The (In)Vulnerability of Ranked Voting Procedures that Are Not Condorcet–Consistent to Various Paradoxes

Abstract

The (in)vulnerability of six ranked voting procedures which are not Condorcet–consistent (Borda count, Alternative vote, Coombs’ procedure, Bucklin’s procedure, Range Voting and Majority Judgment) to 13 paradoxes is examined in this chapter. For those systems that are vulnerable to some voting paradoxes the vulnerability is demonstrated through illustrative examples showing that there are profiles where the paradoxes in question happen when the respective procedures are in use. And for those systems that are invulnerable to some voting paradoxes the invulnerability is explained.

This chapter is partly based on Felsenthal (2012) which contains several examples devised by the second–named author.

All ranked procedures that are not Condorcet–consistent violate Smith’s (1973) Condorcet Principle (cf. Sect. 2.1.3 in Chap. 2). As is demonstrated in Example 6.5.11.1 (in Chap. 6) some of the Condorcet–consistent procedures too may violate this Principle.

All the procedures surveyed in this chapter are invulnerable to the Dependence on Order of Voting (DOV) Paradox (cf. Sect. 2.2.8 in Chap. 2) because under these procedures all candidates are voted upon simultaneously rather than sequentially.

Appendices

Exercises

Problem 5.1

A ranking R of v voters together with its reversal R′ of v voters is called a reversal component by Saari. Show that adding such a component to any 3–alternative profile leaves the Borda ranking of the alternatives unchanged. Show that adding such a component to a 3–alternative profile, may change the Plurality Voting winner. Illustrate using the following profile.

No. of voters	Preference ordering
4	a $\succ$ b $\succ$ c
3	b $\succ$ c $\succ$ a
2	c $\succ$ b $\succ$ a

Problem 5.2

Consider the following procedure: given the profile of voter preference rankings, give each alternative ranked first by a voter 1 point, each alternative ranked second 2 points etc. Now, sum up the points given to each alternative and form the collective ranking on the basis of the sum of scores of the alternatives so that those with larger scores are ranked higher than those with lower scores. How does this method relate to the Borda count? How could this method be modified to end up with a system that always ends up with the same results as the Borda count?

Problem 5.3

Determine the winner according to the Majority Judgment (MJ) procedure given that the following five voters, 1–5, assign ordinal grades on a scale from A (lowest) to F (highest) to three alternatives, x, y, z.

Alternative\voter	1	2	3	4	5
X	C	C	C	F	F
Y	D	A	A	D	D
Z	B	F	B	B	E

Problem 5.4

Assign to grades A–F in Problem 5.3 numerical values (or scores) from 1 to 6, respectively. Then compute the sum of scores for each alternative and determine which alternative is the winner if the winner is the alternative which has the highest sum. How does the result differ from the MJ outcome?

Answers to Exercises

Problem 5.1

Let the ranking R be a $\succ$ b $\succ$ c and let the R′ ranking be c $\succ$ b $\succ$ a. Then a’s total Borda score is added by 2v points, b’s by v + v = 2v points, and c’s by 2v points. Hence the differences between the Borda scores remain the same as before adding the reversal component.

Adding such a component to a 3–alternative profile may change the Plurality Voting winner if before the addition of the reversal component there was a tie between a and b, it is broken in a’s favor when the reversal component is added. Thus, if we add to the (original) profile 4 voters with c $\succ$ a $\succ$ b ranking and 4 voters with its reversal, then the resulting profile gets the following form:

No. of voters	Preference orderings
4	a $\succ$ b $\succ$ c
3	b $\succ$ c $\succ$ a
2	c $\succ$ b $\succ$ a
4	c $\succ$ a $\succ$ b
4	b $\succ$ a $\succ$ c

Thus b is the Plurality Voting winner.

Problem 5.2

This system of assigning points to alternatives results in a social preference ordering which is the reverse of that obtained under Borda’s procedure. To obtain the same social preference ordering as that obtained under Borda’s procedure one must reverse the order of preference in the social preference ordering, i.e., the alternative with the smallest score should be ranked first and that with the largest score should be ranked last.

Problem 5.3

The median grades of alternatives x, y, z are C, D, and B, respectively, so alternative y wins.

Problem 5.4

If one replaces the grades A–F by the numbers 1–6, respectively, then the sum of x’s grades is 21, that of y is 14, and that of z is 17, so x wins.