# [EM] Deterministic tie-breaking in a simple Condorcet election.

Richard Lung voting at ukscientists.com
Mon Dec 20 11:59:08 PST 2021

```The impression I get is a preoccupation with winner-solving methods, under the implicit premis that there are such things as the rightly elected persons.
The preoccupation seems to stem from challenging the "Impossibility" of determinate election solutions. But there may not be definitive answers. There may not be a definitive election result, in truth. (Only probabilities.)

The term Hare RCV is a corruption of what Thomas Hare stood for: at-large stv/pr is the Hare system, without equivocation.
However candidates are eliminated, it is literally unscientific to do so, because it is eliminating preferential information from the count.
The election should not be in till all the preferential information is in. This is what my so-called binomial stv does. Admittedly the hand count version takes a bit of a short-cut, for simplicity at the expense of accuracy.

Regards,
Richard Lung.

On 20 Dec 2021, at 2:42 pm, robert bristow-johnson <rbj at audioimagination.com> wrote:

This is not about cycles and how to resolve them.

I have noticed on the Dominion Democracy Suite software that they have several different RCV methods, unfortunately all are Hare RCV (or "IRV") but they do different things when there is a dead tie at some round and they have to decide which candidate is eliminated.

Let's say we have a pretty straight-forward Condorcet rule.  Something like:

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(3) If no candidate receives a majority of first preferences, a Condorcet-consistent retabulation shall be performed by the presiding election officer. The retabulation shall examine every possible pairing of candidates. Given N as the number of candidates, then the number of possible pairings of candidates is N(N-1)/2. For each possible pairing of candidates, if the number of ballots marked ranking a selected candidate over the other candidate does not exceed the number of ballots marked to the contrary, then the selected candidate is designated as defeated. After all candidate pairs are examined, the candidate who remains not designated as defeated is the Condorcet winner and is elected.

(4) If no Condorcet winner exists in step (3), then the candidate having the plurality of first preferences is elected.  If two or more candidates are tied with a plurality of first preferences, then the candidate among those tied having the plurality of second preferences is elected.

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In the language above, if two candidates are tied in a pairwise comparison, they are *both* marked as defeated and the plurality language in step (4) kicks in if the top two are tied.  I was wondering if a better tie breaker for a pair of candidates that otherwise beat every other candidate (except each other) would be to look at the *smallest* margin each candidate has with other candidates (the margin they have with each other is zero) and elect the candidate with the larger of the smallest margins.  If that ties, then go to the next smallest margin.

Using the most simple language that might have a prayer of getting into legislation, what might be the best way of dealing with dead ties?  Really, the only tie that matters is the one of the "top two" candidates (whatever "top two" means in the Condorcet universe)

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r b-j . _ . _ . _ . _ rbj at audioimagination.com

"Imagination is more important than knowledge."

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