Geometry of run-off elections

Abstract

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.

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.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. $\square$