[EM] Tiebreaking in STV: Lundell-LastDifference

[Stéphane Rouillon]

Thoughts?
Many, essentially about the reproducibility of results to allow fraud detection and prevent hackers. The best approach is to make public all ballots so anyone can run a STV algorithm and obtain a identical winners. But to certify that results will always be the same we need systematic tie-breakers. Thus the paper I presented last year to the MPSA 2018 at Chicago. I'll send it when I get my computer instead of the phone...

Regards.

Envoyé de mon iPhone

> Le 23 mai 2019 à 12:46, John <john.r.moser at gmail.com> a écrit :
> 
> [Not subscribed, CC me on replies]
> 
> Jonathan Lundell proposed a rule for tiebreaking in STV:
> 
> http://www.votingmatters.org.uk/ISSUE22/I22P1.pdf 
> 
> 1. Find the first mention of any member of the tied set of candidates on each ballot, and calculate the total such mentions for each of the candidates, using the transferable weight of each ballot. Ignore ballots that do not mention at least one tied candidate.
> 
> 2. If all n candidates are still tied, exclude one tied candidate at random; finis.
> 
> 3. Otherwise, remove from consideration for exclusion the candidate (or a random choice from the tied set of candidates) with the highest score from step 1.
> 
> 4. If only one candidate remains, exclude that candidate; finis.
> 
> 5. Otherwise, n is now the remaining number of tied candidates (that is, less the reprieved candidates from step 3); continue at step 1.
> 
> Basically, when you're trying to exclude candidates in STV and you have multiple with the same last-place vote count, use the transfer weights of each ballot to perform instant runoff voting between these candidates and eliminate the winner from consideration; repeat until you have one candidate left.  Eliminate THAT candidate from your STV election.
> 
> Lundell cites exclusion of a random candidate in the event of a tie in this algorithm.  I propose using the Last Difference method, by Lundell's own arguments, and only falling back to random exclusion if that fails.
> 
> Lundell's argument for his proposed method is that prior-round tiebreaking encourages insincerity, and that Last Difference is superior to First Difference by O'Neill's arguments, therefor current-round information is even better.
> 
> I observe that Lundell's tiebreaker will run first, and so will dominate over the fallback.  If strategically targeting Last Difference sacrifices Lundell's runoff method, then it will cause losses, and so the strategy is unviable; yet Last Difference, when it produces a break, is better than a random tiebreaker.  The final fallback would be random.
> 
> Last Difference is equivalent to First Difference if the immediate prior round was the first difference.
> 
> Thoughts?
> 
> —John
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