Smith//3Score (was MAM vs Schulze)
(Continued below)
Now, how do we adapt this to general rankings? We assume that equal top
rankings and equal bottom or multiple truncations are allowed.
For each ballot on which a candidate is ranked above bottom but below top
that candidate receives one point. For each ballot on which the candidate
is ranked top or equal top that candidate receives two points.
The Smith candidate with the greatest number of points wins.
[End of definition]
Note that the method does satisfy CD unlike Smith//ImplicitApproval.
Jameson's idea of three slot scores makes it work.
How does it do on burial?
Note that for three candidates and complete rankings the method is just
Smith//Borda. But Smith//Borda is extremely clone dependent. The three
category (top, middle, bottom) point system is key to making the method
clone proof.
For public proposal it is important to have a non-cryptic definition of
Smith.
The smallest set each of whose members pairwise beats every candidate
outside the set is succinct, but probably (by itself) too cryptic for the
non-technical reader.
I think the idea of a beat-path is very intuitive. I remember that ten or
fifteen years ago Michael said that he had success explaining Schulze to
various people on the street from the beat-path perspective.
If A beats B beats C beats D, then that chain of defeats is called a
beat-path from A to D.
The human brain is adept at generalizing this kind of definition to a path
with more or fewer steps..
For the purpose of defining the Smith set we don't even need to talk about
the strength of a beat-path.
Candidate X is a member of Smith iff for each candidate Y (other than X
itself), there is a beat-path from X to Y.
In typical public elections we expect Smith to be a cycle of three. In
that case Smith//3Score is the same as Max Covering 3Score. In general the
latter is preferable because no matter how large Smith may be it continues
to elect an uncovered candidate.
For public consumption it is mainly a question of ease of understanding.
Which is easier to understand a sequence of successive pairwise defeats or
a covering relation.
Another way to elect the same candidate when the top cycle has only three
members is to list the candidates by their point totals (highest total at
the top of the list) and starting with the bottom candidate move up the
list replacing pairwise losers with winners.
In other words the winner will be the highest listed candidate that beats
both the champion from below and every (if any) candidate above it in the
list.
I would be perfectly happy with any of these proposals, though there is an
obvious soft spot in my heart for the covering version.
Another reason I like the covering version is that various strengthenings
of the covering relation can be put to good use to refine this method.
More on that in another post.
Forest
Yes, the method still allows chicken dilemma defection to succeed.
I tried an example in which the B faction is about as large as possible in
comparison to the C faction.
99: A>B
98: B (sincere is B>C)
100: C
This results in a cycle, so everyone is in the Smith-set.
B has more Borda points than anyone else.
It looks as if it isn't possible to have CD in a strategically good,
reliable, uncriticizable. rank method.
ICT isn't good as a ranking method. Only as a 3-slot method in which the
middle rating is used only in Chicken-Dilemma situations.
In ICT, a candidate you rank middle doesn't get much protection from you.
But, in a chicken dilemma situation you don't expect B to be a CWs anyway.
3-Slot ICT is still my favorite, to be used as an Approval version rather
than a ranking method, with the middle rating only for chicken dilemma.
Plain MMPO meets Weak CD, FBC, LNHa, and has wv-like strategy.
Though it fails CD's strong original version, if the defection is a
burial--The method's wv burial defensive advice would warn the A voters to
not rank B, if A is likely to be CWs.
That's a unique, impressive & amazing set of advantages.
But there are some strongly-felt criticisms to it. I've answered them, and
it seems to me that only one of them is a genuine problem:
...the possibility of the perpetual burial fiasco.
But I've told here why there's something a bit mutually contradictory about
that fiasco's requirements.
So I suggest that it doesn't rule out MMPO or MAM, though it makes them
just a little questionable & unreliable.
...but still worth a try because of big advantages.
I'd hoped that Bucklin with conditional votes would be a good CD method.
But it's not as good as I'd hoped, because the conditional votes option can
be strategically taken advantage of, resulting in another chicken dilemma,
differently-caused.
I don't know if that rules out the conditional option, but it supports the
conclusion that a CD rank method always costs.
Michael Ossipoff
On Oct 9, 2016 3:19 PM, "Forest Simmons" fsimmons@pcc.edu wrote:
(Continued below)
Now, how do we adapt this to general rankings? We assume that equal top
rankings and equal bottom or multiple truncations are allowed.
For each ballot on which a candidate is ranked above bottom but below top
that candidate receives one point. For each ballot on which the candidate
is ranked top or equal top that candidate receives two points.
The Smith candidate with the greatest number of points wins.
[End of definition]
Note that the method does satisfy CD unlike Smith//ImplicitApproval.
Jameson's idea of three slot scores makes it work.
How does it do on burial?
Note that for three candidates and complete rankings the method is just
Smith//Borda. But Smith//Borda is extremely clone dependent. The three
category (top, middle, bottom) point system is key to making the method
clone proof.
For public proposal it is important to have a non-cryptic definition of
Smith.
The smallest set each of whose members pairwise beats every candidate
outside the set is succinct, but probably (by itself) too cryptic for the
non-technical reader.
I think the idea of a beat-path is very intuitive. I remember that ten or
fifteen years ago Michael said that he had success explaining Schulze to
various people on the street from the beat-path perspective.
If A beats B beats C beats D, then that chain of defeats is called a
beat-path from A to D.
The human brain is adept at generalizing this kind of definition to a path
with more or fewer steps..
For the purpose of defining the Smith set we don't even need to talk about
the strength of a beat-path.
Candidate X is a member of Smith iff for each candidate Y (other than X
itself), there is a beat-path from X to Y.
In typical public elections we expect Smith to be a cycle of three. In
that case Smith//3Score is the same as Max Covering 3Score. In general the
latter is preferable because no matter how large Smith may be it continues
to elect an uncovered candidate.
For public consumption it is mainly a question of ease of understanding.
Which is easier to understand a sequence of successive pairwise defeats or
a covering relation.
Another way to elect the same candidate when the top cycle has only three
members is to list the candidates by their point totals (highest total at
the top of the list) and starting with the bottom candidate move up the
list replacing pairwise losers with winners.
In other words the winner will be the highest listed candidate that beats
both the champion from below and every (if any) candidate above it in the
list.
I would be perfectly happy with any of these proposals, though there is an
obvious soft spot in my heart for the covering version.
Another reason I like the covering version is that various strengthenings
of the covering relation can be put to good use to refine this method.
More on that in another post.
Forest
Election-Methods mailing list - see http://electorama.com/em for list info
AS Chris Benham and Michael Ossipoff pointed out this Smith//3Score doesn't
disappoint the defecting faction (B) unless the plumping faction (C) is
fairly close to half of the electorate. It only works when C is plumped on
at least 43 percent of the ballots.
To make it work for Mike's example below where C is plumped on only 100 out
of 297 ballots, the middle ranks have to count much les than half of the
equal top ranks. About 2 percent of the equal top value would do.
So with sincere ballots the point totals are ...
99+.02(98) = 100.96 for A,
98+ .02(99)= 99.98 for B, and
100 for C. In this case A is both the CW and the points winner.
The B faction's defection simultaneously promotes C to the Smith set and
reduces A's point total to 99, making C (still with 100 points) the method
winner.
What makes this example hard is that the C faction is just over one third
of the electorate, and that the A and B subfactions are very close in size.
For a defection attempt to succeed under these conditions the B faction
would have to possess very precise information. If the A faction had the
same information it would be easy for them to make a defensive move in the
form of truncating B on a few ballots.
In practice, a point value substantially larger than .02 for the middle
ranks would be adequate.
What would be a reasonable compromise?
How about 1/4 or 25% ?
In that case the A faction could say to the B faction, "You rank me on two
ballots and I will rank you on seven."
The resulting equilibrium would be
92 A
7 A>B
96 B
2 B>A
100 C
Candidate A is elected as the CW.
If B defects from this equilibrium proposal ...
92 A
7 A>B
98 B
100 C
then {A, B, C} forms Smith, and C wins with 100 points, while B gets only
98+7/4 = 99.75 points.
If grade style ballots were in use, the A faction could say I will give you
two B grades in exchange for two D's.
Two B grade points add up to 1.5 which would bring the B candidate up to
99.5, still short of the 100 needed to tie the C candidate.
On the other hand, the two D grades for the A candidate would make her the
Condorcet Winner.
When the plumping faction forms greater than 39 percent of the electorate,
then no strategy would be needed other than for the larger subfaction to
give D grades to the potential defection candidate.
32 A(4), B(1)
31 B(4) (Sincere B>0)
39 C(4)
If eight or more of the B faction give A a positive rating, then A wins as
the only member of the Smith set.
If fewer than eight vote sincerely, then C and B are tied with 156 grade
points each.
So the method I am now proposing is
Smith//GPA
Elect the member of the Smith set with the largest Grade Point Average.
From: Michael Ossipoff
Yes, the method still allows chicken dilemma defection to succeed.
I tried an example in which the B faction is about as large as possible in
comparison to the C faction.
99: A>B
98: B (sincere is B>A)
100: C
This results in a cycle, so everyone is in the Smith-set.
B has more Borda points than anyone else.
It looks as if it isn't possible to have CD in a strategically good,
reliable, uncriticizable. rank method.
ICT isn't good as a ranking method. Only as a 3-slot method in which the
middle rating is used only in Chicken-Dilemma situations.
In ICT, a candidate you rank middle doesn't get much protection from you.
But, in a chicken dilemma situation you don't expect B to be a CWs anyway.
3-Slot ICT is still my favorite, to be used as an Approval version rather
than a ranking method, with the middle rating only for chicken dilemma.
Plain MMPO meets Weak CD, FBC, LNHa, and has wv-like strategy.
Though it fails CD's strong original version, if the defection is a
burial--The method's wv burial defensive advice would warn the A voters to
not rank B, if A is likely to be CWs.
That's a unique, impressive & amazing set of advantages.
But there are some strongly-felt criticisms to it. I've answered them, and
it seems to me that only one of them is a genuine problem:
...the possibility of the perpetual burial fiasco.
But I've told here why there's something a bit mutually contradictory about
that fiasco's requirements.
So I suggest that it doesn't rule out MMPO or MAM, though it makes them
just a little questionable & unreliable.
...but still worth a try because of big advantages.
I'd hoped that Bucklin with conditional votes would be a good CD method.
But it's not as good as I'd hoped, because the conditional votes option can
be strategically taken advantage of, resulting in another chicken dilemma,
differently-caused.
I don't know if that rules out the conditional option, but it supports the
conclusion that a CD rank method always costs.
Michael Ossipoff
On Oct 9, 2016 3:19 PM, "Forest Simmons" fsimmons@pcc.edu wrote:
(Continued below)
Now, how do we adapt this to general rankings? We assume that equal top
rankings and equal bottom or multiple truncations are allowed.
For each ballot on which a candidate is ranked above bottom but below
top
that candidate receives one point. For each ballot on which the
candidate
is ranked top or equal top that candidate receives two points.
The Smith candidate with the greatest number of points wins.
[End of definition]
Note that the method does satisfy CD unlike Smith//ImplicitApproval.
Jameson's idea of three slot scores makes it work.
How does it do on burial?
Forest,
Why do you want to give any points at all for middle ratings?
What is wrong with simply 3-slot Smith//Top Ratings?
Or 3-slot TTR, Top Ratings (aka ICT)?
Chris Benham
On 10/11/2016 9:22 AM, Forest Simmons wrote:
AS Chris Benham and Michael Ossipoff pointed out this Smith//3Score
doesn't disappoint the defecting faction (B) unless the plumping
faction (C) is fairly close to half of the electorate. It only works
when C is plumped on at least 43 percent of the ballots.
To make it work for Mike's example below where C is plumped on only
100 out of 297 ballots, the middle ranks have to count much les than
half of the equal top ranks. About 2 percent of the equal top value
would do.
So with sincere ballots the point totals are ...
99+.02(98) = 100.96 for A,
98+ .02(99)= 99.98 for B, and
100 for C. In this case A is both the CW and the points winner.
The B faction's defection simultaneously promotes C to the Smith set
and reduces A's point total to 99, making C (still with 100 points)
the method winner.
What makes this example hard is that the C faction is just over one
third of the electorate, and that the A and B subfactions are very
close in size.
For a defection attempt to succeed under these conditions the B
faction would have to possess very precise information. If the A
faction had the same information it would be easy for them to make a
defensive move in the form of truncating B on a few ballots.
In practice, a point value substantially larger than .02 for the
middle ranks would be adequate.
What would be a reasonable compromise?
How about 1/4 or 25% ?
In that case the A faction could say to the B faction, "You rank me on
two ballots and I will rank you on seven."
The resulting equilibrium would be
92 A
7 A>B
96 B
2 B>A
100 C
Candidate A is elected as the CW.
If B defects from this equilibrium proposal ...
92 A
7 A>B
98 B
100 C
then {A, B, C} forms Smith, and C wins with 100 points, while B gets only
98+7/4 = 99.75 points.
If grade style ballots were in use, the A faction could say I will
give you two B grades in exchange for two D's.
Two B grade points add up to 1.5 which would bring the B candidate up
to 99.5, still short of the 100 needed to tie the C candidate.
On the other hand, the two D grades for the A candidate would make her
the Condorcet Winner.
When the plumping faction forms greater than 39 percent of the
electorate, then no strategy would be needed other than for the larger
subfaction to give D grades to the potential defection candidate.
32 A(4), B(1)
31 B(4) (Sincere B>0)
39 C(4)
If eight or more of the B faction give A a positive rating, then A
wins as the only member of the Smith set.
If fewer than eight vote sincerely, then C and B are tied with 156
grade points each.
So the method I am now proposing is
Smith//GPA
Elect the member of the Smith set with the largest Grade Point Average.
From: Michael Ossipoff
Yes, the method still allows chicken dilemma defection to succeed.
I tried an example in which the B faction is about as large as
possible in
comparison to the C faction.
99: A>B
98: B (sincere is B>A)
100: C
This results in a cycle, so everyone is in the Smith-set.
B has more Borda points than anyone else.
It looks as if it isn't possible to have CD in a strategically good,
reliable, uncriticizable. rank method.
ICT isn't good as a ranking method. Only as a 3-slot method in
which the
middle rating is used only in Chicken-Dilemma situations.
In ICT, a candidate you rank middle doesn't get much protection
from you.
But, in a chicken dilemma situation you don't expect B to be a CWs
anyway.
3-Slot ICT is still my favorite, to be used as an Approval version
rather
than a ranking method, with the middle rating only for chicken
dilemma.
Plain MMPO meets Weak CD, FBC, LNHa, and has wv-like strategy.
Though it fails CD's strong original version, if the defection is a
burial--The method's wv burial defensive advice would warn the A
voters to
not rank B, if A is likely to be CWs.
That's a unique, impressive & amazing set of advantages.
But there are some strongly-felt criticisms to it. I've answered
them, and
it seems to me that only one of them is a genuine problem:
...the possibility of the perpetual burial fiasco.
But I've told here why there's something a bit mutually
contradictory about
that fiasco's requirements.
So I suggest that it doesn't rule out MMPO or MAM, though it makes
them
just a little questionable & unreliable.
...but still worth a try because of big advantages.
I'd hoped that Bucklin with conditional votes would be a good CD
method.
But it's not as good as I'd hoped, because the conditional votes
option can
be strategically taken advantage of, resulting in another chicken
dilemma,
differently-caused.
I don't know if that rules out the conditional option, but it
supports the
conclusion that a CD rank method always costs.
Michael Ossipoff
On Oct 9, 2016 3:19 PM, "Forest Simmons" <fsimmons@pcc.edu
<mailto:fsimmons@pcc.edu>> wrote:
(Continued below)
Now, how do we adapt this to general rankings? We assume that
equal top
rankings and equal bottom or multiple truncations are allowed.
For each ballot on which a candidate is ranked above bottom but
below top
that candidate receives one point. For each ballot on which
the candidate
is ranked top or equal top that candidate receives two points.
The Smith candidate with the greatest number of points wins.
[End of definition]
Note that the method does satisfy CD unlike
Smith//ImplicitApproval.
Jameson's idea of three slot scores makes it work.
How does it do on burial?
Election-Methods mailing list - see http://electorama.com/em for list info
No virus found in this message.
Checked by AVG - www.avg.com http://www.avg.com
Version: 2016.0.7797 / Virus Database: 4664/13183 - Release Date: 10/10/16
Forrest--
But, when an anti-defection strategy requires a certain percentage to vote
a certain way, then 1 of 2 things is needed: 1) Probabilistic voting by the
subfaction; or 2) organization of voting by neighborhood or by voters'
names.
That complicates voting, and seems to cancel out the advantage of an easily
explained method definition.
A nice thing about plain MMPO is its simple, brief, naturally-motivated
definition, it's Weak CD, & it's wv strategy & FBC.
Michael Ossipoff e
On Oct 10, 2016 3:52 PM, "Forest Simmons" fsimmons@pcc.edu wrote:
AS Chris Benham and Michael Ossipoff pointed out this Smith//3Score
doesn't disappoint the defecting faction (B) unless the plumping faction
(C) is fairly close to half of the electorate. It only works when C is
plumped on at least 43 percent of the ballots.
To make it work for Mike's example below where C is plumped on only 100
out of 297 ballots, the middle ranks have to count much les than half of
the equal top ranks. About 2 percent of the equal top value would do.
So with sincere ballots the point totals are ...
99+.02(98) = 100.96 for A,
98+ .02(99)= 99.98 for B, and
100 for C. In this case A is both the CW and the points winner.
The B faction's defection simultaneously promotes C to the Smith set and
reduces A's point total to 99, making C (still with 100 points) the method
winner.
What makes this example hard is that the C faction is just over one third
of the electorate, and that the A and B subfactions are very close in size.
For a defection attempt to succeed under these conditions the B faction
would have to possess very precise information. If the A faction had the
same information it would be easy for them to make a defensive move in the
form of truncating B on a few ballots.
In practice, a point value substantially larger than .02 for the middle
ranks would be adequate.
What would be a reasonable compromise?
How about 1/4 or 25% ?
In that case the A faction could say to the B faction, "You rank me on two
ballots and I will rank you on seven."
The resulting equilibrium would be
92 A
7 A>B
96 B
2 B>A
100 C
Candidate A is elected as the CW.
If B defects from this equilibrium proposal ...
92 A
7 A>B
98 B
100 C
then {A, B, C} forms Smith, and C wins with 100 points, while B gets only
98+7/4 = 99.75 points.
If grade style ballots were in use, the A faction could say I will give
you two B grades in exchange for two D's.
Two B grade points add up to 1.5 which would bring the B candidate up to
99.5, still short of the 100 needed to tie the C candidate.
On the other hand, the two D grades for the A candidate would make her the
Condorcet Winner.
When the plumping faction forms greater than 39 percent of the electorate,
then no strategy would be needed other than for the larger subfaction to
give D grades to the potential defection candidate.
32 A(4), B(1)
31 B(4) (Sincere B>0)
39 C(4)
If eight or more of the B faction give A a positive rating, then A wins as
the only member of the Smith set.
If fewer than eight vote sincerely, then C and B are tied with 156 grade
points each.
So the method I am now proposing is
Smith//GPA
Elect the member of the Smith set with the largest Grade Point Average.
From: Michael Ossipoff
Yes, the method still allows chicken dilemma defection to succeed.
I tried an example in which the B faction is about as large as possible in
comparison to the C faction.
99: A>B
98: B (sincere is B>A)
100: C
This results in a cycle, so everyone is in the Smith-set.
B has more Borda points than anyone else.
It looks as if it isn't possible to have CD in a strategically good,
reliable, uncriticizable. rank method.
ICT isn't good as a ranking method. Only as a 3-slot method in which the
middle rating is used only in Chicken-Dilemma situations.
In ICT, a candidate you rank middle doesn't get much protection from you.
But, in a chicken dilemma situation you don't expect B to be a CWs anyway.
3-Slot ICT is still my favorite, to be used as an Approval version rather
than a ranking method, with the middle rating only for chicken dilemma.
Plain MMPO meets Weak CD, FBC, LNHa, and has wv-like strategy.
Though it fails CD's strong original version, if the defection is a
burial--The method's wv burial defensive advice would warn the A voters to
not rank B, if A is likely to be CWs.
That's a unique, impressive & amazing set of advantages.
But there are some strongly-felt criticisms to it. I've answered them, and
it seems to me that only one of them is a genuine problem:
...the possibility of the perpetual burial fiasco.
But I've told here why there's something a bit mutually contradictory
about
that fiasco's requirements.
So I suggest that it doesn't rule out MMPO or MAM, though it makes them
just a little questionable & unreliable.
...but still worth a try because of big advantages.
I'd hoped that Bucklin with conditional votes would be a good CD method.
But it's not as good as I'd hoped, because the conditional votes option
can
be strategically taken advantage of, resulting in another chicken dilemma,
differently-caused.
I don't know if that rules out the conditional option, but it supports the
conclusion that a CD rank method always costs.
Michael Ossipoff
On Oct 9, 2016 3:19 PM, "Forest Simmons" fsimmons@pcc.edu wrote:
(Continued below)
Now, how do we adapt this to general rankings? We assume that equal top
rankings and equal bottom or multiple truncations are allowed.
For each ballot on which a candidate is ranked above bottom but below
top
that candidate receives one point. For each ballot on which the
candidate
is ranked top or equal top that candidate receives two points.
The Smith candidate with the greatest number of points wins.
[End of definition]
Note that the method does satisfy CD unlike Smith//ImplicitApproval.
Jameson's idea of three slot scores makes it work.
How does it do on burial?
Election-Methods mailing list - see http://electorama.com/em for list info