FS
Forest Simmons
Wed, Apr 6, 2022 6:57 AM
On each ballot B identify the anti-favorite a'(B) as the candidate a' among
those bottom listed on ballot B that is bottom listed the most on the rest
of the ballots.
[A candidate is bottom listed on a ballot if it is ranked over no candidate
on that ballot]
Elect the candidate X that, on the most ballots B, either is ranked ahead
of or pairwise defeats a'(B).
In other words, elect argmax N(X), where N(X) is the cardinality of the set
of ballots
{B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
The key to monotonicity is that if some candidate X is uniquely raised on
some ballot B, then for any other candidate Y, the difference N(X)-N(Y)
does not decrease.
The only non-trivial case is where X starts in the a'(B) position and swaps
positions with some Z that was ranked above it.
This definitely adds a point to N(X) because X is newly ranked ahead of
a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
In any case, N(X)-N(Y) does not decrease.
Does this method satisfy the Condorcet Criterion?
Well, if X pairwise defeats all other candidates, then N(X) is equal to the
total number of ballots minus the number of ballots on which X is in the a'
position, which must be fewer than half of the ballots, otherwise X would
be the Condorcet loser.
Is it possible for N(Y) to be greater than this N(X) while losing pairwise
to the CW X?
Perhaps ... if so, then this method is not a stand alone Condorcet method.
It would be a valuable task for someone to test its Condorcet efficiency
experimentally.
Thanks!
-Forest
On each ballot B identify the anti-favorite a'(B) as the candidate a' among
those bottom listed on ballot B that is bottom listed the most on the rest
of the ballots.
[A candidate is bottom listed on a ballot if it is ranked over no candidate
on that ballot]
Elect the candidate X that, on the most ballots B, either is ranked ahead
of or pairwise defeats a'(B).
In other words, elect argmax N(X), where N(X) is the cardinality of the set
of ballots
{B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
The key to monotonicity is that if some candidate X is uniquely raised on
some ballot B, then for any other candidate Y, the difference N(X)-N(Y)
does not decrease.
The only non-trivial case is where X starts in the a'(B) position and swaps
positions with some Z that was ranked above it.
This definitely adds a point to N(X) because X is newly ranked ahead of
a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
In any case, N(X)-N(Y) does not decrease.
Does this method satisfy the Condorcet Criterion?
Well, if X pairwise defeats all other candidates, then N(X) is equal to the
total number of ballots minus the number of ballots on which X is in the a'
position, which must be fewer than half of the ballots, otherwise X would
be the Condorcet loser.
Is it possible for N(Y) to be greater than this N(X) while losing pairwise
to the CW X?
Perhaps ... if so, then this method is not a stand alone Condorcet method.
It would be a valuable task for someone to test its Condorcet efficiency
experimentally.
Thanks!
-Forest
JF
James Faran
Wed, Apr 6, 2022 1:55 PM
I may not understand the method, but consider the following with candidates X, Y, Z:
51 X<Y<Z
49 Y>Z>X
Then X is the CW, and X is ranked over or pairwise defeats a'(B) only on those ballots B in the group of 51. Thus N(X)=51.
Meanwhile, Y is ranked over a'(B) on all ballots, so N(Y)=100.
If I've understood the method correctly, it is not Condorcet efficient.
Jim Faran
From: Election-Methods election-methods-bounces@lists.electorama.com on behalf of Forest Simmons forest.simmons21@gmail.com
Sent: Wednesday, April 6, 2022 2:57 AM
To: EM Election-methods@lists.electorama.com; Kristofer Munsterhjelm km_elmet@t-online.de
Subject: [EM] Decloned Copeland Borda Hybrid
On each ballot B identify the anti-favorite a'(B) as the candidate a' among those bottom listed on ballot B that is bottom listed the most on the rest of the ballots.
[A candidate is bottom listed on a ballot if it is ranked over no candidate on that ballot]
Elect the candidate X that, on the most ballots B, either is ranked ahead of or pairwise defeats a'(B).
In other words, elect argmax N(X), where N(X) is the cardinality of the set of ballots
{B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
The key to monotonicity is that if some candidate X is uniquely raised on some ballot B, then for any other candidate Y, the difference N(X)-N(Y) does not decrease.
The only non-trivial case is where X starts in the a'(B) position and swaps positions with some Z that was ranked above it.
This definitely adds a point to N(X) because X is newly ranked ahead of a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
In any case, N(X)-N(Y) does not decrease.
Does this method satisfy the Condorcet Criterion?
Well, if X pairwise defeats all other candidates, then N(X) is equal to the total number of ballots minus the number of ballots on which X is in the a' position, which must be fewer than half of the ballots, otherwise X would be the Condorcet loser.
Is it possible for N(Y) to be greater than this N(X) while losing pairwise to the CW X?
Perhaps ... if so, then this method is not a stand alone Condorcet method.
It would be a valuable task for someone to test its Condorcet efficiency experimentally.
Thanks!
-Forest
I may not understand the method, but consider the following with candidates X, Y, Z:
51 X<Y<Z
49 Y>Z>X
Then X is the CW, and X is ranked over or pairwise defeats a'(B) only on those ballots B in the group of 51. Thus N(X)=51.
Meanwhile, Y is ranked over a'(B) on all ballots, so N(Y)=100.
If I've understood the method correctly, it is not Condorcet efficient.
Jim Faran
________________________________
From: Election-Methods <election-methods-bounces@lists.electorama.com> on behalf of Forest Simmons <forest.simmons21@gmail.com>
Sent: Wednesday, April 6, 2022 2:57 AM
To: EM <Election-methods@lists.electorama.com>; Kristofer Munsterhjelm <km_elmet@t-online.de>
Subject: [EM] Decloned Copeland Borda Hybrid
On each ballot B identify the anti-favorite a'(B) as the candidate a' among those bottom listed on ballot B that is bottom listed the most on the rest of the ballots.
[A candidate is bottom listed on a ballot if it is ranked over no candidate on that ballot]
Elect the candidate X that, on the most ballots B, either is ranked ahead of or pairwise defeats a'(B).
In other words, elect argmax N(X), where N(X) is the cardinality of the set of ballots
{B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
The key to monotonicity is that if some candidate X is uniquely raised on some ballot B, then for any other candidate Y, the difference N(X)-N(Y) does not decrease.
The only non-trivial case is where X starts in the a'(B) position and swaps positions with some Z that was ranked above it.
This definitely adds a point to N(X) because X is newly ranked ahead of a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
In any case, N(X)-N(Y) does not decrease.
Does this method satisfy the Condorcet Criterion?
Well, if X pairwise defeats all other candidates, then N(X) is equal to the total number of ballots minus the number of ballots on which X is in the a' position, which must be fewer than half of the ballots, otherwise X would be the Condorcet loser.
Is it possible for N(Y) to be greater than this N(X) while losing pairwise to the CW X?
Perhaps ... if so, then this method is not a stand alone Condorcet method.
It would be a valuable task for someone to test its Condorcet efficiency experimentally.
Thanks!
-Forest
FS
Forest Simmons
Wed, Apr 6, 2022 7:52 PM
James,
you are absolutely right ... as I stated it the method is basically just
Implicit Approval ... definitely not Condorcet efficient.
I intended to say each candidate ranked above or pairwise beating a' gets a
point, and two points for both.
The monotonicity proof goes through unchanged.... as z goes down, y gains a
point, but so does X as it goes up.
Thanks for taking the time to check this out!
Forest
El mié., 6 de abr. de 2022 6:55 a. m., James Faran jjfaran@buffalo.edu
escribió:
I may not understand the method, but consider the following with
candidates X, Y, Z:
51 X<Y<Z
49 Y>Z>X
Then X is the CW, and X is ranked over or pairwise defeats a'(B) only on
those ballots B in the group of 51. Thus N(X)=51.
Meanwhile, Y is ranked over a'(B) on all ballots, so N(Y)=100.
If I've understood the method correctly, it is not Condorcet efficient.
Jim Faran
From: Election-Methods election-methods-bounces@lists.electorama.com
on behalf of Forest Simmons forest.simmons21@gmail.com
Sent: Wednesday, April 6, 2022 2:57 AM
To: EM Election-methods@lists.electorama.com; Kristofer Munsterhjelm <
km_elmet@t-online.de>
Subject: [EM] Decloned Copeland Borda Hybrid
On each ballot B identify the anti-favorite a'(B) as the candidate a'
among those bottom listed on ballot B that is bottom listed the most on the
rest of the ballots.
[A candidate is bottom listed on a ballot if it is ranked over no
candidate on that ballot]
Elect the candidate X that, on the most ballots B, either is ranked ahead
of or pairwise defeats a'(B).
In other words, elect argmax N(X), where N(X) is the cardinality of the
set of ballots
{B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
The key to monotonicity is that if some candidate X is uniquely raised on
some ballot B, then for any other candidate Y, the difference N(X)-N(Y)
does not decrease.
The only non-trivial case is where X starts in the a'(B) position and
swaps positions with some Z that was ranked above it.
This definitely adds a point to N(X) because X is newly ranked ahead of
a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
In any case, N(X)-N(Y) does not decrease.
Does this method satisfy the Condorcet Criterion?
Well, if X pairwise defeats all other candidates, then N(X) is equal to
the total number of ballots minus the number of ballots on which X is in
the a' position, which must be fewer than half of the ballots, otherwise X
would be the Condorcet loser.
Is it possible for N(Y) to be greater than this N(X) while losing pairwise
to the CW X?
Perhaps ... if so, then this method is not a stand alone Condorcet method.
It would be a valuable task for someone to test its Condorcet efficiency
experimentally.
Thanks!
-Forest
James,
you are absolutely right ... as I stated it the method is basically just
Implicit Approval ... definitely not Condorcet efficient.
I intended to say each candidate ranked above or pairwise beating a' gets a
point, and two points for both.
The monotonicity proof goes through unchanged.... as z goes down, y gains a
point, but so does X as it goes up.
Thanks for taking the time to check this out!
Forest
El mié., 6 de abr. de 2022 6:55 a. m., James Faran <jjfaran@buffalo.edu>
escribió:
> I may not understand the method, but consider the following with
> candidates X, Y, Z:
>
> 51 X<Y<Z
> 49 Y>Z>X
>
> Then X is the CW, and X is ranked over or pairwise defeats a'(B) only on
> those ballots B in the group of 51. Thus N(X)=51.
>
> Meanwhile, Y is ranked over a'(B) on all ballots, so N(Y)=100.
>
> If I've understood the method correctly, it is not Condorcet efficient.
>
> Jim Faran
> ------------------------------
> *From:* Election-Methods <election-methods-bounces@lists.electorama.com>
> on behalf of Forest Simmons <forest.simmons21@gmail.com>
> *Sent:* Wednesday, April 6, 2022 2:57 AM
> *To:* EM <Election-methods@lists.electorama.com>; Kristofer Munsterhjelm <
> km_elmet@t-online.de>
> *Subject:* [EM] Decloned Copeland Borda Hybrid
>
> On each ballot B identify the anti-favorite a'(B) as the candidate a'
> among those bottom listed on ballot B that is bottom listed the most on the
> rest of the ballots.
>
> [A candidate is bottom listed on a ballot if it is ranked over no
> candidate on that ballot]
>
> Elect the candidate X that, on the most ballots B, either is ranked ahead
> of or pairwise defeats a'(B).
>
> In other words, elect argmax N(X), where N(X) is the cardinality of the
> set of ballots
> {B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
>
> The key to monotonicity is that if some candidate X is uniquely raised on
> some ballot B, then for any other candidate Y, the difference N(X)-N(Y)
> does not decrease.
>
> The only non-trivial case is where X starts in the a'(B) position and
> swaps positions with some Z that was ranked above it.
>
> This definitely adds a point to N(X) because X is newly ranked ahead of
> a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
>
> In any case, N(X)-N(Y) does not decrease.
>
> Does this method satisfy the Condorcet Criterion?
>
> Well, if X pairwise defeats all other candidates, then N(X) is equal to
> the total number of ballots minus the number of ballots on which X is in
> the a' position, which must be fewer than half of the ballots, otherwise X
> would be the Condorcet loser.
>
> Is it possible for N(Y) to be greater than this N(X) while losing pairwise
> to the CW X?
>
> Perhaps ... if so, then this method is not a stand alone Condorcet method.
>
> It would be a valuable task for someone to test its Condorcet efficiency
> experimentally.
>
> Thanks!
>
> -Forest
>
>
>
AD
Andy Dienes
Wed, Apr 6, 2022 11:56 PM
Hi all,
I have recently come up with a small change to the way surplus handling is
performed on Allocated Score (AS). It is inspired by the way MES (which you
can read about here
https://proceedings.neurips.cc/paper/2021/hash/69f8ea31de0c00502b2ae571fbab1f95-Abstract.html)
operates on ranked ballots, which is also related to the Expanding
Approvals rule by Aziz.
Basically, the way it works in the reweighting step of AS is:
Set a threshold d such that the total ballot weight of voters who scored
the candidate >= d is at least one quota (using Hare for now, but other
choices are fine). Then, find the minimal amount of voting power that can
be subtracted equally from each ballot such that the total amount taken is
exactly one quota. Note that some ballots may have less than this amount
remaining, so they will be fully exhausted.
It is very similar to the original surplus handling, but rather than
exhausting fully all ballots with score > d and then fractionally ballots
with score = d, it chooses to subtract an equal amount of power from all
ballots above the threshold.
When all scores are 0,1 (i.e. approval ballots) it does not satisfy EJR (in
the same way that AS doesn't), but it does satisfy PJR.
I have already done some simulations and found favorable results, so what I
am mostly looking for is if there are any sneaky ways this can go very
wrong? Of course, every voting method has pathological examples so it's
never good to put too much stock in specific bad scenarios, but I've
already looked at this proposed modification from many other perspectives
so pathological examples are exactly what I'm after here :)
Best,
Andy Dienes
Hi all,
I have recently come up with a small change to the way surplus handling is
performed on Allocated Score (AS). It is inspired by the way MES (which you
can read about here
https://proceedings.neurips.cc/paper/2021/hash/69f8ea31de0c00502b2ae571fbab1f95-Abstract.html)
operates on ranked ballots, which is also related to the Expanding
Approvals rule by Aziz.
Basically, the way it works in the reweighting step of AS is:
Set a threshold d such that the total ballot weight of voters who scored
the candidate >= d is at least one quota (using Hare for now, but other
choices are fine). Then, find the minimal amount of voting power that can
be subtracted equally from each ballot such that the total amount taken is
exactly one quota. Note that some ballots may have less than this amount
remaining, so they will be fully exhausted.
It is very similar to the original surplus handling, but rather than
exhausting fully all ballots with score > d and then fractionally ballots
with score = d, it chooses to subtract an equal amount of power from all
ballots above the threshold.
When all scores are 0,1 (i.e. approval ballots) it does not satisfy EJR (in
the same way that AS doesn't), but it does satisfy PJR.
I have already done some simulations and found favorable results, so what I
am mostly looking for is if there are any sneaky ways this can go very
wrong? Of course, every voting method has pathological examples so it's
never good to put too much stock in specific bad scenarios, but I've
already looked at this proposed modification from many other perspectives
so pathological examples are exactly what I'm after here :)
Best,
Andy Dienes
KV
Kevin Venzke
Thu, Apr 7, 2022 6:44 PM
Hi Forest,
I'm not sure if you're still pursuing this method but as I understand it I don't
think it's monotone.
0.394: B>C>A
0.366: A
0.151: C
0.087: A>B>C
A>B>C>A cycle. Effective last preferences are A, B, A, C. Then C wins.
Switch last faction to
0.087: A>C>B
Cycle remains. Effective last preferences are now A, B, B, B. This causes C to net
lose a few points, and A to gain a lot due to the increased relevance of the A>B
contest, so that A wins.
The mechanism also doesn't seem to ensure majority favorite, but maybe you know that.
I do think the "anti-favorite" concept is interesting.
Kevin
Le mercredi 6 avril 2022, 14:53:01 UTC−5, Forest Simmons forest.simmons21@gmail.com a écrit :
I intended to say each candidate ranked above or pairwise beating a' gets a point, and two points for both.
The monotonicity proof goes through unchanged.... as z goes down, y gains a point, but so does X as it goes up.
On each ballot B identify the anti-favorite a'(B) as the candidate a' among those bottom listed on ballot B that is bottom listed the most on the rest of the ballots.
[A candidate is bottom listed on a ballot if it is ranked over no candidate on that ballot]
Elect the candidate X that, on the most ballots B, either is ranked ahead of or pairwise defeats a'(B).
In other words, elect argmax N(X), where N(X) is the cardinality of the set of ballots
{B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
The key to monotonicity is that if some candidate X is uniquely raised on some ballot B, then for any other candidate Y, the difference N(X)-N(Y) does not decrease.
The only non-trivial case is where X starts in the a'(B) position and swaps positions with some Z that was ranked above it.
This definitely adds a point to N(X) because X is newly ranked ahead of a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
In any case, N(X)-N(Y) does not decrease.
Does this method satisfy the Condorcet Criterion?
Well, if X pairwise defeats all other candidates, then N(X) is equal to the total number of ballots minus the number of ballots on which X is in the a' position, which must be fewer than half of the ballots, otherwise X would be the Condorcet loser.
Is it possible for N(Y) to be greater than this N(X) while losing pairwise to the CW X?
Perhaps ... if so, then this method is not a stand alone Condorcet method.
It would be a valuable task for someone to test its Condorcet efficiency experimentally.
Thanks!
-Forest
Hi Forest,
I'm not sure if you're still pursuing this method but as I understand it I don't
think it's monotone.
0.394: B>C>A
0.366: A
0.151: C
0.087: A>B>C
A>B>C>A cycle. Effective last preferences are A, B, A, C. Then C wins.
Switch last faction to
0.087: A>C>B
Cycle remains. Effective last preferences are now A, B, B, B. This causes C to net
lose a few points, and A to gain a lot due to the increased relevance of the A>B
contest, so that A wins.
The mechanism also doesn't seem to ensure majority favorite, but maybe you know that.
I do think the "anti-favorite" concept is interesting.
Kevin
Le mercredi 6 avril 2022, 14:53:01 UTC−5, Forest Simmons <forest.simmons21@gmail.com> a écrit :
> I intended to say each candidate ranked above or pairwise beating a' gets a point, and two points for both.
>
> The monotonicity proof goes through unchanged.... as z goes down, y gains a point, but so does X as it goes up.
> > On each ballot B identify the anti-favorite a'(B) as the candidate a' among those bottom listed on ballot B that is bottom listed the most on the rest of the ballots.
> >
> > [A candidate is bottom listed on a ballot if it is ranked over no candidate on that ballot]
> >
> > Elect the candidate X that, on the most ballots B, either is ranked ahead of or pairwise defeats a'(B).
> >
> > In other words, elect argmax N(X), where N(X) is the cardinality of the set of ballots
> >
> > {B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
> >
> > The key to monotonicity is that if some candidate X is uniquely raised on some ballot B, then for any other candidate Y, the difference N(X)-N(Y) does not decrease.
> >
> > The only non-trivial case is where X starts in the a'(B) position and swaps positions with some Z that was ranked above it.
> >
> > This definitely adds a point to N(X) because X is newly ranked ahead of a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
> >
> > In any case, N(X)-N(Y) does not decrease.
> >
> > Does this method satisfy the Condorcet Criterion?
> >
> > Well, if X pairwise defeats all other candidates, then N(X) is equal to the total number of ballots minus the number of ballots on which X is in the a' position, which must be fewer than half of the ballots, otherwise X would be the Condorcet loser.
> >
> > Is it possible for N(Y) to be greater than this N(X) while losing pairwise to the CW X?
> >
> > Perhaps ... if so, then this method is not a stand alone Condorcet method.
> >
> > It would be a valuable task for someone to test its Condorcet efficiency experimentally.
> >
> > Thanks!
> >
> > -Forest
RT
Richard, the VoteFair guy
Thu, Apr 7, 2022 10:00 PM
On 4/6/2022 4:56 PM, Andy Dienes wrote:
I've already looked at this proposed modification from many other
perspectives so pathological examples are exactly what I'm
after here :)
How does your version of "surplus handling" handle a ballot on which,
say, 3 candidates are ranked at the same ranking level, and the excess
beyond the quota would give each of the fully supporting
(non-shared-preference) ballots, say, 0.25 (one quarter) of a vote
toward electing the next winner?
Specifically:
-
How much of that shared-preference ballot's influence is lost by
electing a candidate? In other words, how much of that ballot's
influence remains available for electing a second, and even third,
candidate?
-
How are the two kinds of reduced influence interact? How is that
calculated?
I'm interested in your answer because the same concepts are relevant to
any well-designed variation of the single transferable vote (STV).
Thanks!
Richard Fobes
The VoteFair guy
On 4/6/2022 4:56 PM, Andy Dienes wrote:
Hi all,
I have recently come up with a small change to the way surplus handling
is performed on Allocated Score (AS). It is inspired by the way MES
(which you can read about
here https://proceedings.neurips.cc/paper/2021/hash/69f8ea31de0c00502b2ae571fbab1f95-Abstract.html)
operates on ranked ballots, which is also related to the Expanding
Approvals rule by Aziz.
Basically, the way it works in the reweighting step of AS is:
Set a threshold d such that the total ballot weight of voters who scored
the candidate >= d is at least one quota (using Hare for now, but other
choices are fine). Then, find the minimal amount of voting power that
can be subtracted equally from each ballot such that the total amount
taken is exactly one quota. Note that some ballots may have less than
this amount remaining, so they will be fully exhausted.
It is very similar to the original surplus handling, but rather than
exhausting fully all ballots with score > d and then fractionally
ballots with score = d, it chooses to subtract an equal amount of power
from all ballots above the threshold.
When all scores are 0,1 (i.e. approval ballots) it does not satisfy EJR
(in the same way that AS doesn't), but it does satisfy PJR.
I have already done some simulations and found favorable results, so
what I am mostly looking for is if there are any sneaky ways this can go
very wrong? Of course, every voting method has pathological examples so
it's never good to put too much stock in specific bad scenarios, but
I've already looked at this proposed modification from many other
perspectives so pathological examples are exactly what I'm after here :)
Best,
Andy Dienes
Election-Methods mailing list - see https://electorama.com/em for list info
On 4/6/2022 4:56 PM, Andy Dienes wrote:
> I've already looked at this proposed modification from many other
> perspectives so pathological examples are exactly what I'm
> after here :)
How does your version of "surplus handling" handle a ballot on which,
say, 3 candidates are ranked at the same ranking level, and the excess
beyond the quota would give each of the fully supporting
(non-shared-preference) ballots, say, 0.25 (one quarter) of a vote
toward electing the next winner?
Specifically:
* How much of that shared-preference ballot's influence is lost by
electing a candidate? In other words, how much of that ballot's
influence remains available for electing a second, and even third,
candidate?
* How are the two kinds of reduced influence interact? How is that
calculated?
I'm interested in your answer because the same concepts are relevant to
any well-designed variation of the single transferable vote (STV).
Thanks!
Richard Fobes
The VoteFair guy
On 4/6/2022 4:56 PM, Andy Dienes wrote:
> Hi all,
>
> I have recently come up with a small change to the way surplus handling
> is performed on Allocated Score (AS). It is inspired by the way MES
> (which you can read about
> here https://proceedings.neurips.cc/paper/2021/hash/69f8ea31de0c00502b2ae571fbab1f95-Abstract.html)
> operates on ranked ballots, which is also related to the Expanding
> Approvals rule by Aziz.
>
> Basically, the way it works in the reweighting step of AS is:
> Set a threshold d such that the total ballot weight of voters who scored
> the candidate >= d is at least one quota (using Hare for now, but other
> choices are fine). Then, find the minimal amount of voting power that
> can be subtracted equally from each ballot such that the total amount
> taken is exactly one quota. Note that some ballots may have less than
> this amount remaining, so they will be fully exhausted.
>
> It is very similar to the original surplus handling, but rather than
> exhausting fully all ballots with score > d and then fractionally
> ballots with score = d, it chooses to subtract an equal amount of power
> from all ballots above the threshold.
>
> When all scores are 0,1 (i.e. approval ballots) it does not satisfy EJR
> (in the same way that AS doesn't), but it does satisfy PJR.
>
> I have already done some simulations and found favorable results, so
> what I am mostly looking for is if there are any sneaky ways this can go
> very wrong? Of course, every voting method has pathological examples so
> it's never good to put too much stock in specific bad scenarios, but
> I've already looked at this proposed modification from many other
> perspectives so pathological examples are exactly what I'm after here :)
>
> Best,
> Andy Dienes
>
>
> ----
> Election-Methods mailing list - see https://electorama.com/em for list info
>
FS
Forest Simmons
Fri, Apr 8, 2022 5:15 PM
Kevin,
I'm glad you caught that. It looks like we have to go back to explícit
designations of anti-favorites to avoid this "accidental" lowering of B on
a set of ballots resulting from intentional raising of C on other ballots.
The other option would be to count truncated votes fractionally in the
anti-favorite tallies. But that would be messy in hand computations.
The explicit option seems better at distinguishing "don't care" truncations
from strongly felt anti-favorite truncations ... for, e.g. avoiding the
"dark horse" back door that low VSE candidates might otherwise sneak
through.
Also, instead of adding implicit approval points when X is raised from
bottom, it turns out to be smoother to subtract points from anti-favorites
(as we shall see below).
So this method becomes ...
Elect the candidate with the greatest difference between the number of
ballots on which the designated anti-favorite is among those defeated
pairwise by it and the number of ballots on which it (itself) is the
designated anti-favorite.
The dual hybrid method is ...
Elect the candidate with the greatest difference between the number of
ballots on which it (itself) is the designated favorite and the number of
ballots on which it is pairwise defeated by the designated favorite.
Each of these mutually dual methods is a stand alone monotonic clone-free
method with high (but not perfect) Condorcet efficiency.
They can be combined into a full-blown high-falutin hybrid method where the
Borda component is more apparent:
Elect the candidate that maximizes the number of ballots on which it is the
designated favorite Plus the number of ballots on which it pairwise defeats
the designated anti-favorite Minus the number of ballots on which it is the
designated anti-favorite Minus the number of ballots on which it is
pairwise defeated by the designated favorite.
Note that with complete rankings in the three candidate case the first and
third terms of this expression together amount to the Borda Count of the
candidate in question.
It turns out that in such a three candidate case when there are only three
ballot factions and no Condorcet candidate, the Borda Count decides the
finish order.
For example ...
45 A>B>C
35 B> C>A
20 C>A>B
[No need to explicify favorites or anti-favorites where there is no
ambiguity]
The respective candidate scores are the Borda Counts
45-35, 35-20, and 20-45,
while all of the other terms precisely cancel each other.
The non-perfect Condorcet efficiency may well be a blessing in disguise
because it answers a common complaint that "... Condorcet methods have no
mechanism for weeding out low approval/VSE CW's."
Alternatively, modify the method to read ... "if there is no pairwise
undefeated candidate, then elect the candidate that maximizes ...."
-Forest
El jue., 7 de abr. de 2022 11:47 a. m., Kevin Venzke stepjak@yahoo.fr
escribió:
Hi Forest,
I'm not sure if you're still pursuing this method but as I understand it I
don't
think it's monotone.
0.394: B>C>A
0.366: A
0.151: C
0.087: A>B>C
A>B>C>A cycle. Effective last preferences are A, B, A, C. Then C wins.
Switch last faction to
0.087: A>C>B
Cycle remains. Effective last preferences are now A, B, B, B. This causes
C to net
lose a few points, and A to gain a lot due to the increased relevance of
the A>B
contest, so that A wins.
The mechanism also doesn't seem to ensure majority favorite, but maybe you
know that.
I do think the "anti-favorite" concept is interesting.
Kevin
Le mercredi 6 avril 2022, 14:53:01 UTC−5, Forest Simmons <
forest.simmons21@gmail.com> a écrit :
I intended to say each candidate ranked above or pairwise beating a'
gets a point, and two points for both.
The monotonicity proof goes through unchanged.... as z goes down, y
gains a point, but so does X as it goes up.
On each ballot B identify the anti-favorite a'(B) as the candidate a'
among those bottom listed on ballot B that is bottom listed the most on the
rest of the ballots.
[A candidate is bottom listed on a ballot if it is ranked over no
candidate on that ballot]
Elect the candidate X that, on the most ballots B, either is ranked
ahead of or pairwise defeats a'(B).
In other words, elect argmax N(X), where N(X) is the cardinality of
{B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
The key to monotonicity is that if some candidate X is uniquely raised
on some ballot B, then for any other candidate Y, the difference N(X)-N(Y)
does not decrease.
The only non-trivial case is where X starts in the a'(B) position and
swaps positions with some Z that was ranked above it.
This definitely adds a point to N(X) because X is newly ranked ahead
of a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
In any case, N(X)-N(Y) does not decrease.
Does this method satisfy the Condorcet Criterion?
Well, if X pairwise defeats all other candidates, then N(X) is equal
to the total number of ballots minus the number of ballots on which X is in
the a' position, which must be fewer than half of the ballots, otherwise X
would be the Condorcet loser.
Is it possible for N(Y) to be greater than this N(X) while losing
Perhaps ... if so, then this method is not a stand alone Condorcet
It would be a valuable task for someone to test its Condorcet
efficiency experimentally.
Kevin,
I'm glad you caught that. It looks like we have to go back to explícit
designations of anti-favorites to avoid this "accidental" lowering of B on
a set of ballots resulting from intentional raising of C on other ballots.
The other option would be to count truncated votes fractionally in the
anti-favorite tallies. But that would be messy in hand computations.
The explicit option seems better at distinguishing "don't care" truncations
from strongly felt anti-favorite truncations ... for, e.g. avoiding the
"dark horse" back door that low VSE candidates might otherwise sneak
through.
Also, instead of adding implicit approval points when X is raised from
bottom, it turns out to be smoother to subtract points from anti-favorites
(as we shall see below).
So this method becomes ...
Elect the candidate with the greatest difference between the number of
ballots on which the designated anti-favorite is among those defeated
pairwise by it and the number of ballots on which it (itself) is the
designated anti-favorite.
The dual hybrid method is ...
Elect the candidate with the greatest difference between the number of
ballots on which it (itself) is the designated favorite and the number of
ballots on which it is pairwise defeated by the designated favorite.
Each of these mutually dual methods is a stand alone monotonic clone-free
method with high (but not perfect) Condorcet efficiency.
They can be combined into a full-blown high-falutin hybrid method where the
Borda component is more apparent:
Elect the candidate that maximizes the number of ballots on which it is the
designated favorite Plus the number of ballots on which it pairwise defeats
the designated anti-favorite Minus the number of ballots on which it is the
designated anti-favorite Minus the number of ballots on which it is
pairwise defeated by the designated favorite.
Note that with complete rankings in the three candidate case the first and
third terms of this expression together amount to the Borda Count of the
candidate in question.
It turns out that in such a three candidate case when there are only three
ballot factions and no Condorcet candidate, the Borda Count decides the
finish order.
For example ...
45 A>B>C
35 B> C>A
20 C>A>B
[No need to explicify favorites or anti-favorites where there is no
ambiguity]
The respective candidate scores are the Borda Counts
45-35, 35-20, and 20-45,
while all of the other terms precisely cancel each other.
The non-perfect Condorcet efficiency may well be a blessing in disguise
because it answers a common complaint that "... Condorcet methods have no
mechanism for weeding out low approval/VSE CW's."
Alternatively, modify the method to read ... "if there is no pairwise
undefeated candidate, then elect the candidate that maximizes ...."
-Forest
El jue., 7 de abr. de 2022 11:47 a. m., Kevin Venzke <stepjak@yahoo.fr>
escribió:
> Hi Forest,
>
> I'm not sure if you're still pursuing this method but as I understand it I
> don't
> think it's monotone.
>
> 0.394: B>C>A
> 0.366: A
> 0.151: C
> 0.087: A>B>C
>
> A>B>C>A cycle. Effective last preferences are A, B, A, C. Then C wins.
>
> Switch last faction to
> 0.087: A>C>B
>
> Cycle remains. Effective last preferences are now A, B, B, B. This causes
> C to net
> lose a few points, and A to gain a lot due to the increased relevance of
> the A>B
> contest, so that A wins.
>
> The mechanism also doesn't seem to ensure majority favorite, but maybe you
> know that.
>
> I do think the "anti-favorite" concept is interesting.
>
> Kevin
>
>
> Le mercredi 6 avril 2022, 14:53:01 UTC−5, Forest Simmons <
> forest.simmons21@gmail.com> a écrit :
> > I intended to say each candidate ranked above or pairwise beating a'
> gets a point, and two points for both.
> >
> > The monotonicity proof goes through unchanged.... as z goes down, y
> gains a point, but so does X as it goes up.
>
> > > On each ballot B identify the anti-favorite a'(B) as the candidate a'
> among those bottom listed on ballot B that is bottom listed the most on the
> rest of the ballots.
> > >
> > > [A candidate is bottom listed on a ballot if it is ranked over no
> candidate on that ballot]
> > >
> > > Elect the candidate X that, on the most ballots B, either is ranked
> ahead of or pairwise defeats a'(B).
> > >
> > > In other words, elect argmax N(X), where N(X) is the cardinality of
> the set of ballots
> > >
> > > {B | X is ranked ahead of or pairwise defeats a'(B) (or both)}.
> > >
> > > The key to monotonicity is that if some candidate X is uniquely raised
> on some ballot B, then for any other candidate Y, the difference N(X)-N(Y)
> does not decrease.
> > >
> > > The only non-trivial case is where X starts in the a'(B) position and
> swaps positions with some Z that was ranked above it.
> > >
> > > This definitely adds a point to N(X) because X is newly ranked ahead
> of a'(B). This will also add a point to Y if Z is pairwise defeated by Y.
> > >
> > > In any case, N(X)-N(Y) does not decrease.
> > >
> > > Does this method satisfy the Condorcet Criterion?
> > >
> > > Well, if X pairwise defeats all other candidates, then N(X) is equal
> to the total number of ballots minus the number of ballots on which X is in
> the a' position, which must be fewer than half of the ballots, otherwise X
> would be the Condorcet loser.
> > >
> > > Is it possible for N(Y) to be greater than this N(X) while losing
> pairwise to the CW X?
> > >
> > > Perhaps ... if so, then this method is not a stand alone Condorcet
> method.
> > >
> > > It would be a valuable task for someone to test its Condorcet
> efficiency experimentally.
> > >
> > > Thanks!
> > >
> > > -Forest
>
>
KM
Kristofer Munsterhjelm
Fri, Apr 8, 2022 10:30 PM
On 08.04.2022 19:15, Forest Simmons wrote:
Kevin,
I'm glad you caught that. It looks like we have to go back to explícit
designations of anti-favorites to avoid this "accidental" lowering of B
on a set of ballots resulting from intentional raising of C on other
ballots.
The other option would be to count truncated votes fractionally in the
anti-favorite tallies. But that would be messy in hand computations.
There's a third option, I think, which involves counting both B and C.
E.g. in my three-candidate method enumeration, the IRV scoring function is
f(A) = -fpC, f(B) = -fpA, f(C) = -fpB
This is nonmonotone because raising A can lower B's first preferences
and thus increase f(C), making C win instead of A. But introducing an
fpA term
f(A) = fpA - fpC
makes sure that A's score always increases when A is moved first,
canceling out any potential improvement to C's score.
Your method seems similar enough (based on last preferences rather than
first ones) that a similar kind of fix could be employed. But perhaps
you've already anticipated that by your description:
Elect the candidate with the greatest difference between the number of
ballots on which it (itself) is the designated favorite and the number
of ballots on which it is pairwise defeated by the designated favorite.
I'm not sure, I thought I should mention it anyway :-)
-km
On 08.04.2022 19:15, Forest Simmons wrote:
> Kevin,
>
> I'm glad you caught that. It looks like we have to go back to explícit
> designations of anti-favorites to avoid this "accidental" lowering of B
> on a set of ballots resulting from intentional raising of C on other
> ballots.
>
> The other option would be to count truncated votes fractionally in the
> anti-favorite tallies. But that would be messy in hand computations.
There's a third option, I think, which involves counting both B and C.
E.g. in my three-candidate method enumeration, the IRV scoring function is
f(A) = -fpC, f(B) = -fpA, f(C) = -fpB
This is nonmonotone because raising A can lower B's first preferences
and thus increase f(C), making C win instead of A. But introducing an
fpA term
f(A) = fpA - fpC
makes sure that A's score always increases when A is moved first,
canceling out any potential improvement to C's score.
Your method seems similar enough (based on last preferences rather than
first ones) that a similar kind of fix could be employed. But perhaps
you've already anticipated that by your description:
> Elect the candidate with the greatest difference between the number of
> ballots on which it (itself) is the designated favorite and the number
> of ballots on which it is pairwise defeated by the designated favorite.
I'm not sure, I thought I should mention it anyway :-)
-km
FS
Forest Simmons
Fri, Apr 8, 2022 11:10 PM
Yes ... there are several possibilities ... still exploring pros and cons
of each.
Thanks!
El vie., 8 de abr. de 2022 3:30 p. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
On 08.04.2022 19:15, Forest Simmons wrote:
Kevin,
I'm glad you caught that. It looks like we have to go back to explícit
designations of anti-favorites to avoid this "accidental" lowering of B
on a set of ballots resulting from intentional raising of C on other
ballots.
The other option would be to count truncated votes fractionally in the
anti-favorite tallies. But that would be messy in hand computations.
There's a third option, I think, which involves counting both B and C.
E.g. in my three-candidate method enumeration, the IRV scoring function is
f(A) = -fpC, f(B) = -fpA, f(C) = -fpB
This is nonmonotone because raising A can lower B's first preferences
and thus increase f(C), making C win instead of A. But introducing an
fpA term
f(A) = fpA - fpC
makes sure that A's score always increases when A is moved first,
canceling out any potential improvement to C's score.
Your method seems similar enough (based on last preferences rather than
first ones) that a similar kind of fix could be employed. But perhaps
you've already anticipated that by your description:
Elect the candidate with the greatest difference between the number of
ballots on which it (itself) is the designated favorite and the number
of ballots on which it is pairwise defeated by the designated favorite.
I'm not sure, I thought I should mention it anyway :-)
-km
Yes ... there are several possibilities ... still exploring pros and cons
of each.
Thanks!
El vie., 8 de abr. de 2022 3:30 p. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
> On 08.04.2022 19:15, Forest Simmons wrote:
> > Kevin,
> >
> > I'm glad you caught that. It looks like we have to go back to explícit
> > designations of anti-favorites to avoid this "accidental" lowering of B
> > on a set of ballots resulting from intentional raising of C on other
> > ballots.
> >
> > The other option would be to count truncated votes fractionally in the
> > anti-favorite tallies. But that would be messy in hand computations.
>
> There's a third option, I think, which involves counting both B and C.
>
> E.g. in my three-candidate method enumeration, the IRV scoring function is
>
> f(A) = -fpC, f(B) = -fpA, f(C) = -fpB
>
> This is nonmonotone because raising A can lower B's first preferences
> and thus increase f(C), making C win instead of A. But introducing an
> fpA term
>
> f(A) = fpA - fpC
>
> makes sure that A's score always increases when A is moved first,
> canceling out any potential improvement to C's score.
>
> Your method seems similar enough (based on last preferences rather than
> first ones) that a similar kind of fix could be employed. But perhaps
> you've already anticipated that by your description:
>
> > Elect the candidate with the greatest difference between the number of
> > ballots on which it (itself) is the designated favorite and the number
> > of ballots on which it is pairwise defeated by the designated favorite.
>
> I'm not sure, I thought I should mention it anyway :-)
>
> -km
>
KM
Kristofer Munsterhjelm
Tue, Apr 12, 2022 8:30 AM
On 07.04.2022 01:56, Andy Dienes wrote:
Hi all,
I have recently come up with a small change to the way surplus handling
is performed on Allocated Score (AS). It is inspired by the way MES
(which you can read about
here https://proceedings.neurips.cc/paper/2021/hash/69f8ea31de0c00502b2ae571fbab1f95-Abstract.html
https://proceedings.neurips.cc/paper/2021/hash/69f8ea31de0c00502b2ae571fbab1f95-Abstract.html)
operates on ranked ballots, which is also related to the Expanding
Approvals rule by Aziz.
Basically, the way it works in the reweighting step of AS is:
Set a threshold d such that the total ballot weight of voters who scored
the candidate >= d is at least one quota (using Hare for now, but other
choices are fine). Then, find the minimal amount of voting power that
can be subtracted equally from each ballot such that the total amount
taken is exactly one quota. Note that some ballots may have less than
this amount remaining, so they will be fully exhausted.
It is very similar to the original surplus handling, but rather than
exhausting fully all ballots with score > d and then fractionally
ballots with score = d, it chooses to subtract an equal amount of power
from all ballots above the threshold.
When all scores are 0,1 (i.e. approval ballots) it does not satisfy EJR
(in the same way that AS doesn't), but it does satisfy PJR.
I have already done some simulations and found favorable results, so
what I am mostly looking for is if there are any sneaky ways this can go
very wrong? Of course, every voting method has pathological examples so
it's never good to put too much stock in specific bad scenarios, but
I've already looked at this proposed modification from many other
perspectives so pathological examples are exactly what I'm after here :)
I don't think there should be a problem as long as you make sure to
never elect more than one candidate in one go. E.g. if both A and B are
above the quota, apply the tiebreaker of your choice (say it picks A),
then elect A, deweight the voters, and then check if there's a quota
for B. Otherwise a majority might get too many candidates elected.
Your change might also reduce free-riding. Let's say my honest vote is
A: 5, B: 4. I know some other voters are also voting A:5, so it's
tempting for me to vote A: 4 instead so that my voting weight is only
exhausted fractionally. With the original reweighting (as you describe
it), I may get exhausted fully with A: 5 but only partially with A: 4.
But with your change, I would get exhausted equally.
In general, as long as you reasonably fairly reweight the votes, then
just how you distribute the surpluses shouldn't matter much. You'll get
quota proportionality anyway, as long as you reweight everybody who
contributes to getting a candidate elected, in such a way that their
total weight afterwards is equal to the surplus.
For instance, I constructed a more strategy resistant variant of EAR,
https://electowiki.org/wiki/Maximum_Constrained_Approval_Bucklin, which
maximally delays the decision of just what ballot weights to attenuate.
So although it's very hard to reason about just whose votes will count
less, it still passes Droop proportionality, because it does the
reweighting as soon as a candidate is backed by a quota.
-km
On 07.04.2022 01:56, Andy Dienes wrote:
> Hi all,
>
> I have recently come up with a small change to the way surplus handling
> is performed on Allocated Score (AS). It is inspired by the way MES
> (which you can read about
> here https://proceedings.neurips.cc/paper/2021/hash/69f8ea31de0c00502b2ae571fbab1f95-Abstract.html
> <https://proceedings.neurips.cc/paper/2021/hash/69f8ea31de0c00502b2ae571fbab1f95-Abstract.html>)
> operates on ranked ballots, which is also related to the Expanding
> Approvals rule by Aziz.
>
> Basically, the way it works in the reweighting step of AS is:
> Set a threshold d such that the total ballot weight of voters who scored
> the candidate >= d is at least one quota (using Hare for now, but other
> choices are fine). Then, find the minimal amount of voting power that
> can be subtracted equally from each ballot such that the total amount
> taken is exactly one quota. Note that some ballots may have less than
> this amount remaining, so they will be fully exhausted.
>
> It is very similar to the original surplus handling, but rather than
> exhausting fully all ballots with score > d and then fractionally
> ballots with score = d, it chooses to subtract an equal amount of power
> from all ballots above the threshold.
>
> When all scores are 0,1 (i.e. approval ballots) it does not satisfy EJR
> (in the same way that AS doesn't), but it does satisfy PJR.
>
> I have already done some simulations and found favorable results, so
> what I am mostly looking for is if there are any sneaky ways this can go
> very wrong? Of course, every voting method has pathological examples so
> it's never good to put too much stock in specific bad scenarios, but
> I've already looked at this proposed modification from many other
> perspectives so pathological examples are exactly what I'm after here :)
I don't think there should be a problem as long as you make sure to
never elect more than one candidate in one go. E.g. if both A and B are
above the quota, apply the tiebreaker of your choice (say it picks A),
then elect A, deweight the voters, and *then* check if there's a quota
for B. Otherwise a majority might get too many candidates elected.
Your change might also reduce free-riding. Let's say my honest vote is
A: 5, B: 4. I know some other voters are also voting A:5, so it's
tempting for me to vote A: 4 instead so that my voting weight is only
exhausted fractionally. With the original reweighting (as you describe
it), I may get exhausted fully with A: 5 but only partially with A: 4.
But with your change, I would get exhausted equally.
In general, as long as you reasonably fairly reweight the votes, then
just how you distribute the surpluses shouldn't matter much. You'll get
quota proportionality anyway, as long as you reweight everybody who
contributes to getting a candidate elected, in such a way that their
total weight afterwards is equal to the surplus.
For instance, I constructed a more strategy resistant variant of EAR,
https://electowiki.org/wiki/Maximum_Constrained_Approval_Bucklin, which
maximally delays the decision of just what ballot weights to attenuate.
So although it's very hard to reason about just whose votes will count
less, it still passes Droop proportionality, because it does the
reweighting as soon as a candidate is backed by a quota.
-km
