Duncan Proposal Draft
Are you referring to how equal-ranking is counted in Borda?
Yes, mainly truncated ballots. Do you consider that truncation is the
same thing as ranking equal bottom?
I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.
Can you clarify the exact meaning of the phrase "in the ranking"? Is
this a hint that your answer to my last
question is "no"?
And why is that what you advocate? In some online discussion I saw it
was mentioned (I think by N.Tideman) that
the Baldwin method doesn't meet Condorcet unless the Fractional (i.e.
based on the symmetrically completed ballots)
version of Borda count is used.
So I've tended to assume that that is the "correct" way of doing Borda
counts.
Say there are three candidates A,B,C. Say 46 ballots bullet-vote A.
In the fractional version these votes are counted the same as
23 A>B, 23 A>C, giving 92 points to A and 23 points each to B and C.
Is that the same as what you advocate, or would you have those
truncating ballots give 2 points to A and zero points to both of B and C?
Or something else?
/So another way of putting it is: "If there is no CW, elect the member of //the Smith set with the second-worst score". ///
Would you mind telling why that’s another way of putting it?
By definition the members of the Smith set all pairwise beat all the
non-member candidates, and also have a pairwise defeat at the hands
of one of the other members (assuming there are three members).
So the Smith set member with the highest score must be pairwise beaten
by one with a lower score. That much is clear.
To be honest I may have confused myself again as to further details. It
seems to be possible for the lowest-scored Smith-set member to win.
Say the top cycle is (in terms of score order) Middle > High > Low >
Middle.
Then High is disqualified by being pairwise beaten by Middle and Middle
is disqualified by being pairwise beaten by Low.
So Low wins.
Remember that sincere top-cycles are vanishingly rare.
What mind-reading technology have you accessed to determine that? I
don't know how you can know that.
Simple scenarios with lots of truncation and a top cycle look very
plausible to me, so I don't know why I should
accept your assurance that such a thing would be "vanishingly rare".
Also it seems to me you are giving yourself a marketing problem:
"Ok, we admit that Condorcet methods are very vulnerable to Burial
strategy, so in a effort to combat that we have
to employ this apparently nonsensical, anti-monotonic, anti-intuitive
completion method.
But don't worry, probably we'll never have to use it."
Chris B.
Michael Ossipoffemail9648742 at gmail.com
mailto:election-methods%40lists.electorama.com?Subject=Re%3A%20%5BEM%5D%20Duncan%20Proposal%20Draft&In-Reply-To=%3CCAOKDY5D-CUNWajrZxSi8VauLY8FAys8vnt6SvX%3D7KiQFeEpeJw%40mail.gmail.com%3E
/Sat Oct 21 12:38:58 PDT 2023/
On Sat, Oct 21, 2023 at 11:05 C.Benham <cbenham at adam.com.au http://lists.electorama.com/listinfo.cgi/election-methods-electorama.com> wrote:
"Am I right in assuming that the Borda counts are based on the symmetrically
completed ballots?"
Are you referring to how equal-ranking is counted in Borda? I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.
/Duncan Definition: />//>/In the vast majority of the cases ... those in which the pairwise counts />/of the ballots unambiguously identify the candidate that pairbeats
each of />/the others ... elect that candidate. />//>/Otherwise, elect the highest score candidate that pairbeats every />/candidate with lower score. />//>//>/So another way of putting it is: "If there is no CW, elect the member of />/the Smith set with the second-worst score". />//
Would you mind telling why that’s another way of putting it?
//>/To put it bluntly, that is bound to have monotonicity problems and
doesn't />/fly philosophically. />//
How doesn’t it fly philosophically?
//>//>/Trying to deter or frustrate order-reversal Burial strategy is fine, but />/the algorithm should "appear fair" and be able to be justified when />/we assume that all the votes are sincere (or even just all equally
likely />/to be sincere). />//
Remember that sincere top-cycles are vanishingly rare. That’s why
Sequential-Pairwise’s Pareto failure isn’t important, & it’s why
MinMax(wv)’s Condorcet Loser failure isn’t important.
It matters much more what happens in a strategic cycle. Does the method
reward or penalize burial? …or neither?
Duncan tends to often do neither, because it will likely disqualify both
Bus & BF. (I defined those usages when I defined CTE.
I prefer CTE to Duncan, because it more often penalizes, instead of merely
not rewarding.
But I remind you that these methods are intended to deter
probabilisticallly, but aren’t claimed to penalize burial in every possible
example.
//>//
//>/That breaks (at least one version of) "Double Defeat". B is pairwise />/beaten by a candidate with a higher "score". />//>//>/Chris B. />//>//>/On 14/10/2023 4:43 am, Michael Ossipoff wrote: />//>/Yes, I like Duncan because burying the CW in an attempt to help your />/favorite won’t help hir when it causes hir disqualification, as it
probably />/will. />//>/…& Duncan is remarkably briefly-defined, needing only a very slight />/modification of Black’s method. />
On Mon, Oct 23, 2023 at 06:01 C.Benham cbenham@adam.com.au wrote:
Are you referring to how equal-ranking is counted in Borda?
Yes, mainly truncated ballots. Do you consider that truncation is the
same thing as ranking equal bottom?
Yes.
I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.
Can you clarify the exact meaning of the phrase "in the ranking"? Is this
a hint that your answer to my last
question is "no"?
“In the ranking” mean written in the ranking :-)
The name of the candidate is written in the ranking, at some rank-position.
:-)
It isn’t a hint to anything other than what it says.
And why is that what you advocate?
Because burying someone better raises the score of everyone under or with
whom you burohir.
In some online discussion I saw it was mentioned (I think by N.Tideman)
that
the Baldwin method doesn't meet Condorcet unless the Fractional (i.e.
based on the symmetrically completed ballots)
version of Borda count is used.
Then it’s a good thing that we aren’t proposing Baldwin.
So I've tended to assume that that is the "correct" way of doing Borda
counts.
Depends on the Borda count’s purpose & use in a particular method.
Say there are three candidates A,B,C. Say 46 ballots bullet-vote A. In
the fractional version these votes are counted the same as
23 A>B, 23 A>C, giving 92 points to A and 23 points each to B and C.
Is that the same as what you advocate[?]
is that what I said? :-)
No.
or would you have those truncating ballots give 2 points to A and zero
points to both of B and C?
Is that what I said? :-)
No.
Or something else?
How about what I said?
Of the candidates that your ballot ranks, 3 aren’t ranked over A.
A gets 3 points.
Of the candidates that your ballot ranks, 2 aren’t ranked over B, or over C.
Therefore, by what I said, B & C each get 2 points.
Yes that’s different from any usual Borda count. So it would be better to
add the word “other”, & say:
The ranking gives to each candidate a score equal to the number of other
candidates (candidates other than itself) in the ranking whom the ranking
doesn’t rank over it.
That change obviously doesn’t change the differences in candidates’ scores.
(That’s unnecessary wordy, but I want to make it as explicit as possible.)
- So another way of putting it is: "If there is no CW, elect the member of
** the Smith set with the second-worst score".
Would you mind telling why that’s another way of putting it?
By definition the members of the Smith set all pairwise beat all the
non-member candidates, and also have a pairwise defeat at the hands
of one of the other members (assuming there are three members).
So the Smith set member with the highest score must be pairwise beaten by
one with a lower score. That much is clear.
To be honest I may have confused myself again as to further details. It
seems to be possible for the lowest-scored Smith-set member to win.
Suit yourself & reword it as you wish.
Say the top cycle is (in terms of score order) Middle > High > Low >
Middle.
Then High is disqualified by being pairwise beaten by Middle and Middle is
disqualified by being pairwise beaten by Low.
So Low wins.
Bingo !
Low wins.
Low is probably CW. Low is unlikely to be BF.
The burial didn’t succeed. CW won.
Remember that sincere top-cycles are vanishingly rare.
What mind-reading technology have you accessed to determine that? I don't
know how you can know that.
I used a crystal-ball, along with every Condorcet poll that I’ve heard of,
participated in or conducted.
Simple scenarios with lots of truncation and a top cycle look very
plausible to me, so I don't know why I should
accept your assurance that such a thing would be "vanishingly rare".
Howabout experience.
Also it seems to me you are giving yourself a marketing problem:
"Ok, we admit that Condorcet methods are very vulnerable to Burial strategy
Margins Condorcet is.
wv Condorcet isn’t, if people use defensive truncation.
I believe that has been discussed here. :-)
But probabilistically-auto-deterrent methods protect against burial,
preferably by deterrence due to penalty, without anyone using defensive
strategy.
As I said I prefer CTE to Duncan, because CTE often deters when Duncan
would merely not reward.
so in a effort to combat that we have
to employ this apparently nonsensical, anti-monotonic, anti-intuitive
completion method.
The auto-deterrent methods aren’t designed for cosmetics in sincere
top-cycles, in which there’s no CW to protect. …& which are unheard-of in
Condorcet polls.
Admittedly not all methods are monotonic like IRV. :-)
But don't worry, probably we'll never have to use it."
Is that what your crystal-ball says?
Chris B.
The auto-deterrent methods likely aren’t as obvious, briefly-defined,
simply-explained, or familiar as MinMax(wv), & so are maybe less likely to
be proposed as a 1st single-winner reform.
But it’s desirable that defensive strategy not be needed.
When proposing single-winner reform, I offer a list of proposals. It will
include CTE. What the Condorcet Enactment-Committee proposes will be their
choice, but I prefer to offer them a variety, including the familiar
briefly defined MinMax(wv), CW,Approval, SP, & the auto-deterrent CTE.
Michael Ossipoff email9648742 at gmail.com
<election-methods%40lists.electorama.com?Subject=Re%3A%20%5BEM%5D%20Duncan%20Proposal%20Draft&In-Reply-To=%3CCAOKDY5D-CUNWajrZxSi8VauLY8FAys8vnt6SvX%3D7KiQFeEpeJw%40mail.gmail.com%3E> Sat
Oct 21 12:38:58 PDT 2023
On Sat, Oct 21, 2023 at 11:05 C.Benham <cbenham at adam.com.au http://lists.electorama.com/listinfo.cgi/election-methods-electorama.com> wrote:
"Am I right in assuming that the Borda counts are based on the symmetrically
completed ballots?"
Are you referring to how equal-ranking is counted in Borda? I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.
- Duncan Definition:
>> In the vast majority of the cases ... those in which the pairwise counts
> of the ballots unambiguously identify the candidate that pairbeats each of
> the others ... elect that candidate.
>> Otherwise, elect the highest score candidate that pairbeats every
> candidate with lower score.
>>> So another way of putting it is: "If there is no CW, elect the member of
> the Smith set with the second-worst score".
*>
Would you mind telling why that’s another way of putting it?
- To put it bluntly, that is bound to have monotonicity problems and doesn't
> fly philosophically.
*>
How doesn’t it fly philosophically?
- Trying to deter or frustrate order-reversal Burial strategy is fine, but
> the algorithm should "appear fair" and be able to be justified when
> we assume that all the votes are sincere (or even just all equally likely
> to be sincere).
*>
Remember that sincere top-cycles are vanishingly rare. That’s why
Sequential-Pairwise’s Pareto failure isn’t important, & it’s why
MinMax(wv)’s Condorcet Loser failure isn’t important.
It matters much more what happens in a strategic cycle. Does the method
reward or penalize burial? …or neither?
Duncan tends to often do neither, because it will likely disqualify both
Bus & BF. (I defined those usages when I defined CTE.
I prefer CTE to Duncan, because it more often penalizes, instead of merely
not rewarding.
But I remind you that these methods are intended to deter
probabilisticallly, but aren’t claimed to penalize burial in every possible
example.
- That breaks (at least one version of) "Double Defeat". B is pairwise
> beaten by a candidate with a higher "score".
>>> Chris B.
>>> On 14/10/2023 4:43 am, Michael Ossipoff wrote:
>> Yes, I like Duncan because burying the CW in an attempt to help your
> favorite won’t help hir when it causes hir disqualification, as it probably
> will.
>> …& Duncan is remarkably briefly-defined, needing only a very slight
> modification of Black’s method.
*>
BTW, Forest has suggested a small modification of the CTE that I posted
here:
Instead of eliminating the lowest Borda-scorer, eliminate the pair-loser of
the 2 lowest Borda-scorers.
Even with only 3 candidates, that won’t elect BF unless CW is the highest
Borda-scorer.
If CW is the lowest Borda-scorer, which is most likely, Bus wins & the
burial is penalized.
With more candidates, & therefore likely more Buses, it becomes even less
likely for the winner to not be a Bus.
On Mon, Oct 23, 2023 at 06:01 C.Benham cbenham@adam.com.au wrote:
Are you referring to how equal-ranking is counted in Borda?
Yes, mainly truncated ballots. Do you consider that truncation is the
same thing as ranking equal bottom?
I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.
Can you clarify the exact meaning of the phrase "in the ranking"? Is this
a hint that your answer to my last
question is "no"?
And why is that what you advocate? In some online discussion I saw it was
mentioned (I think by N.Tideman) that
the Baldwin method doesn't meet Condorcet unless the Fractional (i.e.
based on the symmetrically completed ballots)
version of Borda count is used.
So I've tended to assume that that is the "correct" way of doing Borda
counts.
Say there are three candidates A,B,C. Say 46 ballots bullet-vote A. In
the fractional version these votes are counted the same as
23 A>B, 23 A>C, giving 92 points to A and 23 points each to B and C.
Is that the same as what you advocate, or would you have those truncating
ballots give 2 points to A and zero points to both of B and C?
Or something else?
- So another way of putting it is: "If there is no CW, elect the member of
** the Smith set with the second-worst score".
Would you mind telling why that’s another way of putting it?
By definition the members of the Smith set all pairwise beat all the
non-member candidates, and also have a pairwise defeat at the hands
of one of the other members (assuming there are three members).
So the Smith set member with the highest score must be pairwise beaten by
one with a lower score. That much is clear.
To be honest I may have confused myself again as to further details. It
seems to be possible for the lowest-scored Smith-set member to win.
Say the top cycle is (in terms of score order) Middle > High > Low >
Middle.
Then High is disqualified by being pairwise beaten by Middle and Middle is
disqualified by being pairwise beaten by Low.
So Low wins.
Remember that sincere top-cycles are vanishingly rare.
What mind-reading technology have you accessed to determine that? I don't
know how you can know that.
Simple scenarios with lots of truncation and a top cycle look very
plausible to me, so I don't know why I should
accept your assurance that such a thing would be "vanishingly rare".
Also it seems to me you are giving yourself a marketing problem:
"Ok, we admit that Condorcet methods are very vulnerable to Burial
strategy, so in a effort to combat that we have
to employ this apparently nonsensical, anti-monotonic, anti-intuitive
completion method.
But don't worry, probably we'll never have to use it."
Chris B.
Michael Ossipoff email9648742 at gmail.com
<election-methods%40lists.electorama.com?Subject=Re%3A%20%5BEM%5D%20Duncan%20Proposal%20Draft&In-Reply-To=%3CCAOKDY5D-CUNWajrZxSi8VauLY8FAys8vnt6SvX%3D7KiQFeEpeJw%40mail.gmail.com%3E> Sat
Oct 21 12:38:58 PDT 2023
On Sat, Oct 21, 2023 at 11:05 C.Benham <cbenham at adam.com.au http://lists.electorama.com/listinfo.cgi/election-methods-electorama.com> wrote:
"Am I right in assuming that the Borda counts are based on the symmetrically
completed ballots?"
Are you referring to how equal-ranking is counted in Borda? I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.
- Duncan Definition:
>> In the vast majority of the cases ... those in which the pairwise counts
> of the ballots unambiguously identify the candidate that pairbeats each of
> the others ... elect that candidate.
>> Otherwise, elect the highest score candidate that pairbeats every
> candidate with lower score.
>>> So another way of putting it is: "If there is no CW, elect the member of
> the Smith set with the second-worst score".
*>
Would you mind telling why that’s another way of putting it?
- To put it bluntly, that is bound to have monotonicity problems and doesn't
> fly philosophically.
*>
How doesn’t it fly philosophically?
- Trying to deter or frustrate order-reversal Burial strategy is fine, but
> the algorithm should "appear fair" and be able to be justified when
> we assume that all the votes are sincere (or even just all equally likely
> to be sincere).
*>
Remember that sincere top-cycles are vanishingly rare. That’s why
Sequential-Pairwise’s Pareto failure isn’t important, & it’s why
MinMax(wv)’s Condorcet Loser failure isn’t important.
It matters much more what happens in a strategic cycle. Does the method
reward or penalize burial? …or neither?
Duncan tends to often do neither, because it will likely disqualify both
Bus & BF. (I defined those usages when I defined CTE.
I prefer CTE to Duncan, because it more often penalizes, instead of merely
not rewarding.
But I remind you that these methods are intended to deter
probabilisticallly, but aren’t claimed to penalize burial in every possible
example.
- That breaks (at least one version of) "Double Defeat". B is pairwise
> beaten by a candidate with a higher "score".
>>> Chris B.
>>> On 14/10/2023 4:43 am, Michael Ossipoff wrote:
>> Yes, I like Duncan because burying the CW in an attempt to help your
> favorite won’t help hir when it causes hir disqualification, as it probably
> will.
>> …& Duncan is remarkably briefly-defined, needing only a very slight
> modification of Black’s method.
*>
I should add that a rule has been added to eliminate all candidates not in
Smith set before starting the count.
On Mon, Oct 23, 2023 at 06:01 C.Benham cbenham@adam.com.au wrote:
Are you referring to how equal-ranking is counted in Borda?
Yes, mainly truncated ballots. Do you consider that truncation is the
same thing as ranking equal bottom?
I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.
Can you clarify the exact meaning of the phrase "in the ranking"? Is this
a hint that your answer to my last
question is "no"?
And why is that what you advocate? In some online discussion I saw it was
mentioned (I think by N.Tideman) that
the Baldwin method doesn't meet Condorcet unless the Fractional (i.e.
based on the symmetrically completed ballots)
version of Borda count is used.
So I've tended to assume that that is the "correct" way of doing Borda
counts.
Say there are three candidates A,B,C. Say 46 ballots bullet-vote A. In
the fractional version these votes are counted the same as
23 A>B, 23 A>C, giving 92 points to A and 23 points each to B and C.
Is that the same as what you advocate, or would you have those truncating
ballots give 2 points to A and zero points to both of B and C?
Or something else?
- So another way of putting it is: "If there is no CW, elect the member of
** the Smith set with the second-worst score".
Would you mind telling why that’s another way of putting it?
By definition the members of the Smith set all pairwise beat all the
non-member candidates, and also have a pairwise defeat at the hands
of one of the other members (assuming there are three members).
So the Smith set member with the highest score must be pairwise beaten by
one with a lower score. That much is clear.
To be honest I may have confused myself again as to further details. It
seems to be possible for the lowest-scored Smith-set member to win.
Say the top cycle is (in terms of score order) Middle > High > Low >
Middle.
Then High is disqualified by being pairwise beaten by Middle and Middle is
disqualified by being pairwise beaten by Low.
So Low wins.
Remember that sincere top-cycles are vanishingly rare.
What mind-reading technology have you accessed to determine that? I don't
know how you can know that.
Simple scenarios with lots of truncation and a top cycle look very
plausible to me, so I don't know why I should
accept your assurance that such a thing would be "vanishingly rare".
Also it seems to me you are giving yourself a marketing problem:
"Ok, we admit that Condorcet methods are very vulnerable to Burial
strategy, so in a effort to combat that we have
to employ this apparently nonsensical, anti-monotonic, anti-intuitive
completion method.
But don't worry, probably we'll never have to use it."
Chris B.
Michael Ossipoff email9648742 at gmail.com
<election-methods%40lists.electorama.com?Subject=Re%3A%20%5BEM%5D%20Duncan%20Proposal%20Draft&In-Reply-To=%3CCAOKDY5D-CUNWajrZxSi8VauLY8FAys8vnt6SvX%3D7KiQFeEpeJw%40mail.gmail.com%3E> Sat
Oct 21 12:38:58 PDT 2023
On Sat, Oct 21, 2023 at 11:05 C.Benham <cbenham at adam.com.au http://lists.electorama.com/listinfo.cgi/election-methods-electorama.com> wrote:
"Am I right in assuming that the Borda counts are based on the symmetrically
completed ballots?"
Are you referring to how equal-ranking is counted in Borda? I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.
- Duncan Definition:
>> In the vast majority of the cases ... those in which the pairwise counts
> of the ballots unambiguously identify the candidate that pairbeats each of
> the others ... elect that candidate.
>> Otherwise, elect the highest score candidate that pairbeats every
> candidate with lower score.
>>> So another way of putting it is: "If there is no CW, elect the member of
> the Smith set with the second-worst score".
*>
Would you mind telling why that’s another way of putting it?
- To put it bluntly, that is bound to have monotonicity problems and doesn't
> fly philosophically.
*>
How doesn’t it fly philosophically?
- Trying to deter or frustrate order-reversal Burial strategy is fine, but
> the algorithm should "appear fair" and be able to be justified when
> we assume that all the votes are sincere (or even just all equally likely
> to be sincere).
*>
Remember that sincere top-cycles are vanishingly rare. That’s why
Sequential-Pairwise’s Pareto failure isn’t important, & it’s why
MinMax(wv)’s Condorcet Loser failure isn’t important.
It matters much more what happens in a strategic cycle. Does the method
reward or penalize burial? …or neither?
Duncan tends to often do neither, because it will likely disqualify both
Bus & BF. (I defined those usages when I defined CTE.
I prefer CTE to Duncan, because it more often penalizes, instead of merely
not rewarding.
But I remind you that these methods are intended to deter
probabilisticallly, but aren’t claimed to penalize burial in every possible
example.
- That breaks (at least one version of) "Double Defeat". B is pairwise
> beaten by a candidate with a higher "score".
>>> Chris B.
>>> On 14/10/2023 4:43 am, Michael Ossipoff wrote:
>> Yes, I like Duncan because burying the CW in an attempt to help your
> favorite won’t help hir when it causes hir disqualification, as it probably
> will.
>> …& Duncan is remarkably briefly-defined, needing only a very slight
> modification of Black’s method.
*>
On Mon, Oct 23, 2023, 1:52 PM Michael Ossipoff email9648742@gmail.com
wrote:
I should add that a rule has been added to eliminate all candidates not in
Smith set before starting the count.
Because it is easy to show that the buried CW is never covered ... in fact
the only candidates that beat the buried CW are buses ... which are beaten
by CW in two step beatpaths via the BF candidate.
On Mon, Oct 23, 2023 at 06:01 C.Benham cbenham@adam.com.au wrote:
Are you referring to how equal-ranking is counted in Borda?
Yes, mainly truncated ballots. Do you consider that truncation is the
same thing as ranking equal bottom?
I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.
Can you clarify the exact meaning of the phrase "in the ranking"? Is
this a hint that your answer to my last
question is "no"?
And why is that what you advocate? In some online discussion I saw it
was mentioned (I think by N.Tideman) that
the Baldwin method doesn't meet Condorcet unless the Fractional (i.e.
based on the symmetrically completed ballots)
version of Borda count is used.
So I've tended to assume that that is the "correct" way of doing Borda
counts.
Say there are three candidates A,B,C. Say 46 ballots bullet-vote A. In
the fractional version these votes are counted the same as
23 A>B, 23 A>C, giving 92 points to A and 23 points each to B and C.
Is that the same as what you advocate, or would you have those truncating
ballots give 2 points to A and zero points to both of B and C?
Or something else?
- So another way of putting it is: "If there is no CW, elect the member of
** the Smith set with the second-worst score".
Would you mind telling why that’s another way of putting it?
By definition the members of the Smith set all pairwise beat all the
non-member candidates, and also have a pairwise defeat at the hands
of one of the other members (assuming there are three members).
So the Smith set member with the highest score must be pairwise beaten by
one with a lower score. That much is clear.
To be honest I may have confused myself again as to further details. It
seems to be possible for the lowest-scored Smith-set member to win.
Say the top cycle is (in terms of score order) Middle > High > Low >
Middle.
Then High is disqualified by being pairwise beaten by Middle and Middle
is disqualified by being pairwise beaten by Low.
So Low wins.
Remember that sincere top-cycles are vanishingly rare.
What mind-reading technology have you accessed to determine that? I
don't know how you can know that.
Simple scenarios with lots of truncation and a top cycle look very
plausible to me, so I don't know why I should
accept your assurance that such a thing would be "vanishingly rare".
Also it seems to me you are giving yourself a marketing problem:
"Ok, we admit that Condorcet methods are very vulnerable to Burial
strategy, so in a effort to combat that we have
to employ this apparently nonsensical, anti-monotonic, anti-intuitive
completion method.
But don't worry, probably we'll never have to use it."
Chris B.
Michael Ossipoff email9648742 at gmail.com
<election-methods%40lists.electorama.com?Subject=Re%3A%20%5BEM%5D%20Duncan%20Proposal%20Draft&In-Reply-To=%3CCAOKDY5D-CUNWajrZxSi8VauLY8FAys8vnt6SvX%3D7KiQFeEpeJw%40mail.gmail.com%3E> Sat
Oct 21 12:38:58 PDT 2023
On Sat, Oct 21, 2023 at 11:05 C.Benham <cbenham at adam.com.au http://lists.electorama.com/listinfo.cgi/election-methods-electorama.com> wrote:
"Am I right in assuming that the Borda counts are based on the symmetrically
completed ballots?"
Are you referring to how equal-ranking is counted in Borda? I advocate
that, in these methods’ Borda-count, a ranking give each of its
ranked-candidates a number of points equal to the number of candidates in
the ranking who aren’t ranked over hir.
- Duncan Definition:
>> In the vast majority of the cases ... those in which the pairwise counts
> of the ballots unambiguously identify the candidate that pairbeats each of
> the others ... elect that candidate.
>> Otherwise, elect the highest score candidate that pairbeats every
> candidate with lower score.
>>> So another way of putting it is: "If there is no CW, elect the member of
> the Smith set with the second-worst score".
*>
Would you mind telling why that’s another way of putting it?
- To put it bluntly, that is bound to have monotonicity problems and doesn't
> fly philosophically.
*>
How doesn’t it fly philosophically?
- Trying to deter or frustrate order-reversal Burial strategy is fine, but
> the algorithm should "appear fair" and be able to be justified when
> we assume that all the votes are sincere (or even just all equally likely
> to be sincere).
*>
Remember that sincere top-cycles are vanishingly rare. That’s why
Sequential-Pairwise’s Pareto failure isn’t important, & it’s why
MinMax(wv)’s Condorcet Loser failure isn’t important.
It matters much more what happens in a strategic cycle. Does the method
reward or penalize burial? …or neither?
Duncan tends to often do neither, because it will likely disqualify both
Bus & BF. (I defined those usages when I defined CTE.
I prefer CTE to Duncan, because it more often penalizes, instead of merely
not rewarding.
But I remind you that these methods are intended to deter
probabilisticallly, but aren’t claimed to penalize burial in every possible
example.
- That breaks (at least one version of) "Double Defeat". B is pairwise
> beaten by a candidate with a higher "score".
>>> Chris B.
>>> On 14/10/2023 4:43 am, Michael Ossipoff wrote:
>> Yes, I like Duncan because burying the CW in an attempt to help your
> favorite won’t help hir when it causes hir disqualification, as it probably
> will.
>> …& Duncan is remarkably briefly-defined, needing only a very slight
> modification of Black’s method.
*>
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