, 30, respectively. So candidate e is eliminated at the end of the second counting round and the revised Borda scores of a, b, c, d, are 26, 24, 23, 23, respectively. So at the end of the third counting round both c and d are eliminated and in the fourth counting round a beats b (9:7) thus remaining the BER winner without violating monotonicity.
Example 13
The following example of 6-candidate election under variable electorate shows that, ceteris paribus, monotonicity is violated by NER but not by BER when the initial winner under these two rules is the same (and is not a Condorcet Winner).
Suppose that the original situation is the same as the original situation in Example 11, i.e., under both NER and BER candidate a is elected. Now suppose that, ceteris paribus, an additional voter whose preference ordering is a > b > c > e > d > f joins the electorate—thus increasing a’s support.
As a result the preference orderings of the 17 voters can now be depicted by the following pairwise comparison matrix:
a | b | c | d | e | f | Sum | |
|---|---|---|---|---|---|---|---|
a | – | 10 | 8 | 10 | 8 | 11 | 47 |
b | 7 | – | 8 | 11 | 11 | 9 | 46 |
c | 9 | 9 | – | 6 | 10 | 10 | 44 |
d | 7 | 6 | 11 | – | 8 | 10 | 42 |
e | 9 | 6 | 7 | 9 | – | 10 | 41 |
f | 6 | 8 | 7 | 7 | 7 | – | 35 |
255 |
As the average Borda score is 42.5 (255:6), candidates d, e, f are eliminated according to NER at the end of the first counting round. The revised Borda scores of candidates a, b, c are 18, 15, and 18, respectively, so at the end of the second counting round candidate b is eliminated and thereafter candidate c beats candidate a (9:8) and is elected—thus demonstrating monotonicity failure according to NER.
However, ceteris paribus, according to BER candidate f is eliminated after the first counting round and the revised Borda scores of a, b, c, d, e are 36, 37, 34, 32, 31, respectively. So candidate e is eliminated at the end of the second counting round and the revised Borda scores of a, b, c, d, are 28, 26, 24, 24, respectively. So at the end of the third counting round both c and d are eliminated and in the fourth counting round a beats b (10:7) thus remaining the BER winner without violating monotonicity.
Example 13A
We have already shown (in Examples 3–4) that BER is vulnerable to downward monotonicity failures. So to provide the reader with just one example demonstrating that NER too is vulnerable to downward monotonicity failure in n-candidate elections (n > 3), the following example of 4-candidate election under variable electorate shows the same phenomenon shown by Example 13 with one difference: here NER (but not BER) demonstrates downward monotonicity failure when the initial winner under these two rules is the same (and is not a Condorcet Winner).
Suppose there are 11 voters whose preference orderings among four candidates, a,b,c,d, are as follows:
No. of voters | Preference ordering |
|---|---|
5 | b > c > d > a |
4 | c > d > a > b |
2 | a > d > b > c |
The initial Borda scores of candidates a ,b, c, d are 10, 17, 22, and 17, respectively, and the average score is therefore 16.5. Consequently candidate a is eliminated according to both BER and NER, and thereafter the revised Borda scores of candidates b, c, and d, are 12, 13, and 8, respectively, with an average score of 11. So according to NER and BER candidate d is eliminated, and in the contest between b and c in the third counting round b beats c (7:4) and is therefore declared the winner according to both NER and BER.
As candidate c has not been elected, suppose now that, ceteris paribus, one additional voter with preference ordering d > a > b > c joins the electorate thereby decreasing c’s support further. As a result the Borda scores of candidates a, b, c, d are 12, 18, 22, and 20, respectively, and the average score is therefore 18. Consequently both candidates a and b are eliminated according to NER in the first counting round, and in the contest between c and d in the second counting round c beats d (9:3) and is declared the winner according to NER—in violation of monotonicity. However, ceteris paribus, according to BER only a is eliminated in the first counting round. The revised Borda scores of b,c, and d are 13, 13, and 10, respectively, so d is now eliminated according to BER, and in the contest between b and c in the third counting round b beats c (8:4) and is declared the winner (as in the original situation)—without violation of monotonicity.
Example 14
The following example of 4-candidate election under variable electorate shows that, ceteris paribus, monotonicity is violated by BER but not by NER when the initial winners under these two rules are the same (and there is no Condorcet Winner).
Suppose there are 13 voters whose preference orderings among four candidates, a, b, c, d, are as follows:
No. of voters | Preference ordering |
|---|---|
5 | a > b > c > d |
4 | b > c > d > a |
4 | c > a > d > b |
The Borda scores of candidates a, b, c, d, are 23, 22, 25, and 8, respectively, and the average score is therefore 19.5. Consequently candidate d is eliminated according to both BER and NER, and thereafter the revised Borda scores of candidates a, b, and c, are 14, 13, and 12, respectively, with an average score of 13. So according to NER both candidates b and c are eliminated and hence candidate a is elected, whereas according to BER candidate c is eliminated, and in the contest between a and b in the third counting round a beats b (9:4) and is therefore declared the winner according to BER too.
Suppose now that, ceteris paribus, two voters with preference ordering a > c > b > d join the electorate, thereby increasing a’s support. As a result the Borda scores of a, b, c, and d are now 29, 24, 29, and 8, respectively, with an average of 22.5. So at the end of the first counting round d is eliminated according to both NER and BER. After the elimination of d the revised Borda scores of a, b, and c are 18, 13 and 14 with an average of 15. So according to NER both b and c are eliminated and hence a remains the winner according to NER without violating monotonicity. However, according to BER candidate b is eliminated after the elimination of d, and in the contest between a and c in the third counting round c beats a (8:7) and becomes the BER winner—in violation of monotonicity.
Example 15
The following example of 4-candidate election under both fixed and variable electorates shows that, ceteris paribus, monotonicity is violated by both BER and by NER when no Condorcet Winner exists and the initial winners under these two rules are the same.
Suppose there are 36 voters whose preference ordering among 4 candidates, a, b, c, d, are as follows:
No. of voters | Preference ordering |
|---|---|
1 | a > b > c > d |
1 | a > b > d > c |
2 | a > c > b > d |
2 | a > c > d > b |
1 | a > d > b > c |
2 | a > d > c > b |
2 | b > a > c > d |
2 | b > a > d > c |
1 | b > c > a > d |
1 | b > c > d > a |
2 | b > d > a > c |
1 | b > d > c > a |
2 | c > a > b > d |
1 | c > a > d > b |
3 | c > b > a > d |
2 | c > b > d > a |
1 | c > d > a > b |
2 | c > d > b > a |
1 | d > a > b > c |
1 | d > a > c > b |
2 | d > b > a > c |
1 | d > c > a > b |
2 | d > c > b > a |
This preference list can be transformed into the following pairwise comparison matrix:
a | b | c | d | Sum | |
|---|---|---|---|---|---|
a | – | 16 | 19 | 20 | 55 |
b | 20 | – | 15 | 20 | 55 |
c | 17 | 21 | – | 20 | 58 |
d | 16 | 16 | 16 | – | 48 |
216 |
As can be seen from this matrix, the initial Borda scores of a, b, c, d, are 55, 55, 58, and 48, respectively, with an average of 54. So according to NER d is eliminated at the end of the first counting round. The revised Borda scores of a, b, c, are 35, 35, and 38, respectively, with an average of 36. So according to NER both a and b are eliminated and c is declared the winner.
According to BER d is eliminated at the end of the first counting round, both a and b are eliminated at the end of the second counting round, so c is the BER winner too.
Now suppose that, ceteris paribus, the first voter changes his/her preference ordering from a > b > c > d to a > c > b > d—thereby increasing c’s support. As a result the Borda scores of a, b, c, d, are now 55, 54, 59, and 48, respectively, with an average of 54. So according to NER both b and d are eliminated at the end of the first counting round, and in the second counting round a beats c (19:17) and becomes the NER winner—thus demonstrating monotonicity failure.
According to BER d is eliminated at the end of the first counting round, b is eliminated at the end of the second counting round, and a beats c (19:17) in the third counting round and becomes the BER winner too—thus also demonstrating monotonicity failure.
The same example can serve also to demonstrate monotonicity failure by both NER and BER under variable electorate. In this case suppose that instead of the first voter changing his/her preference ordering (thus increasing c’s support), that, ceteris paribus, two additional voters join the electorate whose preference orderings are c > a > b > d. In this case the Borda scores of a, b, c, d are 59, 57, 64, and 48, respectively, with an average of 57. So according to NER both b and d are eliminated at the end of the first counting round, and in the second counting round a and c are tied (19:19). So if ties are broken either randomly or lexicographically (i.e., in favor of the candidate who is denoted by a letter nearer to the beginning of the alphabet), a may become the NER winner—thus demonstrating monotonicity failure.
According to BER d is eliminated at the end of the first counting round, b is eliminated at the end of the second counting round, and a beats c (19:17) in the third counting round and becomes the BER winner too—thus also demonstrating monotonicity failure.
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Felsenthal, D.S., Nurmi, H. Monotonicity Violations by Borda’s Elimination and Nanson’s Rules: A Comparison. Group Decis Negot 27, 637–664 (2018). https://doi.org/10.1007/s10726-018-9580-z
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DOI: https://doi.org/10.1007/s10726-018-9580-z