Re: [EM] An ABE solution
Mike, De : MIKE OSSIPOFF nkk...@hotmail.com À : election-meth...@electorama.com Envoyé le : Samedi 26 Novembre 2011 13h39 Objet : [EM] An ABE solution This was answered in the first part of the paragraph that you're quoting. What Woodall calls a preferential election rule is by definition a rank method. Kevin Kevin said: By definition an election method doesn't use approval ballots. [endquote] Whose definition? Election-Methods mailing list - see http://electorama.com/em for list info
[EM] An ABE solution
Kevin said: By definition an election method doesn't use approval ballots. [endquote] Whose definition? Do you think that if you hold an election by Approval, you aren't using an election method? Mike Ossipoff Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] An ABE solution
2011/11/24 Chris Benham cbenha...@yahoo.com.au Jameson, Your range scores are a little bit wrong,.. I've re-checked them and I don't see how. I gave each candidate 2 points for a top-rating, 1 for a middle-rating and zero for a bottom rating (or truncation). So in the initial sincere scenario for example C has 9 top-ratings and 1 middle-rating to make a score of 19, I count 9 and 2, for a total of 20. B has 8 top-ratings and 1 middle-rating to make a score of 17, I count 8 and 2, for a total of 18. and A has 5 top-ratings and 2 middle-ratings to make a score of 12. I count 5 and 5, for a total of 15. My counts differed from yours in the other cases; in particular, in the second failed strategy case, I found a tie between A and B, which is why I said you need an extra half-voter plumping B for the scenario to work as you claim. Again, this is mostly irrelevant, because I do agree that with this minor modification your scenario proves what you say it does. Jameson Chris Benham *From:* Jameson Quinn jameson.qu...@gmail.com *Sent:* Friday, 25 November 2011 5:39 AM *Subject:* Re: An ABE solution Chris: Your range scores are a little bit wrong, so you have to add half a B vote for the example to work (or double all factions and add one B vote if you discriminate against fractional people), but yes, this is at heart a valid example where the method fails FBC. Note that in my tendentious terminology this is only a defensive failure, that is, it starts from a position of a sincere condorcet cycle, which I believe will be rare enough in real elections to be discountable. In particular, this failure does not result in a stable two-party-lesser-evil-strategy self-reinforcing equilibrium. Jameson 2011/11/24 Chris Benham cbenha...@yahoo.com.au Forest, In reference to your new Condorcet method suggestion (pasted at the bottom), which elects an uncovered candidate and if there is none one-at-time disqualifies the Range loser until a remaining candidate X covers all the other remaining candidates and then elects X, you wrote: Indeed, the three slot case does appear to satisfy the FBC... No. Here is my example, based on that Kevin Venke proof you didn't like. Say sincere is 3: BA 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, B17, A12. CB 8-5, BA 10-5, AC 7-6. C wins. Now we focus on the 3 BA preferrers. Suppose (believing the method meets the FBC) they vote B=A. 3: B=A (sincere is BA) 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, B17, A15. CB 8-5, BA 7-5, AC 7-6. C still wins. Now suppose they instead rate their sincere favourite Middle: 3: AB (sincere is BA) 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, A15, B12. AB 8-7, AC 7-6,CB 8-5 Now those 3 voters get a result they prefer, the election of their compromise candidate A. Since it is clear they couldn't have got a result for themselves as good or better by voting BA or B=A or B or BC or B=C this is a failure of the FBC. Chris Benham *From:* fsimm...@pcc.edu fsimm...@pcc.edu *Sent:* Wednesday, 23 November 2011 9:01 AM *Subject:* Re: An ABE solution You are right that although the method is defined for any number of slots, I suggested three slots as most practical. So my example of two slots was only to disprove the statement the assertion that the method cannot be FBC compliant, since it is obviously compliant in that case. Furthermore something must be wrong with the quoted proof (of the incompatibility of the FBC and the CC) because the winner of the two slot case can be found entirely on the basis of the pairwise matrix. The other escape hatch is to say that two slots are not enough to satisfy anything but the voted ballots version of the Condorcet Criterion. But this applies equally well to the three slot case. Either way the cited therorem is not good enough to rule out compliance with the FBC by this new method. Indeed, the three slot case does appear to satisfy the FBC as well. It is an open question. I did not assert that it does. But I did say that IF it is strategically equivalent to Approval (as Range is, for example) then for practical purposes it satisfies the FBC. Perhaps not the letter of the law, but the spirit of the law. Indeed, in a non-stratetgical environment nobody worries about the FBC, i.e. only strategic voters will betray their favorite. If optimal strategy is approval strategy, and approval strategy requires you to top rate your favorite, then why would you do otherwise? Forest - Original Message - From: Chris Benham Forest, When the range ballots have only two slots, the method is simply Approval, which does satisfy the FBC. When you introduced the method you suggested that 3-slot ballots be used for simplicity. I thought you might be open to say 4-6 slots,
Re: [EM] An ABE solution
Forest, In reference to your new Condorcet method suggestion (pasted at the bottom), which elects an uncovered candidate and if there is none one-at-time disqualifies the Range loser until a remaining candidate X covers all the other remaining candidates and then elects X, you wrote: Indeed, the three slot case does appear to satisfy the FBC... No. Here is my example, based on that Kevin Venke proof you didn't like. Say sincere is 3: BA 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, B17, A12. CB 8-5, BA 10-5, AC 7-6. C wins. Now we focus on the 3 BA preferrers. Suppose (believing the method meets the FBC) they vote B=A. 3: B=A (sincere is BA) 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, B17, A15. CB 8-5, BA 7-5, AC 7-6. C still wins. Now suppose they instead rate their sincere favourite Middle: 3: AB (sincere is BA) 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, A15, B12. AB 8-7, AC 7-6, CB 8-5 Now those 3 voters get a result they prefer, the election of their compromise candidate A. Since it is clear they couldn't have got a result for themselves as good or better by voting BA or BC or B this is a failure of the FBC. Chris Benham From: fsimm...@pcc.edu fsimm...@pcc.edu Sent: Wednesday, 23 November 2011 9:01 AM Subject: Re: An ABE solution voters to avoid the middle slot. Then the method reduces to Approval, which does satisfy the FBC. The FBC doesn't stipulate that all the voters use optimal strategy, so that isn't relavent. http://wiki.electorama.com/wiki/FBC http://nodesiege.tripod.com/elections/#critfbc Chris Benham Forest Simmons wrote (17 Nov 2011): Here’s my current favorite deterministic proposal: Ballots are Range Style, say three slot for simplicity. When the ballots are collected, the pairwise win/loss/tie relations are determined among the candidates. The covering relations are also determined. Candidate X covers candidate Y if X beats Y as well as every candidate that Y beats. In other words row X of the win/loss/tie matrix dominates row Y. Then starting with the candidates with the lowest Range scores, they are disqualified one by one until one of the remaining candidates X covers any other candidates that might remain. Elect X. You are right that although the method is defined for any number of slots, I suggested three slots as most practical. So my example of two slots was only to disprove the statement the assertion that the method cannot be FBC compliant, since it is obviously compliant in that case. Furthermore something must be wrong with the quoted proof (of the incompatibility of the FBC and the CC) because the winner of the two slot case can be found entirely on the basis of the pairwise matrix. The other escape hatch is to say that two slots are not enough to satisfy anything but the voted ballots version of the Condorcet Criterion. But this applies equally well to the three slot case. Either way the cited therorem is not good enough to rule out compliance with the FBC by this new method. Indeed, the three slot case does appear to satisfy the FBC as well. It is an open question. I did not assert that it does. But I did say that IF it is strategically equivalent to Approval (as Range is, for example) then for practical purposes it satisfies the FBC. Perhaps not the letter of the law, but the spirit of the law. Indeed, in a non-stratetgical environment nobody worries about the FBC, i.e. only strategic voters will betray their favorite. If optimal strategy is approval strategy, and approval strategy requires you to top rate your favorite, then why would you do otherwise? Forest - Original Message - From: Chris Benham Forest, When the range ballots have only two slots, the method is simply Approval, which does satisfy the FBC. When you introduced the method you suggested that 3-slot ballots be used for simplicity. I thought you might be open to say 4-6 slots, but a complicated algorithm on 2-slot ballots that is equivalent to Approval ?? Now consider the case of range ballots with three slots: and suppose that optimal strategy requires the Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] An ABE solution
Chris: Your range scores are a little bit wrong, so you have to add half a B vote for the example to work (or double all factions and add one B vote if you discriminate against fractional people), but yes, this is at heart a valid example where the method fails FBC. Note that in my tendentious terminology this is only a defensive failure, that is, it starts from a position of a sincere condorcet cycle, which I believe will be rare enough in real elections to be discountable. In particular, this failure does not result in a stable two-party-lesser-evil-strategy self-reinforcing equilibrium. Jameson 2011/11/24 Chris Benham cbenha...@yahoo.com.au Forest, In reference to your new Condorcet method suggestion (pasted at the bottom), which elects an uncovered candidate and if there is none one-at-time disqualifies the Range loser until a remaining candidate X covers all the other remaining candidates and then elects X, you wrote: Indeed, the three slot case does appear to satisfy the FBC... No. Here is my example, based on that Kevin Venke proof you didn't like. Say sincere is 3: BA 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, B17, A12. CB 8-5, BA 10-5, AC 7-6. C wins. Now we focus on the 3 BA preferrers. Suppose (believing the method meets the FBC) they vote B=A. 3: B=A (sincere is BA) 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, B17, A15. CB 8-5, BA 7-5, AC 7-6. C still wins. Now suppose they instead rate their sincere favourite Middle: 3: AB (sincere is BA) 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, A15, B12. AB 8-7, AC 7-6,CB 8-5 Now those 3 voters get a result they prefer, the election of their compromise candidate A. Since it is clear they couldn't have got a result for themselves as good or better by voting BA or BC or B this is a failure of the FBC. Chris Benham *From:* fsimm...@pcc.edu fsimm...@pcc.edu *Sent:* Wednesday, 23 November 2011 9:01 AM *Subject:* Re: An ABE solution You are right that although the method is defined for any number of slots, I suggested three slots as most practical. So my example of two slots was only to disprove the statement the assertion that the method cannot be FBC compliant, since it is obviously compliant in that case. Furthermore something must be wrong with the quoted proof (of the incompatibility of the FBC and the CC) because the winner of the two slot case can be found entirely on the basis of the pairwise matrix. The other escape hatch is to say that two slots are not enough to satisfy anything but the voted ballots version of the Condorcet Criterion. But this applies equally well to the three slot case. Either way the cited therorem is not good enough to rule out compliance with the FBC by this new method. Indeed, the three slot case does appear to satisfy the FBC as well. It is an open question. I did not assert that it does. But I did say that IF it is strategically equivalent to Approval (as Range is, for example) then for practical purposes it satisfies the FBC. Perhaps not the letter of the law, but the spirit of the law. Indeed, in a non-stratetgical environment nobody worries about the FBC, i.e. only strategic voters will betray their favorite. If optimal strategy is approval strategy, and approval strategy requires you to top rate your favorite, then why would you do otherwise? Forest - Original Message - From: Chris Benham Forest, When the range ballots have only two slots, the method is simply Approval, which does satisfy the FBC. When you introduced the method you suggested that 3-slot ballots be used for simplicity. I thought you might be open to say 4-6 slots, but a complicated algorithm on 2-slot ballots that is equivalent to Approval ?? Now consider the case of range ballots with three slots: and suppose that optimal strategy requires the voters to avoid the middle slot. Then the method reduces to Approval, which does satisfy the FBC. The FBC doesn't stipulate that all the voters use optimal strategy, so that isn't relavent. http://wiki.electorama.com/wiki/FBC http://nodesiege.tripod.com/elections/#critfbc Chris Benham Forest Simmons wrote (17 Nov 2011): Here’s my current favorite deterministic proposal: Ballots are Range Style, say three slot for simplicity. When the ballots are collected, the pairwise win/loss/tie relations are determined among the candidates. The covering relations are also determined. Candidate X covers candidate Y if X beats Y as well as every candidate that Y beats. In other words row X of the win/loss/tie matrix dominates row Y. Then starting with the candidates with the lowest Range scores, they are disqualified one by one until one of the remaining candidates X covers any other candidates that might remain. Elect X. Election-Methods mailing
Re: [EM] An ABE solution
Jameson, Your range scores are a little bit wrong,.. I've re-checked them and I don't see how. I gave each candidate 2 points for a top-rating, 1 for a middle-rating and zero for a bottom rating (or truncation). So in the initial sincere scenario for example C has 9 top-ratings and 1 middle-rating to make a score of 19, B has 8 top-ratings and 1 middle-rating to make a score of 17, and A has 5 top-ratings and 2 middle-ratings to make a score of 12. Chris Benham From: Jameson Quinn jameson.qu...@gmail.com Sent: Friday, 25 November 2011 5:39 AM Subject: Re: An ABE solution Chris: Your range scores are a little bit wrong, so you have to add half a B vote for the example to work (or double all factions and add one B vote if you discriminate against fractional people), but yes, this is at heart a valid example where the method fails FBC. Note that in my tendentious terminology this is only a defensive failure, that is, it starts from a position of a sincere condorcet cycle, which I believe will be rare enough in real elections to be discountable. In particular, this failure does not result in a stable two-party-lesser-evil-strategy self-reinforcing equilibrium. Jameson 2011/11/24 Chris Benham cbenha...@yahoo.com.au Forest, In reference to your new Condorcet method suggestion (pasted at the bottom), which elects an uncovered candidate and if there is none one-at-time disqualifies the Range loser until a remaining candidate X covers all the other remaining candidates and then elects X, you wrote: Indeed, the three slot case does appear to satisfy the FBC... No. Here is my example, based on that Kevin Venke proof you didn't like. Say sincere is 3: BA 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, B17, A12. CB 8-5, BA 10-5, AC 7-6. C wins. Now we focus on the 3 BA preferrers. Suppose (believing the method meets the FBC) they vote B=A. 3: B=A (sincere is BA) 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, B17, A15. CB 8-5, BA 7-5, AC 7-6. C still wins. Now suppose they instead rate their sincere favourite Middle: 3: AB (sincere is BA) 3: A=C 3: B=C 2: AC 2: BA 2: CB 1: C Range (0,1,2) scores: C19, A15, B12. AB 8-7, AC 7-6, CB 8-5 Now those 3 voters get a result they prefer, the election of their compromise candidate A. Since it is clear they couldn't have got a result for themselves as good or better by voting BA or B=A or B or BC or B=C this is a failure of the FBC. Chris Benham From: fsimm...@pcc.edu fsimm...@pcc.edu Sent: Wednesday, 23 November 2011 9:01 AM Subject: Re: An ABE solution You are right that although the method is defined for any number of slots, I suggested three slots as most practical. So my example of two slots was only to disprove the statement the assertion that the method cannot be FBC compliant, since it is obviously compliant in that case. Furthermore something must be wrong with the quoted proof (of the incompatibility of the FBC and the CC) because the winner of the two slot case can be found entirely on the basis of the pairwise matrix. The other escape hatch is to say that two slots are not enough to satisfy anything but the voted ballots version of the Condorcet Criterion. But this applies equally well to the three slot case. Either way the cited therorem is not good enough to rule out compliance with the FBC by this new method. Indeed, the three slot case does appear to satisfy the FBC as well. It is an open question. I did not assert that it does. But I did say that IF it is strategically equivalent to Approval (as Range is, for example) then for practical purposes it satisfies the FBC. Perhaps not the letter of the law, but the spirit of the law. Indeed, in a non-stratetgical environment nobody worries about the FBC, i.e. only strategic voters will betray their favorite. If optimal strategy is approval strategy, and approval strategy requires you to top rate your favorite, then why would you do otherwise? Forest - Original Message - From: Chris Benham Forest, When the range ballots have only two slots, the method is simply Approval, which does satisfy the FBC. When you introduced the method you suggested that 3-slot ballots be used for simplicity. I thought you might be open to say 4-6 slots, but a complicated algorithm on 2-slot ballots that is equivalent to Approval ?? Now consider the case of range ballots with three slots: and suppose that optimal strategy requires the voters to avoid the middle slot. Then the method reduces to Approval, which does satisfy the FBC. The FBC doesn't stipulate that all the voters use optimal strategy, so that isn't relavent. http://wiki.electorama.com/wiki/FBC http://nodesiege.tripod.com/elections/#critfbc Chris Benham Forest Simmons wrote (17 Nov 2011): Here’s my current favorite
Re: [EM] An ABE solution
I agree it's silly to create complicated rules for a two-slot ballot. But,
though Forest didn't quite say so, I also think that FBC and (voted ballot)
Condorcet are not incompatible for a 3-slot ballot.
Jameson
2011/11/22 Chris Benham cbenha...@yahoo.com.au
Forest,
When the range ballots have only two slots, the method is simply
Approval, which does satisfy the
FBC.
When you introduced the method you suggested that 3-slot ballots be used
for simplicity.
I thought you might be open to say 4-6 slots, but a complicated algorithm
on 2-slot ballots
that is equivalent to Approval ??
Now consider the case of range ballots with three slots: and suppose that
optimal strategy requires the
voters to avoid the middle slot. Then the method reduces to Approval,
which does satisfy the FBC.
The FBC doesn't stipulate that all the voters use optimal strategy, so
that isn't relavent.
http://wiki.electorama.com/wiki/FBC
http://nodesiege.tripod.com/elections/#critfbc
Chris Benham
*From:* fsimm...@pcc.edu fsimm...@pcc.edu
*To:* C.Benham cbenha...@yahoo.com.au
*Cc:* em election-meth...@electorama.com; MIKE OSSIPOFF
nkk...@hotmail.com
*Sent:* Tuesday, 22 November 2011 11:11 AM
*Subject:* Re: An ABE solution
From: C.Benham
Forest Simmons, responding to questions from Mike Ossipff, wrote
(19 Nov
2011):
4. How does it do by FBC? And by the criteria that bother some
people here about MMPO (Kevin's MMPO bad-example) and MDDTR
(Mono-Add-Plump)?
I think it satisfies the FBC.
Forest's definition of the method being asked about:
Here’s my current favorite deterministic proposal: Ballots are
Range
Style, say three slot for simplicity.
When the ballots are collected, the pairwise win/loss/tie
relations are
determined among the candidates.
The covering relations are also determined. Candidate X covers
candidate Y if X
beats Y as well as every candidate that Y beats. In other
words row X
of the
win/loss/tie matrix dominates row Y.
Then starting with the candidates with the lowest Range
scores, they are
disqualified one by one until one of the remaining candidates
X covers
any other
candidates that might remain. Elect X.
Forest,
Doesn't this method meet the Condorcet criterion? Compliance
with
Condorcet is incompatible with FBC, so
why do you think it satisfies FBC?
When the range ballots have only two slots, the method is simply Approval,
which does satisfy the
FBC. Does Approval satisfy the Condorcet Criterion? I would say no, but
it does satisfy the votes only
Condorcet Criterion. which means that the Approval winner X pairwise
beats every other candidate Y
according to the ballots, i.e. X is rated above Y on more ballots than Y
is rated above X.
Now consider the case of range ballots with three slots: and suppose that
optimal strategy requires the
voters to avoid the middle slot. Then the method reduces to Approval,
which does satisfy the FBC.
http://lists.electorama.com/pipermail/election-methods-
electorama.com/2005-June/016410.html
Hello,
This is an attempt to demonstrate that Condorcet and FBC are
incompatible. I modified Woodall's proof that Condorcet and
LNHarm are incompatible.
(Douglas R. Woodall, Monotonicity of single-seat preferential
election rules,
Discrete Applied Mathematics 77 (1997), pages 86 and 87.)
I've suggested before that in order to satisfy FBC, it must be
the case
that increasing the votes for A over B in the pairwise matrix
can never
increase the probability that the winner comes from {a,b};
that is, it
must
not move the win from some other candidate C to A. This is
necessary
because
if sometimes it were possible to move the win from C to A by
increasing v[a,b], the voter with the preference order BAC
would have incentive to
reverse B and A in his ranking (and equal ranking would be
inadequate).
I won't presently try to argue that this requirement can't be
avoided
somehow.
I'm sure it can't be avoided when the method's result is
determined solely
from the pairwise matrix.
Note that in our method the Cardinal Ratings order (i.e. Range Order) is
needed in addition to the
pairwise matrix; the covering information comes from the pairwise matrix,
but candidates are dropped
from the bottom of the range order.
In the two slot case can the approval order be determined from the
pairwise matrix? If so, then this is a
counterexample to the last quoted sentence above in the attempted proof of
the incompatibility of the CC
and the FBC.
Forest
Election-Methods mailing list - see http://electorama.com/em for list info
Election-Methods mailing list - see http://electorama.com/em for list info
Re: [EM] An ABE solution
Forest,
When the range ballots have only two slots, the method is simply Approval,
which does satisfy the
FBC.
When you introduced the method you suggested that 3-slot ballots be used for
simplicity.
I thought you might be open to say 4-6 slots, but a complicated algorithm on
2-slot ballots
that is equivalent to Approval ??
Now consider the case of range ballots with three slots: and suppose that
optimal strategy requires the
voters to avoid the middle slot. Then the method reduces to Approval, which
does satisfy the FBC.
The FBC doesn't stipulate that all the voters use optimal strategy, so that
isn't relavent.
http://wiki.electorama.com/wiki/FBC
http://nodesiege.tripod.com/elections/#critfbc
Chris Benham
From: fsimm...@pcc.edu fsimm...@pcc.edu
To: C.Benham cbenha...@yahoo.com.au
Cc: em election-meth...@electorama.com; MIKE OSSIPOFF nkk...@hotmail.com
Sent: Tuesday, 22 November 2011 11:11 AM
Subject: Re: An ABE solution
From: C.Benham
Forest Simmons, responding to questions from Mike Ossipff, wrote
(19 Nov
2011):
4. How does it do by FBC? And by the criteria that bother some
people here about MMPO (Kevin's MMPO bad-example) and MDDTR
(Mono-Add-Plump)?
I think it satisfies the FBC.
Forest's definition of the method being asked about:
Here’s my current favorite deterministic proposal: Ballots are
Range
Style, say three slot for simplicity.
When the ballots are collected, the pairwise win/loss/tie
relations are
determined among the candidates.
The covering relations are also determined. Candidate X covers
candidate Y if X
beats Y as well as every candidate that Y beats. In other
words row X
of the
win/loss/tie matrix dominates row Y.
Then starting with the candidates with the lowest Range
scores, they are
disqualified one by one until one of the remaining candidates
X covers
any other
candidates that might remain. Elect X.
Forest,
Doesn't this method meet the Condorcet criterion? Compliance
with
Condorcet is incompatible with FBC, so
why do you think it satisfies FBC?
When the range ballots have only two slots, the method is simply Approval,
which does satisfy the
FBC. Does Approval satisfy the Condorcet Criterion? I would say no, but it
does satisfy the votes only
Condorcet Criterion. which means that the Approval winner X pairwise beats
every other candidate Y
according to the ballots, i.e. X is rated above Y on more ballots than Y is
rated above X.
Now consider the case of range ballots with three slots: and suppose that
optimal strategy requires the
voters to avoid the middle slot. Then the method reduces to Approval, which
does satisfy the FBC.
http://lists.electorama.com/pipermail/election-methods-
electorama.com/2005-June/016410.html
Hello,
This is an attempt to demonstrate that Condorcet and FBC are
incompatible. I modified Woodall's proof that Condorcet and
LNHarm are incompatible.
(Douglas R. Woodall, Monotonicity of single-seat preferential
election rules,
Discrete Applied Mathematics 77 (1997), pages 86 and 87.)
I've suggested before that in order to satisfy FBC, it must be
the case
that increasing the votes for A over B in the pairwise matrix
can never
increase the probability that the winner comes from {a,b};
that is, it
must
not move the win from some other candidate C to A. This is
necessary
because
if sometimes it were possible to move the win from C to A by
increasing v[a,b], the voter with the preference order BAC
would have incentive to
reverse B and A in his ranking (and equal ranking would be
inadequate).
I won't presently try to argue that this requirement can't be
avoided
somehow.
I'm sure it can't be avoided when the method's result is
determined solely
from the pairwise matrix.
Note that in our method the Cardinal Ratings order (i.e. Range Order) is needed
in addition to the
pairwise matrix; the covering information comes from the pairwise matrix, but
candidates are dropped
from the bottom of the range order.
In the two slot case can the approval order be determined from the pairwise
matrix? If so, then this is a
counterexample to the last quoted sentence above in the attempted proof of the
incompatibility of the CC
and the FBC.
Forest
Election-Methods mailing list - see http://electorama.com/em for list info
[EM] An ABE solution
Mike, thanks for your comments. I'll respond in line below. From: MIKE OSSIPOFF Hi Forest-- Thanks for answering my question about MTA vs MCA. Your argument on that question is convincing, and answers my question about the strategy difference between those two methods. Certainly, electing C in the ABE avoids the ABE problem. I'd been hoping that the election of C can be attained without diverging from Plurality's results enough to upset some people, as MMPO and MDDTR seem to do. So the method that you describe might avoid the public relations (non)problems of methods that elect A. I have a few questions about the method that you describe: 1. What name do you give to it? In this post I'll call it Range till cover-winner or RCW Good Idea. 2. The covering relation doesn't look at pairwise ties? Different variants handle ties differently, but I favor using dominance in the win/loss/tie matrix, in which the entry in row x column y is respectively one, minus one, or zero depending on whether candidate x beats, is beaten by, or ties with candidate y, respectively. We consider a candidate to be tied with itself, so for each x, the diagonal (x, x) entry is zero. Note that this matrix can be gotten by subtracting the transpose of the pairwise matrix from itself, and then replacing each entry by its sign, where sign(t) is 1, -1, or 0, depending on whether t is positive, negativve, or zero. In other words, replace each entry in the pairwise margins matrix with its sign. Row x dominates row y iff the two rows are not identical and every entry in row x is at least as great as the corresponding entry in row y. Candidate x covers candidate y iff row x dominates row y in the above sense. 3. Does the ballot ask the voter for cardinal ratings of the candidates, or is the range score calculated a la Borda? Cardinal ratings are assumed, and for public proposal I contemplate three slots. If the ballots are ordinal, you can transform them (clone free and monotonically) to cardinal ballots via my algorithm under the heading Borda Done Right. 4. How does it do by FBC? And by the criteria that bother some people here about MMPO (Kevin's MMPO bad-example) and MDDTR (Mono-Add-Plump)? I think it satisfies the FBC. In fact, it reduces to Approval in the two slot version or if all voters rate only at the extremes., Approval strategy is probably near near optimal. It satisfies Mono-Add-Plump. Proof: First note that addition of a ballot that truncates all of the candidates doesn't change the winner since it doesn't change the covering relations nor does it change the range score (cardinal rating) order. Now, on such a ballot raise the winner from trunctated status to non-truncated status, and leave the rest of the candidates at the bottom. By the monotonicity of the method, the winner is preserved. I believe it satisfies the Plurality criterion. At least in Kevin's MMPO bad example it ties the two outside candidates. There's much hope that, by electing C instead of A, RCW can avoid those criticisms. I'm also interested in how it does by 1CM and 3P, but I'll look at that, instead of asking you to do everything for me, especially since I'm the one promoting those two criteria. I don't hve those two criteria on the tip of my tongue, but I'll look them up and see if I can help you on that. Mike Forest Here?s my current favorite deterministic proposal: Ballots are Range Style, say three slot for simplicity. When the ballots are collected, the pairwise win/loss/tie relations are determined among the candidates. The covering relations are also determined. Candidate X covers candidate Y if X beats Y as well as every candidate that Y beats. In other words row X of the win/loss/tie matrix dominates row Y. Then starting with the candidates with the lowest Range scores, they are disqualified one by one until one of the remaining candidates X covers any other candidates that might remain. Elect X. For practical purposes this method is the same as Smith//Range. Where they differ, the member of Smith with the highest range score is covered by some other Smith member with a range score not far behind. This method resolves the ABE (approval bad example) in the following way: Suppose that the ballots are 49 C 27 A(top)B(middle) 24 B No candidate covers any other candidate. The range order is CBA. Both A and B are removed before reaching candidate C, which is not covered by any remaining candidate. So the Smith//Range candidate C wins. If the ballots are sincere, then nobody can say that the Range winner was a horrible choice. But more to the point, if the ballots are sincere, the A supporters have a way of rescuing B: just rate hir equal top with A. Suppose, on the other hand that the B supporters like A better than C and the A supporters know this. Then
[EM] An ABE solution
Forest Simmons, responding to questions from Mike Ossipff, wrote (19 Nov
2011):
4. How does it do by FBC? And by the criteria that bother some
people here about MMPO (Kevin's MMPO bad-example) and MDDTR
(Mono-Add-Plump)?
I think it satisfies the FBC.
Forest's definition of the method being asked about:
Here’s my current favorite deterministic proposal: Ballots are Range
Style, say three slot for simplicity.
When the ballots are collected, the pairwise win/loss/tie relations are
determined among the candidates.
The covering relations are also determined. Candidate X covers
candidate Y if X
beats Y as well as every candidate that Y beats. In other words row X
of the
win/loss/tie matrix dominates row Y.
Then starting with the candidates with the lowest Range scores, they are
disqualified one by one until one of the remaining candidates X covers
any other
candidates that might remain. Elect X.
Forest,
Doesn't this method meet the Condorcet criterion? Compliance with
Condorcet is incompatible with FBC, so
why do you think it satisfies FBC?
http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-June/016410.html
Hello,
This is an attempt to demonstrate that Condorcet and FBC are incompatible.
I modified Woodall's proof that Condorcet and LNHarm are incompatible.
(Douglas R. Woodall, Monotonicity of single-seat preferential
election rules,
Discrete Applied Mathematics 77 (1997), pages 86 and 87.)
I've suggested before that in order to satisfy FBC, it must be the case
that increasing the votes for A over B in the pairwise matrix can never
increase the probability that the winner comes from {a,b}; that is, it
must
not move the win from some other candidate C to A. This is necessary
because
if sometimes it were possible to move the win from C to A by increasing
v[a,b], the voter with the preference order BAC would have incentive to
reverse B and A in his ranking (and equal ranking would be inadequate).
I won't presently try to argue that this requirement can't be avoided
somehow.
I'm sure it can't be avoided when the method's result is determined solely
from the pairwise matrix.
Suppose a method satisfies this property, and also Condorcet. Consider
this
scenario:
a=b 3
a=c 3
b=c 3
ac 2
ba 2
cb 2
There is an ACBA cycle, and the scenario is symmetrical, as based on
the submitted rankings, the candidates can't be differentiated. This means
that an anonymous and neutral method has to elect each candidate with
33.33%
probability.
Now suppose the a=b voters change their vote to ab (thereby
increasing v[a,b]).
This would turn A into the Condorcet winner, who would have to win
with 100%
probability due to Condorcet.
But the probability that the winner comes from {a,b} has increased
from 66.67%
to 100%, so the first property is violated.
Thus the first property and Condorcet are incompatible, and I contend
that FBC
requires the first property.
Thoughts?
Kevin Venzke
Chris Benham
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[EM] An ABE solution.
Hi Forest-- Thanks for answering my question about MTA vs MCA. Your argument on that question is convincing, and answers my question about the strategy difference between those two methods. Certainly, electing C in the ABE avoids the ABE problem. I'd been hoping that the election of C can be attained without diverging from Plurality's results enough to upset some people, as MMPO and MDDTR seem to do. So the method that you describe might avoid the public relations (non)problems of methods that elect A. I have a few questions about the method that you describe: 1. What name do you give to it? In this post I'll call it Range till cover-winner or RCW 2. The covering relation doesn't look at pairwise ties? 3. Does the ballot ask the voter for cardinal ratings of the candidates, or is the range score calculated a la Borda? 4. How does it do by FBC? And by the criteria that bother some people here about MMPO (Kevin's MMPO bad-example) and MDDTR (Mono-Add-Plump)? There's much hope that, by electing C instead of A, RCW can avoid those criticisms. I'm also interested in how it does by 1CM and 3P, but I'll look at that, instead of asking you to do everything for me, especially since I'm the one promoting those two criteria. Mike Here’s my current favorite deterministic proposal: Ballots are Range Style, say three slot for simplicity. When the ballots are collected, the pairwise win/loss/tie relations are determined among the candidates. The covering relations are also determined. Candidate X covers candidate Y if X beats Y as well as every candidate that Y beats. In other words row X of the win/loss/tie matrix dominates row Y. Then starting with the candidates with the lowest Range scores, they are disqualified one by one until one of the remaining candidates X covers any other candidates that might remain. Elect X. For practical purposes this method is the same as Smith//Range. Where they differ, the member of Smith with the highest range score is covered by some other Smith member with a range score not far behind. This method resolves the ABE (approval bad example) in the following way: Suppose that the ballots are 49 C 27 A(top)B(middle) 24 B No candidate covers any other candidate. The range order is CBA. Both A and B are removed before reaching candidate C, which is not covered by any remaining candidate. So the Smith//Range candidate C wins. If the ballots are sincere, then nobody can say that the Range winner was a horrible choice. But more to the point, if the ballots are sincere, the A supporters have a way of rescuing B: just rate hir equal top with A. Suppose, on the other hand that the B supporters like A better than C and the A supporters know this. Then the threat of C being elected will deter B faction defection, and they will rationally vote A in the middle: 49 C 27 A(top)B(middle) 24 B(top)A(middle) Now A covers both other candidates, so no matter the Range score order A wins. This completely resolves the ABE to my satisfaction. The method also allows for easy defense against burial of the CW. In the case 40 AB (sincere ACB) 30 BC 30 CA where C is the sincere CW, the C supporters can defend C's win by truncating A. Then the Nash equilibrium is 40 A 30 BC 30 C in which C is the ballot CW, and so is elected. Now for another topic... MTA vs. MCA I like MTA better than MCA because in the case where they differ (two or more candidates with majorities of top preferences) the MCA decision is made only by the voters whose ballots already had the effect of getting the ”finalists” into the final round, while the MTA decision reaches for broader support. Because of this, in MTA there is less incentive to top rate a lesser evil. If you don’t believe the fake polls about how hot the lesser evil is, you can take a wait and see attitude by voting her in the middle slot. If it turns out that she did end up as a finalist (against the greater evil) then your ballot will give her full support in the final round. Election-Methods mailing list - see http://electorama.com/em for list info
[EM] An ABE solution.
Here’s my current favorite deterministic proposal: Ballots are Range Style, say three slot for simplicity. When the ballots are collected, the pairwise win/loss/tie relations are determined among the candidates. The covering relations are also determined. Candidate X covers candidate Y if X beats Y as well as every candidate that Y beats. In other words row X of the win/loss/tie matrix dominates row Y. Then starting with the candidates with the lowest Range scores, they are disqualified one by one until one of the remaining candidates X covers any other candidates that might remain. Elect X. For practical purposes this method is the same as Smith//Range. Where they differ, the member of Smith with the highest range score is covered by some other Smith member with a range score not far behind. This method resolves the ABE (approval bad example) in the following way: Suppose that the ballots are 49 C 27 A(top)B(middle) 24 B No candidate covers any other candidate. The range order is CBA. Both A and B are removed before reaching candidate C, which is not covered by any remaining candidate. So the Smith//Range candidate C wins. If the ballots are sincere, then nobody can say that the Range winner was a horrible choice. But more to the point, if the ballots are sincere, the A supporters have a way of rescuing B: just rate hir equal top with A. Suppose, on the other hand that the B supporters like A better than C and the A supporters know this. Then the threat of C being elected will deter B faction defection, and they will rationally vote A in the middle: 49 C 27 A(top)B(middle) 24 B(top)A(middle) Now A covers both other candidates, so no matter the Range score order A wins. This completely resolves the ABE to my satisfaction. The method also allows for easy defense against burial of the CW. In the case 40 AB (sincere ACB) 30 BC 30 CA where C is the sincere CW, the C supporters can defend C's win by truncating A. Then the Nash equilibrium is 40 A 30 BC 30 C in which C is the ballot CW, and so is elected. Now for another topic... MTA vs. MCA I like MTA better than MCA because in the case where they differ (two or more candidates with majorities of top preferences) the MCA decision is made only by the voters whose ballots already had the effect of getting the ”finalists” into the final round, while the MTA decision reaches for broader support. Because of this, in MTA there is less incentive to top rate a lesser evil. If you don’t believe the fake polls about how hot the lesser evil is, you can take a wait and see attitude by voting her in the middle slot. If it turns out that she did end up as a finalist (against the greater evil) then your ballot will give her full support in the final round. Election-Methods mailing list - see http://electorama.com/em for list info
