The paradox of grading systems

Because ties for highest median grade are common when relatively few grades can be assigned, Balinski and Laraki (2011) propose a method for breaking ties in a system they call “majority judgment.” Cumulative voting, in which voters can allocate a fixed number of votes to one or more candidates, effectively allows voters to grade candidates, though it is not usually thought of as a grading system. It is used to elect multiple winners, unlike the systems analyzed here, and affords parties or factions the opportunity to elect candidates in proportion to their share in the electorate. For analyses of its properties and experience with it, see Brams (2003 [1975]) and Bowler et al. (2003).

Merrill and Nagel (1987) distinguish a balloting method, such as one that allows two grades, from a method for aggregating the ballots (what they call “decision rules”), such as one that elects the candidate with more superior grades in pairwise comparisons—or, equivalently, in the case of AV the highest average grade. As we will show, using a different aggregation method can produce a winner different from the AG-SG winner, even when only two grades are possible (the Hare system is an example). However, systems that choose as winners the candidates with the highest median grade or the highest Borda score duplicate AV in always selecting AG–SG winners for the case of two grades.

For any g, the method breaks ties as follows: Order the grades for each candidate from low to high; for an even number of grades, treat the grade just below the middle, rather than the average of the two middle grades, as the median; delete each candidate’s (tied) median grade; obtain the median for each candidate from the new (smaller) sets of grades; and if a tie still exists, continue to delete grades successively, one at a time, in the same way as before, until the tie is broken. As an example for g = 2, suppose the ordered grades from six voters are 0 0 1 1 1 1 for candidate X and 0 1 1 1 1 1 for Y, with the remaining candidates receiving more 0s than 1s (the medians, as just defined, are underscored). Deleting both underscored medians yields 0 0 1 1 1 for X and 0 1 1 1 1 for Y. The new medians are again tied, so they are deleted, giving 0 0 1 1 for X and 0 1 1 1 for Y. The tie is now broken in favor of Y as the median-grade winner, who indeed did receive more 1s.

Taft, also, would have benefited from the approval of both his supporters and Roosevelt supporters, but probably not to the extent of Roosevelt for two reasons: Roosevelt (i) edged him out in the plurality vote and (ii) probably would have drawn more support from Wilson voters, and even supporters of Eugene Debs, the Socialist candidate who received 6 % of the vote, because he generally was perceived to be less conservative than Taft.

If a voter’s ranking is not strict, the usual convention is to give all tied candidates the mean of the ranks as if the ranking were strict. To illustrate in Example 2, with ranks of 2, 1, and 0 for the three candidates, the Borda scores of the first voter would be (2, ½, ½), of the second voter (0, 1½, 1½), and of the third voter (0, 1½, 1½). In this example, B and C would be the tied winners, and no paradox of grading systems would occur, because these candidates are preferred by two of the three voters. In general, when translating ranks into grades or grades into ranks, ranking systems and grading systems may yield different outcomes.

Such a dichotomization is effectively what AV forces, and not just for two candidates. In the presence of two candidates only, plurality voting suffices to enable a voter to vote for just his or her preferred candidate.

This was also true in Example 2, but “all except one” meant only 2 of the 3 voters rather than, as in Example 4, 4 of the 5 voters.

We are grateful to Sean J. Vasquez for writing the computer program to do the simulation.

The sampling is carried out with replacement, so in general it may include duplicate combinations, but that is unlikely for this example.

For n Bernoulli trials each with success probability p (the binomial distribution), the standard error of the sample proportion of successes is \( \sqrt {p(1 - p)/n} \), which is maximized at p = ½.

We did not calculate the probability of a strong paradox. It seemed to us that the possibility, not the certainty, of different AG and SG winners was the first question to address in inquiring whether a discrepancy posed a serious problem.

Regenwetter et al. (2006) make a similar point about the probability of the Condorcet paradox, showing that the paradox turns up much less often than its theoretical probability, based on the impartial-culture assumption.

The first column of Table 3 when g = 3 repeats probabilities from Table  2 .

Some uses of, and experience with, AV in elections are discussed in Brams and Fishburn (2005), Laslier and Sanver (2010), Felsenthal and Machover (2012), Baujard et al. (2014), and references therein. Brams (2008), Brams and Sanver (2009), Sanver (2010), and Camps et al. (2014) suggest ways in which ranking and grading systems can be combined or otherwise reconciled.