[EM] Condorcet Loser, and equivalents

Toby Pereira tdp201b at yahoo.co.uk
Tue Jun 4 15:03:08 PDT 2019

As I understand it you're saying is that if you can pick off Condorcet losers one by one (so with no cycles involved), then none of these candidates should be elected. But this seems a very specific requirement. You might have a Condorcet loser, a Condorcet second-loser and a Condorcet third-loser. Under your criterion none of these should be elected. But what if there was a three-way loser cycle? Does it then become more acceptable to elect one of these? I don't see why it should.


      From: John <john.r.moser at gmail.com>
 To: election-methods at electorama.com 
 Sent: Tuesday, 4 June 2019, 19:51
 Subject: [EM] Condorcet Loser, and equivalents
[Not subscribed, please CC me]
Andy Montroll beats Bob Kiss.Andy Montroll beats Kurt Wright.Bob Kiss beats Kurt Wright.Montroll, Kiss, and Wright beat Dan Smith.Montroll, Kiss, Wright, and Smith beat James Simpson.
James Simpson is the Condorcet Loser.
…is that right?
In the race {Montroll, Kiss, Wright}, Kurt Wright is the Condorcet Loser.
Let's say candidate D appears.  D signs up, becomes a candidate, is on the ballot, and never campaigns.
D loses horribly, of course.
D is now the Condorcet Loser.  Kurt Wright isn't.
Does this change the nature of the candidate, Kurt Wright, or the social choice made?
Think about it.  Without Simpson, Smith is the Condorcet Loser.  Without Smith, Wright is the Condorcet Loser.  You have a chain of absolute Condorcet Loser until you have a tie or a strongly connected component containing more than one candidate which is not part of the Smith or Schwartz set.
This is important.
We say Instant Runoff Voting passes the Condorcet Loser criterion, but can elect the second-place Condorcet Loser.  That means Kurt Wright is the Condorcet Loser and IRV can't elect Wright; but in theory, you can add Candidate D and get Kurt Wright elected.
In practice, between two candidates, the loser is the Condorcet loser.  Montroll beats Kiss, so Kiss is the Condorcet Loser.  By adding Kurt Wright, you have a new Condorcet Loser.  This eliminates Montroll and, being that Kurt Wright is now the Condorcet Loser and is one-on-one with another candidate, Bob Kiss wins.
It seems to me the Condorcet Loser criterion is incomplete and inexact:  a single Condorcet Loser is meaningless.  The proper criterion should be ALL Condorcet Losers, such that eliminating the single Condorcet Loser leaves you with exactly one Condorcet Loser, thus both of them are the least-optimal set.
I suppose we can call this the Generalized Least-Optimal Alternative Theorem, unless somebody else (probably Markus Schulze) came up with it before I did.  It's the property I systematically manipulate when breaking IRV.
Thoughts?  Has this been done before?  Does this generalize not just to IRV, but to all systems which specifically pass the Condorcet Loser criterion proper (i.e. they have no special property like Smith-efficiency that implies Condorcet Loser criterion, but CAN elect the second-place Condorcet loser)?  That last one seems like it must be true.----
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