Empathy List Archives

EB

Etjon Basha

Sat, Sep 21, 2024 11:53 AM

Dear gentlemen,

A while ago I did write here about the Iterated Bucklin
https://electowiki.org/wiki/Iterated_Bucklin method on which I’ve
recently had a chance to think and generalize about a bit more. Maybe some
of the below could be novel or otherwise of interest.

First, and for our purposes today, let's define the Serious Candidates Set
in the context of a ranked ballot, to include those candidates who would
win an approval count if they served as the approval cutoff across all
ballots.

In the [2:A>B, 3:C>A, 4:A>B] election as an example, the Set would include
A and B only, as applying the cutoff at C would still elect B.

I’ve been checking some random simulations from Kevin Venzke’s
votingmethods.net, and here are some properties of this Set that I
suspect:

   If there is a Condorcet Winner, this Set should always include

them.

   Otherwise, this Set should always partially overlap with the

Smith Set.

Now, quite a few methods emerge once the Serious Candidate Set is isolated
(by actually checking the approval winner once every candidate is used as a
cutoff). The five below allow truncation and equal ranking, and have been
checked (again courtesy of votingmethods.net) to ensure that they are
different from one-another and the 40-odd other methods Kevin has
aggregated over there.

So, which member of the Serious Candidate Set should be elected?

   Electing the Serious Candidate that wins their cutoff count by

the most approvals. Rather obvious but not too much of an improvement over
Approval (if any at all). Terrible Later No Harm failures, though this is
in the context where truncation is allowed. Fails Condorcet.

   Electing the Serious Candidate that wins their cutoff count by

the least approvals. A bit counterintuitive, but winning by the least
means that the winner had to “dip” the least into each approver’s rankings.
If this is not compliant with Later No Harm, it should at least fail
rarely. It would fail Later No Help spectacularly though, indeed having a
huge incentive to always rank your least favorite candidate that is still
likely to win last, instead of leaving them unranked. Unfortunately, I’ve
seen it elect the Condorcet Loser at least once.

   Electing the Serious Candidate that wins their cutoff count by

the most approvals compared to the runner up. May fail Condorcet the
least. Likely the most sensible of the bunch.

   Iterated Bucklin (now fitting into this generalised family) will

always elect a member of the Set, but it seems to be neither of the three
above with consistency. I cannot seem to find the pattern the method lands
on.

   Electing the Serious Candidate that wins the election if the

cutoff is set at the FPP winner. If the FPP winner is in the Set to begin
with, they will be elected. Otherwise, again a method that elects a winner
from the set through no obvious pattern. Of particular interest to me since
it’s the only method in here that can be hand-counted with relative ease
(it’s just an FPP count and an approval count after that).

For reference, standard Bucklin may not always elect members of the Set so
cannot be retconned into this tree. I've tested quite a few other methods,
and there are some for which I'm still to find a failure to elect from the
Serious Candidates Set, including Borda and, unsurprisingly, many approval
variations and Approval-Condorcet hybrids.

Just some preliminary thoughts above, hopefully of some interest.

Best regards,

Etjon Basha

Dear gentlemen, A while ago I did write here about the Iterated Bucklin <https://electowiki.org/wiki/Iterated_Bucklin> method on which I’ve recently had a chance to think and generalize about a bit more. Maybe some of the below could be novel or otherwise of interest. First, and for our purposes today, let's define the *Serious Candidates Set* in the context of a ranked ballot, to include those candidates who would win an approval count if they served as the approval cutoff across all ballots. In the [2:A>B, 3:C>A, 4:A>B] election as an example, the Set would include A and B only, as applying the cutoff at C would still elect B. I’ve been checking some random simulations from Kevin Venzke’s votingmethods.net, and here are some properties of this Set that I *suspect*: 1. If there is a Condorcet Winner, this Set should always include them. 2. Otherwise, this Set should always partially overlap with the Smith Set. Now, quite a few methods emerge once the Serious Candidate Set is isolated (by actually checking the approval winner once every candidate is used as a cutoff). The five below allow truncation and equal ranking, and have been checked (again courtesy of votingmethods.net) to ensure that they are different from one-another and the 40-odd other methods Kevin has aggregated over there. So, which member of the Serious Candidate Set should be elected? 1. Electing the Serious Candidate that wins their cutoff count by the most approvals. Rather obvious but not too much of an improvement over Approval (if any at all). Terrible Later No Harm failures, though this is in the context where truncation is allowed. Fails Condorcet. 2. Electing the Serious Candidate that wins their cutoff count by the *least* approvals. A bit counterintuitive, but winning by the least means that the winner had to “dip” the least into each approver’s rankings. If this is not compliant with Later No Harm, it should at least fail rarely. It would fail Later No Help spectacularly though, indeed having a huge incentive to always rank your least favorite candidate that is still likely to win last, instead of leaving them unranked. Unfortunately, I’ve seen it elect the Condorcet Loser at least once. 3. Electing the Serious Candidate that wins their cutoff count by the most approvals *compared to the runner up*. May fail Condorcet the least. Likely the most sensible of the bunch. 4. Iterated Bucklin (now fitting into this generalised family) will always elect a member of the Set, but it seems to be neither of the three above with consistency. I cannot seem to find the pattern the method lands on. 5. Electing the Serious Candidate that wins the election if the cutoff is set at the FPP winner. If the FPP winner is in the Set to begin with, they will be elected. Otherwise, again a method that elects a winner from the set through no obvious pattern. Of particular interest to me since it’s the only method in here that can be hand-counted with relative ease (it’s just an FPP count and an approval count after that). For reference, standard Bucklin may not always elect members of the Set so cannot be retconned into this tree. I've tested quite a few other methods, and there are some for which I'm still to find a failure to elect from the Serious Candidates Set, including Borda and, unsurprisingly, many approval variations and Approval-Condorcet hybrids. Just some preliminary thoughts above, hopefully of some interest. Best regards, Etjon Basha

KM

Kristofer Munsterhjelm

Thu, Sep 26, 2024 4:46 PM

On 2024-09-21 13:53, Etjon Basha wrote:

Dear gentlemen,

A while ago I did write here about the Iterated Bucklin
https://electowiki.org/wiki/Iterated_Bucklin method on which I’ve
recently had a chance to think and generalize about a bit more. Maybe
some of the below could be novel or otherwise of interest.

First, and for our purposes today, let's define the Serious Candidates
Set in the context of a ranked ballot, to include those candidates who
would win an approval count if they served as the approval cutoff across
all ballots.

In the [2:A>B, 3:C>A, 4:A>B] election as an example, the Set would
include A and B only, as applying the cutoff at C would still elect B.

I’ve been checking some random simulations from Kevin Venzke’s
votingmethods.net http://votingmethods.net, and here are some
properties of this Set that I suspect:

1.If there is a Condorcet Winner, this Set should always include them.

I think that should hold, at least with strict ranks. Suppose A beats B
pairwise. Then if we set the approval cutoff to A inclusive, every
ballot that has B>A will give a point to B and A, whereas every ballot
that has A>B will give a point only to A. Since A beats B pairwise, the
number of A>B ballots exceed the number of B>A ballots. Thus A must have
a higher approval score than B.

Since A is the CW, this reasoning holds for all other B. A must obtain a
higher approval score than B for any other B. Hence A is the winner and
belongs to the SCS.

2.Otherwise, this Set should always partially overlap with the Smith Set.

The reasoning above gets us part of the way. Suppose A is in the Smith
set. For any non-Smith member B, A must obtain more points than B for
the same reason as above. So the non-Smith members can't win.

What we have to show is that it's impossible for e.g. A to be the winner
when the approval cutoff is set at B, B be the winner when it's set at
C, and C be the winner when it's set at A. I'm not sure how to prove
that, though.

-km

On 2024-09-21 13:53, Etjon Basha wrote: > Dear gentlemen, > > > A while ago I did write here about the Iterated Bucklin > <https://electowiki.org/wiki/Iterated_Bucklin> method on which I’ve > recently had a chance to think and generalize about a bit more. Maybe > some of the below could be novel or otherwise of interest. > > > First, and for our purposes today, let's define the *Serious Candidates > Set* in the context of a ranked ballot, to include those candidates who > would win an approval count if they served as the approval cutoff across > all ballots. > > > In the [2:A>B, 3:C>A, 4:A>B] election as an example, the Set would > include A and B only, as applying the cutoff at C would still elect B. > > > I’ve been checking some random simulations from Kevin Venzke’s > votingmethods.net <http://votingmethods.net>, and here are some > properties of this Set that I *suspect*: > > 1.If there is a Condorcet Winner, this Set should always include them. I think that should hold, at least with strict ranks. Suppose A beats B pairwise. Then if we set the approval cutoff to A inclusive, every ballot that has B>A will give a point to B and A, whereas every ballot that has A>B will give a point only to A. Since A beats B pairwise, the number of A>B ballots exceed the number of B>A ballots. Thus A must have a higher approval score than B. Since A is the CW, this reasoning holds for all other B. A must obtain a higher approval score than B for any other B. Hence A is the winner and belongs to the SCS. > 2.Otherwise, this Set should always partially overlap with the Smith Set. The reasoning above gets us part of the way. Suppose A is in the Smith set. For any non-Smith member B, A must obtain more points than B for the same reason as above. So the non-Smith members can't win. What we have to show is that it's impossible for e.g. A to be the winner when the approval cutoff is set at B, B be the winner when it's set at C, and C be the winner when it's set at A. I'm not sure how to prove that, though. -km

EB

Etjon Basha

Sat, Sep 28, 2024 11:06 AM

Thanks Kristofer,

On why the CW must be in the Serious Candidate Set: once explained in
those terms, it really seems quite intuitive indeed, and I should have
thought a bit harder before posting.

I also think that explanation is sufficient to model the observed
relationship between the Smith and SC Sets: using your argument, setting
the approval cut-off at a Smith Set member will elect some Smith Set
member. Setting to the cutoff at some non-member may elect any candidate,
inside or outside the Smith Set.

These two suffice to explain what I've observed by random simulation: The
Smith and Serious Candidate Sets will share at least one member in common
and, beyond this, it seems like we may be able to say no further.

Regards,

Etjon Basha

On Fri, Sep 27, 2024 at 2:46 AM Kristofer Munsterhjelm <
km-elmet@munsterhjelm.no> wrote: