[EM] A family of easy-to-explain Condorcet methods
dcarrera at gmail.com
Tue Jun 29 14:18:20 PDT 2021
On Tue, Jun 29, 2021 at 7:16 AM Markus Schulze <markus.schulze8 at gmail.com>
> Dear Daniel,
> > Step 1: Sort candidates according to your favourite rule.
> > Step 2: Pick the bottom two candidates. Remove the pairwise loser.
> > Step 3: Repeat until only 1 candidate is left.
> > Every method in this category is Smith-efficient, so it automatically
> > meets many important rules like Condorcet loser, Mutual Majority,
> > and ISDA.
> This proposal doesn't automatically meet ISDA.
Hmm... yes, I see that. My bad. I think that if you change Step 2 so that
the bottom candidate is compared against every other candidate *AND* you
pick a sorting rule that puts the Smith set on top, then I think that would
meet IDSA. For example:
Step 1: Rank candidates by number of pairwise wins (Copeland sorting). Use
first-place votes to break ties.
Step 2: Pick the bottom candidate and compare him against every other
candidate. If he loses any match, kick him out.
Step 3: Repeat until you find a Condorcet winner or only 1 candidate is
I think *this* version would meet ISDA. Am I right?
Someone on Reddit has made the case to try to make the system resistant to
DH3. The only way I can think of to avoid DH3 is to sort by first-place
votes, but then I suspect that that wouldn't meet ISDA.
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