[EM] Condorcet Loser, and equivalents
John
john.r.moser at gmail.com
Thu Jun 6 15:24:10 PDT 2019
That's a good question.
With the straight-line loser cycle, it's obvious: you can keep adding
Condorcet losers. {A,B,C} is the smith set, D is the Condorcet loser. If
you add E, such that {A,B,C,D} defeat E, E is the Condorcet loser. Nothing
has changed, so what is different now about D?
To extend this, if you add {E,F} such that {A,B,C} is the Smith set and
{D,E,F} would be the Smith set if we excluded {A,B,C} (thus there are two
cycles), we reach your concern: we have now constructed a situation where
Condorcet Loser D is not the Condorcet Loser, and can be elected without
violating the Condorcet Loser criterion.
This sort of implies that Condorcet Loser is a meaningless criterion.
Consider IRV, with candidates {A,B,C} such that the first-round vote totals
are A>B>C, thus C is eliminated. C votes go to B, A is eliminated, B wins.
Well now we add Candidate D. D takes some votes from B, and now the vote
totals are A>C>D>B. B is eliminated, some of B votes go to C, some go to
D. It's now C>D>A, A is eliminated. C defeats D one-on-one, C wins.
We can contrive this such that B is able to defeat A, C, or D one-on-one,
and is the Condorcet candidate. As well, perhaps A defeats C, so C is the
Condorcet Loser. Adding D eliminates B and captures votes from A, and this
causes the defeat of A, and then C defeats D.
It doesn't actually need to be so contrived, though.
Consider that B is the Condorcet Winner. in A vs. B, B has a simple
majority, and A is the Condorcet Loser.
Add C and C becomes the Condorcet Loser: B defeats A, B defeats C. Under
IRV, B is not eliminated first round, so B wins.
Now add Candidate D. B splits votes with C and D, causing B to lose.
At this point, you have a string of progressively-added Condorcet losers.
The elections {A,B}, {A,B,C}, and {A,B,C,D} all have B as the Condorcet
Winner; yet by adding the two extra candidates {C,D}, we have split the
vote for B, causing ANY OTHER CANDIDATE to win.
So the Condorcet Loser criteria may actually be pointless entirely, and may
be analogous to Smith-efficiency. There's a real mathematical effect here
where we can't elect the absolute loser because e.g. IRV runs down to a
two-way race, but you can circumvent this by adding another candidate (or
several others) and eliminating those candidates who defeat a given
candidate one-on-one.
In other words: a criterion which banks on the qualities of who we don't
elect might inherently be vulnerable to the strategy of eliminating that
individual.
On Tue, Jun 4, 2019 at 6:03 PM Toby Pereira <tdp201b at yahoo.co.uk> wrote:
> 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.
>
> Toby
>
>
> ------------------------------
> *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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>
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