[EM] Condorcet Loser, and equivalents

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[EM] Condorcet Loser, and equivalents

John
[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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Re: [EM] Condorcet Loser, and equivalents

robert bristow-johnson



---------------------------- Original Message ----------------------------
Subject: [EM] Condorcet Loser, and equivalents
From: "John" <[hidden email]>
Date: Tue, June 4, 2019 11:50 am
To: [hidden email]
--------------------------------------------------------------------------

> [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?

yes.  and he was also an absent candidate.  most of us didn't even know he was a candidate until election day.  he got only about a dozen voters ranking him at all.

however, Dan Smith was a serious candidate.  he is now the president of some small Vermont college, i think.

now it looks like he's going back to Burlington.

https://www.vermontcf.org/NewsEvents/ViewArticle/tabid/96/ArticleId/161/Vermont-Community-Foundation-Hires-Dan-Smith-as-President-CEO.aspx


> 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.

you mean like James Simpson?

> D loses horribly, of course.
>
> D is now the Condorcet Loser. Kurt Wright isn't.

you mean like James Simpson?


> 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.

I've always said (maybe not in these circles) that even though the Burlington 2009 race was close enough to have one Plurality Winner (Wright), a different IRV Winner (Kiss), and a yet different Condorcet Winner (Montroll), that the outcome was well ordered, from a Condorcet POV.

Andy was unambiguously preferred over all of the candidates.

Remove Andy from the slate and Bob is preferred over all of the remaining candidates.

Remove Andy and Bob from the slate and Kurt is preferred over all of the remaining candidates.

Remove Andy and Bob and Kurt from the slate and Dan is preferred over the only other remaining candidate.

But there were three plausible winners.  There was a spoiler (Kurt) but usually the "spoiler" is someone who is way below the top two candidates, yet has enough votes to be far more than the margin of the top two.

Ironically, in the 2014 governor's race in Vermont, we had a serious spoiler candidate, but this time it was the Republican candidate who got screwed.  The spoiler was a Libertarian who got 3 times more votes than the margin between the GOP and Dem.  Since there was no majority, it went to the Vermont Legislature who routinely picks the plurality winner.  the guv was reelected.

that's when i started lecturing the GOP that sometimes they get hoisted by their own petard.


> 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.

Only if Candidate D can get into the final round with Kurt.  If either Andy or Bob do, they beat Kurt in the final round.

we had a Candidate D (his name is Dan).

 


> 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.

oh, so you're saying that Candidate D has more first-choice votes than either Andy or Bob?

doesn't sound like a "spoiler" to me.  sounds like a "winner".


> 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.

i don't worry too much about the Condorcet Loser.


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Re: [EM] Condorcet Loser, and equivalents

Toby Pereira
In reply to this post by John
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 <[hidden email]>
To: [hidden email]
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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Re: [EM] Condorcet Loser, and equivalents

John
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 <[hidden email]> 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 <[hidden email]>
To: [hidden email]
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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