Abstract
20 voting procedures for electing a single candidate are introduced and briefly commented upon. The procedures fall into three classes in terms of the type of voter input and Condorcet consistency: non-ranked procedures, ranked procedures that are not Condorcet-consistent and ranked ones that are Condorcet-consistent. The first class consists of four procedures, the second consists of seven procedures and the third class consists of nine procedures.
This chapter is largely based on Felsenthal (2012, pp. 24–31) and on Felsenthal and Nurmi (2018, Chap. 3).
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Notes
- 1.
However, it is unclear how a tie between two candidates, say a and b, ought to be broken under Bucklin’s procedure when both a and b are supported in the same counting round by the same number of voters and this number constitutes a majority of the voters. If one tries to break the tie between a and b in such an eventuality by performing the next counting round in which all other candidates are also allowed to participate, then it is possible that the number of (cumulated) votes of another candidate, c, will exceed that of a and b.
To see this, consider the following simple example. Suppose there are 18 voters who must elect one candidate under Bucklin’s procedure and whose preference orderings among four candidates, a, b, c, d are as follows: seven voters with preference ordering a > b > c > d, eight voters with preference ordering b > a > c > d, one voter with preference ordering d > c > a > b, and two voters with preference ordering d > c > b > a. None of the candidates constitutes the top preference of a majority of the voters. However, both a and b constitute the top + second preference by a majority of voters (15). If one tries to break the tie between a and b by performing the next (third) counting round in which c and d are also allowed to participate, then c will be elected (with 18 votes), but if only a and b are allowed to participate in this counting round then b will be elected (with 17 votes).
So which candidate ought to be elected in this example under Bucklin’s procedure? As far as we know, Bucklin did not supply an answer to this question.
- 2.
This is the preference ordering of the majority of the voters with respect to any pair of alternatives.
- 3.
Young (1977, p. 349) prefers to call this procedure ‘Minimax function’.
- 4.
Although Nanson’s procedure satisfies the strong Condorcet condition, i.e., it always elects a candidate who beats every other candidate in paired comparisons, this procedure may not satisfy the weak Condorcet condition which requires that if there exist(s) candidate(s) who is (are) unbeaten by any other candidate then this (these) candidate(s)—and only this (these) candidate(s)—ought to be elected. For an example of violation of the weak Condorcet condition by Nanson’s procedure see Niou (1987). Niou shows that when the set of Nanson winners consists of two candidates, one of them may not satisfy the weak Condorcet condition, while the other Nanson winner does. The following profile (where the symbol > means “is preferred to”) shows that the Nanson winner may be distinct from those candidates that satisfy the weak Condorcet condition (Nurmi, 1989, p. 202): one voter: a > b > c > d > e, one voter: a > d > b > c > e, one voter: a > d > e > b > c, one voter: b > c > e > d > a, two voters: c > e > d > b > a. Here the Nanson winner is c, but the only candidate satisfying the weak Condorcet condition is a.
- 5.
Tideman (2006, pp. 187–189) proposes two heuristic procedures that simplify the need to examine all n! preference orderings.
- 6.
According to Kemeny (1959) the distance between two preference orderings, R and R′, is the number of pairs of candidates (alternatives) on which they differ. For example, if R = a > b > c > d and R′ = d > a > b > c, then the distance between R and R′ is 3, because they agree on three pairs [(a > b), (a > c), (b > c)] but differ on the remaining three pairs, i.e., on the preference ordering between a and d, b and d, and between c and d. Similarly, if R″ is c > d > a > b then the distance between R and R″ is 4 and the distance between R′ and R″ is 3. According to Kemeny’s procedure the most likely social preference ordering is that R such that the sum of distances of the voters’ preference orderings from R is minimized. Because this R has the properties of the median central measure in statistics it is called the median preference ordering. The median preference ordering (but not the mean preference ordering which is that R which minimizes the sum of the squared differences between R and the voters’ preference orderings) will be identical to the possible social preference ordering W which maximizes the sum of voters that agree with all paired comparisons implied by W.
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Appendices
Exercises for Chapter 2
- Problem 2.1 :
-
Consider the following profile:
No. of voters | Preference ordering |
|---|---|
5 | a > c > b > d |
4 | b > d > a > c |
2 | c > d > a > b |
Determine the winners according to Bucklin’s, Minimax, Plurality with Runoff, and Copeland’s procedures.
- Problem 2.2 :
-
Consider the following profile:
No. of voters | Preference orderings |
|---|---|
10 | d > a > b > c |
7 | b > c > a > d |
7 | c > a > b > d |
4 | d > c > a > b |
1 | b > a > c > d |
Determine the winners according to Dodgson’s, Minimax, Nanson’s, and Kemeny’s procedures.
-
Problem 2.3
-
Construct a profile where the Plurality, Plurality with Runoff and Copeland winners differ from each other.
-
Problem 2.4
Show by way of an example that the Borda winner may differ from the winner ensuing from Copeland’s procedure.
-
Problem 2.5
An Absolute Winner is a candidate who is ranked first by more than 50% of the voters. Determine whether each of the following voting procedures always elects the Absolute Winner when one exists: Successive Elimination, Coombs’s procedure, Borda count, Plurality Voting, Alternative Vote, Bucklin’s procedure.
Answers to Exercises of Chapter 2
-
Problem 2.1
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The Bucklin Winner is c (on the second round of computing with 7 voters placing it either first or second);
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The Minimax Winner is a (with a maximum of 6 votes against it);
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The Plurality with Runoff Winner is a (in the runoff with b);
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The Copeland Winners are a and c (each defeats two other candidates).
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Problem 2.2
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The winner according to Dodgson’s procedure is d (needing just 3 preference inversions to become a Condorcet Winner);
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The Minimax winner is also d whose maximal loss (15) is smallest;
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The winner according to Nanson’s procedure is c (which beats a 18:11 in the second count);
-
The most likely (transitive) social preference orderings according to Kemeny are a > b > c > d and c > a > b > d (each supported by the largest number (95) of pairs fitting these social preference orderings), so here according to Kemeny there is a tie between a and c.
-
Problem 2.3
-
- 4 voters::
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x > z > y
- 3 voters::
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y > z > x
- 2 voters::
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z > y > x
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Plurality Voting elects x, the Plurality with Runoff elects y (after the runoff between x and y) and Copeland’s procedure elects z since it defeats x 5:4 and y 6:3, while y defeats only x and x defeats no candidate.
-
Problem 2.4
-
Consider, e.g., the following profile:
-
- 5 voters::
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x > y > z
- 4 voters::
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y > z > x
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Here the Borda scores of x, y and z are 10, 13 and 4, respectively. Copeland’s procedure results in x since it defeats both y and z, while y defeats z and z defeats no other candidate. Thus, y wins under the Borda count, while x wins when Copeland’s procedure is in use.
-
Problem 2.5
-
Of the procedures listed only one, viz., the Borda count, may not end up with an Absolute Winner when one exists. An example of this is shown in the answer to Problem 2.4.
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Felsenthal, D.S., Nurmi, H. (2019). 20 Voting Procedures Designed to Elect a Single Candidate. In: Voting Procedures Under a Restricted Domain. SpringerBriefs in Economics. Springer, Cham. https://doi.org/10.1007/978-3-030-12627-8_2
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