Geometry of run-off elections

We present a geometric representation of the method of run-off voting. With this representation we can observe the non-monotonicity of the method and its susceptibility to the no-show paradox. The geometry allows us easily to identify a novel compromise rule between run-off voting and plurality voting that is monotonic.

1. We can check this by examining profiles \(\pi\) and \(\pi ^{\prime }\). For some other pair of profiles that generate the same plurality points that \(\pi\) and \(\pi ^{\prime }\) generate, the outcome of a run-off region could very well change.

2. An example of monotonicity failure requires a pair of profiles. By moving back and forth between the two profiles in the pair we will see that one is vulnerable to DMF and the other vulnerable to UMF. However, Leppelley et al. and Miller show that no single-peaked profile can be vulnerable to DMF. Hence, there can be no pair of single-peaked profiles that serve as an example of monotonicity failure. Run-off voting is also monotonic on a single-peaked domain when there are four candidates, though not when there are five or more.

3. In 1969, in the United States, a proposed constitutional amendment to abolish the Electoral College and replace it with a direct popular election of the president and vice-president, using this 40% rule, passed with strong bipartisan support in the House of Representatives, but was not approved by Senate.

4. Note that the method of run-off voting is complicated by the fact that ties may occur in the first round of voting as well as in the run-off.

I am extremely grateful to Nick Miller, Ashley Piggins, Bill Zwicker and two anonymous reviewers for many helpful comments and suggestions.

Authors and Affiliations

1. J.E. Cairnes School of Business and Economics and the Whitaker Institute, National University of Ireland Galway, University Road, Galway, Ireland

   Conal Duddy

Corresponding author

Correspondence to Conal Duddy.

Appendix

Proof of Proposition 1

To make matters more precise and to allow for the possibility of random tie-breaking, an outcome is a probability distribution over the candidates, indicating each candidate’s probability of winning.^{Footnote 4} If no random tie-breaking is required, then a probability of one is assigned to one candidate. A voting rule associates an outcome to every profile. Given a voting rule f and a profile \(\pi\), we write \(f(\pi )(A)\) for the probability of candidate A being elected. Monotonicity requires then that \(\pi (A^{\uparrow })\pi ^{\prime }\) implies \(f(\pi )(A)\le f(\pi ^{\prime })(A)\).

Take any candidate A and profiles \(\pi\) and \(\pi ^{\prime }\) with \(\pi (A^{\uparrow })\pi ^{\prime }\). Let f be the monotonic run-off rule. We must prove that \(f(\pi )(A)\le f(\pi ^{\prime })(A)\). If \(f(\pi )(A)=0\), then it follows immediately that \(f(\pi )(A)\le f(\pi ^{\prime })(A)\). So let us assume that \(f(\pi )(A)>0\).

Let us assume, by way of contradiction, that \(f(\pi )(A)>f(\pi ^{\prime })(A)\). Profile \(\pi ^{\prime }\) is obtained from \(\pi\) by promoting A on a ballot. We can see that this must be a promotion from second to first place on that ballot. Otherwise, no change in the tally of first-round votes materializes. That would in turn imply a contradiction since the promotion of A on a ballot cannot cause the outcome of a pairwise comparison with another candidate to become less favorable to A. So let us assume that A is promoted from second to first place on a ballot to obtain \(\pi ^{\prime }\) from \(\pi\).

Since \(f(\pi )(A)>0\), it must be that, at profile \(\pi\), (1) A is elected without a run-off, (2) A is in the top two in the first-round tally of votes, with both A and another candidate being above the midpoint between first and third place, or (3) A is one of three of more candidates in joint first place in the first-round tally.

Case (1) :

If A is elected without a run-off at \(\pi\), then A uniquely has the most votes at \(\pi\). It follows that an additional vote for A in the first round, at the expense of another candidate, cannot cause the midpoint between first and third place to fall. Since none of the other candidates were above that threshold at \(\pi\), and since they have not gained any votes, it follows that no other candidate is above the threshold at \(\pi ^{\prime }\). Hence, A is elected without a run-off at \(\pi ^{\prime }\). So we can rule out this case.

Case (2) :

A is in the top two in the first round and above the midpoint between first and third place. The promotion of A to the top of a ballot means that A must still be in the top two and must still be above the midpoint between first and third place. The other candidate who was in the top two at \(\pi\) must either still be in the top two, or in joint second place with one or more other candidates. Therefore, we either have the same run-off at \(\pi ^{\prime }\) as we had at \(\pi\), or else A is elected at \(\pi ^{\prime }\) without a run-off. So we can rule out this case also.

Case (3) :

If A is one of three or more candidates in joint first place in the first round at \(\pi\) then A must be uniquely first in the first round at \(\pi ^{\prime }\). It must also be true that the candidate in second place at \(\pi ^{\prime }\) is either equidistant between first and third place, or is closer to third than to first. In either case, A is elected without a run-off. Thus, we can rule out this final case. We conclude that the monotonic run-off rule is monotonic. \(\square\)

Reprints and permissions