[EM] Condorcet Loser, and equivalents

[C.Benham]

John,

> It seems to me the Condorcet Loser criterion is incomplete and inexact:  a
> single Condorcet Loser is meaningless.
The Condorcet Loser criterion is "incomplete and inexact" for what 
purpose?  The criterion just
refers to one desirable property that IRV (and of course some other 
methods) have that not all
other (used or proposed or plausible) methods have.

Isn't the pairwise winner out of two random candidates clearly a bit 
better than simply Random Candidate?

> 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.
>
>
As someone else pointed out, what if there is no single Condorcet loser 
but rather a bottom cycle (an upside-down
"Smith set")?  Saying that it shouldn't be eliminated but a single 
Condorcet loser should implies violating Clone Independence.

And if you say that we should keep eliminating the CL or the bottom 
cycle candidates I don't see how that is different from
saying that the winner has to be from the top cycle (Smith set).

But yes, just as Smith implies Condorcet but (because not vice versa) 
they are still two separate criteria, I suppose you
could have a bottom version of "Smith" to go along with CL but not to 
replace it.

Chris Benham




> *John* john.r.moser at gmail.com 
> <mailto:election-methods%40lists.electorama.com?Subject=Re%3A%20%5BEM%5D%20Condorcet%20Loser%2C%20and%20equivalents&In-Reply-To=%3CCAKYQpdvUv1iq%3DJRsd-vtKfN9%3Db3OJxTm5aV7ryb99GGoqtE2nA%40mail.gmail.com%3E>
> /Tue Jun 4 11:50:48 PDT 2019/
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>
> 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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