Empathy List Archives

V

VoteFair

Wed, Jul 21, 2021 1:19 AM

Version 2 of the Ranked Choice Including Pairwise Elimination (RCIPE)
voting method is now coded.

Here's a link to the scatter plot that shows its Clone Independence and
IIA (Independence of Irrelevant Alternatives) success rates compared to
other single-winner methods:

https://www.rankedchoiceoregon.org/img/clone_iia_success_rates.jpg

Based on Kristofer's feedback on the scatter plot, I've moved the IRV
(instant-runoff voting) data points to the top to show that IRV has a
zero Clone Independence failure rate. (The CI measurements yield a
non-zero CI failure rate because IRV does not specify a tie breaker
method and there are lots of lower-level ties in these "stress test"
measurements.) I've removed the IRV-BTR data because it's so similar to
IRV.

Here is how RCIPE version 2 is counted:

Eliminate the "pairwise losing candidate" (Condorcet loser) when there
is one.
Calculate a "pairwise support count" for each candidate. Each ballot
contributes to that count the number of remaining(!) candidates who are
ranked lower than the specified candidate.
Eliminate the candidate who has the smallest pairwise support count.
Repeat the loop to eliminate the next candidate. The last candidate
remaining is the winner.
If there is a tie for the smallest pairwise support count, use IRV
(instant-runoff voting) as the tie-breaker. Specifically, eliminate the
tied(!) candidate who is ranked highest -- compared to the other tied
candidates -- on the fewest ballots.
If there is still a tie, eliminate the candidate who has the smallest
one-on-one pairwise count against any other tied candidate. (This
addition is based on recent suggestions here.)

That's it!

The change from RCIPE v1 to v2 consists of:

Reversing the order of the "IRV" step and the tie-breaker step that
involves the pairwise support counts
Ignoring the pairwise opposition counts

This RCIPE version 2 method "fails" lots of fairness criteria. However,
the success rates for the most important criteria are high, and similar
to -- and for CI even better than -- the Condorcet-Kemeny method. For
example, where C-K indicates the Condorcet-Kemeny method, and IPE
indicates Instant Pairwise Elimination:

Condorcet compliance
Method, 2 choices, 6 choices, 9 choices
C-K, 100, 100, 100
IPE, 100, 100, 100
RCIPEv2, 100, 100, 99.6
RCIPEv1, 100, 99.1, 96.5

Clone Independence
Method, 2 choices, 6 choices, 9 choices
C-K, 100, 94.2, 94.4
IPE, 100, 94.7, 93.7
RCIPEv2, 100, 98.4, 97.3
RCIPEv1, 100, 99.8, 99.6

Independence of Irrelevant Alternatives
Method, 6 choices, 9 choices
C-K, 83.7, 78.2
IPE, 83, 76.9
RCIPEv2, 78.5, 70.7
RCIPEv1, 77.5, 69.2

In other words, as is well-known, getting smaller failure rates for one
criterion requires accepting larger failure rates for other criteria.

This method is a nice compromise regarding failure rates.

The big advantages are:

Simple for voters to understand how the counting is done. Whereas
STAR voting gets a candidate's score from the ballot's column heading,
RCIPE v2 gets a candidate's "score" from the ballot by counting how many
remaining candidates are ranked lower than that candidate.
Pairwise losing candidates are identified using the same pairwise counts.
Eliminates one candidate at a time, which many voters find to be
easier to understand and trust.
Avoids using fractions or decimal numbers to allow two or more
candidates to be ranked at the same preference level, which was needed
for RCIPE v1.
Is precinct summable!

As a bonus, the ballots can be counted manually. Or the calculations
can be done using a calculator and paper and pen.

Is there anything I'm overlooking? If not, I'll update the RCIPE
article on Electowiki. Of course I'll move the version 1 description to
the history section.

Thanks for any feedback.

Richard Fobes
The VoteFair guy

Version 2 of the Ranked Choice Including Pairwise Elimination (RCIPE) voting method is now coded. Here's a link to the scatter plot that shows its Clone Independence and IIA (Independence of Irrelevant Alternatives) success rates compared to other single-winner methods: https://www.rankedchoiceoregon.org/img/clone_iia_success_rates.jpg Based on Kristofer's feedback on the scatter plot, I've moved the IRV (instant-runoff voting) data points to the top to show that IRV has a zero Clone Independence failure rate. (The CI measurements yield a non-zero CI failure rate because IRV does not specify a tie breaker method and there are lots of lower-level ties in these "stress test" measurements.) I've removed the IRV-BTR data because it's so similar to IRV. Here is how RCIPE version 2 is counted: * Eliminate the "pairwise losing candidate" (Condorcet loser) when there is one. * Calculate a "pairwise support count" for each candidate. Each ballot contributes to that count the number of remaining(!) candidates who are ranked lower than the specified candidate. * Eliminate the candidate who has the smallest pairwise support count. * Repeat the loop to eliminate the next candidate. The last candidate remaining is the winner. * If there is a tie for the smallest pairwise support count, use IRV (instant-runoff voting) as the tie-breaker. Specifically, eliminate the tied(!) candidate who is ranked highest -- compared to the other tied candidates -- on the fewest ballots. * If there is still a tie, eliminate the candidate who has the smallest one-on-one pairwise count against any other tied candidate. (This addition is based on recent suggestions here.) That's it! The change from RCIPE v1 to v2 consists of: * Reversing the order of the "IRV" step and the tie-breaker step that involves the pairwise support counts * Ignoring the pairwise opposition counts This RCIPE version 2 method "fails" lots of fairness criteria. However, the success rates for the most important criteria are high, and similar to -- and for CI even better than -- the Condorcet-Kemeny method. For example, where C-K indicates the Condorcet-Kemeny method, and IPE indicates Instant Pairwise Elimination: Condorcet compliance Method, 2 choices, 6 choices, 9 choices C-K, 100, 100, 100 IPE, 100, 100, 100 RCIPEv2, 100, 100, 99.6 RCIPEv1, 100, 99.1, 96.5 Clone Independence Method, 2 choices, 6 choices, 9 choices C-K, 100, 94.2, 94.4 IPE, 100, 94.7, 93.7 RCIPEv2, 100, 98.4, 97.3 RCIPEv1, 100, 99.8, 99.6 Independence of Irrelevant Alternatives Method, 6 choices, 9 choices C-K, 83.7, 78.2 IPE, 83, 76.9 RCIPEv2, 78.5, 70.7 RCIPEv1, 77.5, 69.2 In other words, as is well-known, getting smaller failure rates for one criterion requires accepting larger failure rates for other criteria. This method is a nice compromise regarding failure rates. The big advantages are: * Simple for voters to understand how the counting is done. Whereas STAR voting gets a candidate's score from the ballot's column heading, RCIPE v2 gets a candidate's "score" from the ballot by counting how many remaining candidates are ranked lower than that candidate. * Pairwise losing candidates are identified using the same pairwise counts. * Eliminates one candidate at a time, which many voters find to be easier to understand and trust. * Avoids using fractions or decimal numbers to allow two or more candidates to be ranked at the same preference level, which was needed for RCIPE v1. * Is precinct summable! As a bonus, the ballots can be counted manually. Or the calculations can be done using a calculator and paper and pen. Is there anything I'm overlooking? If not, I'll update the RCIPE article on Electowiki. Of course I'll move the version 1 description to the history section. Thanks for any feedback. Richard Fobes The VoteFair guy

KM

Kristofer Munsterhjelm

Wed, Jul 21, 2021 11:41 AM

On 21.07.2021 03:19, VoteFair wrote:

Version 2 of the Ranked Choice Including Pairwise Elimination (RCIPE)
voting method is now coded.

Here's a link to the scatter plot that shows its Clone Independence and
IIA (Independence of Irrelevant Alternatives) success rates compared to
other single-winner methods:

https://www.rankedchoiceoregon.org/img/clone_iia_success_rates.jpg

Based on Kristofer's feedback on the scatter plot, I've moved the IRV
(instant-runoff voting) data points to the top to show that IRV has a
zero Clone Independence failure rate. (The CI measurements yield a
non-zero CI failure rate because IRV does not specify a tie breaker
method and there are lots of lower-level ties in these "stress test"
measurements.) I've removed the IRV-BTR data because it's so similar to
IRV.

Here is how RCIPE version 2 is counted:

Eliminate the "pairwise losing candidate" (Condorcet loser) when there
is one.
Calculate a "pairwise support count" for each candidate. Each ballot
contributes to that count the number of remaining(!) candidates who are
ranked lower than the specified candidate.

I would of course suggest moving the tiebreaker up here to replace the
current elimination condition :-)

"Eliminate the candidate who has the smallest one-on-one pairwise count
against any other remaining candidate."

You need the pairwise matrix to determine the Condorcet loser anyway, so
there's not much of a gain in complexity.

-km

On 21.07.2021 03:19, VoteFair wrote: > Version 2 of the Ranked Choice Including Pairwise Elimination (RCIPE) > voting method is now coded. > > Here's a link to the scatter plot that shows its Clone Independence and > IIA (Independence of Irrelevant Alternatives) success rates compared to > other single-winner methods: > > https://www.rankedchoiceoregon.org/img/clone_iia_success_rates.jpg > > Based on Kristofer's feedback on the scatter plot, I've moved the IRV > (instant-runoff voting) data points to the top to show that IRV has a > zero Clone Independence failure rate. (The CI measurements yield a > non-zero CI failure rate because IRV does not specify a tie breaker > method and there are lots of lower-level ties in these "stress test" > measurements.) I've removed the IRV-BTR data because it's so similar to > IRV. > > Here is how RCIPE version 2 is counted: > > * Eliminate the "pairwise losing candidate" (Condorcet loser) when there > is one. > > * Calculate a "pairwise support count" for each candidate. Each ballot > contributes to that count the number of remaining(!) candidates who are > ranked lower than the specified candidate. I would of course suggest moving the tiebreaker up here to replace the current elimination condition :-) "Eliminate the candidate who has the smallest one-on-one pairwise count against any other remaining candidate." You need the pairwise matrix to determine the Condorcet loser anyway, so there's not much of a gain in complexity. -km

V

VoteFair

Wed, Jul 21, 2021 3:25 PM

On 7/21/2021 4:41 AM, Kristofer Munsterhjelm wrote:

I would of course suggest moving the tiebreaker up here to replace the
current elimination condition :-)

"Eliminate the candidate who has the smallest one-on-one pairwise count
against any other remaining candidate."

I tried placing a higher priority on this kind of elimination because it
avoids eliminating the Condorcet loser. However it causes odd and
undesirable behavior regarding Clone Independence and IIA failures.

Using IRV as the tie breaker protects against clone independence
failures -- which means it protects against money-based vote splitting
tactics.

You need the pairwise matrix to determine the Condorcet loser anyway, so
there's not much of a gain in complexity.

A major goal of this redesign is to make the method easier to
demonstrate using an animated video.

Introducing the idea of a pairwise matrix would not work well in an
animated video. It's much too difficult for most voters to understand.

Instead, this method can be animated by having upward-pointing arrows
emerge from the marks on a ranked-choice ballot.

For each candidate, the number of those arrows equals the number of
marks to the right of the mark for that candidate. This assumes that
unmarked candidates are marked by default at the far-right, lowest
ranking level.

I'm intending to use candidate names that are readily associated with a
single letter, such as:

Jay (J), Kay (K), Cici (C), Gigi (G)

Each arrow will have the pairwise winner's initial at the top of the
arrow and the pairwise loser's initial at the bottom of the arrow.

That will allow:

Removing arrows that are associated with the candidate who was just
eliminated
Using the same arrows both for checking for Condorcet losers -- when
the arrows are grouped into subgroups (one group per candidate, one
subgroup per arrow letter pair) -- and checking for the smallest
pairwise support count -- when the arrows are not subgrouped.

I'm learning that what's easy to describe in words is not necessarily
easy to explain using graphics.

With fewer and fewer people reading, and more and more people expecting
to learn from videos (which are now easy to access, unlike a few decades
ago), I'm trying to make that shift in what I create.

As always, thank you Kristofer for your wise feedback!

Richard Fobes
The VoteFair guy

On 7/21/2021 4:41 AM, Kristofer Munsterhjelm wrote:

On 21.07.2021 03:19, VoteFair wrote:

Version 2 of the Ranked Choice Including Pairwise Elimination (RCIPE)
voting method is now coded.

Here's a link to the scatter plot that shows its Clone Independence and
IIA (Independence of Irrelevant Alternatives) success rates compared to
other single-winner methods:

https://www.rankedchoiceoregon.org/img/clone_iia_success_rates.jpg

Based on Kristofer's feedback on the scatter plot, I've moved the IRV
(instant-runoff voting) data points to the top to show that IRV has a
zero Clone Independence failure rate. (The CI measurements yield a
non-zero CI failure rate because IRV does not specify a tie breaker
method and there are lots of lower-level ties in these "stress test"
measurements.) I've removed the IRV-BTR data because it's so similar to
IRV.

Here is how RCIPE version 2 is counted:

Eliminate the "pairwise losing candidate" (Condorcet loser) when there
is one.
Calculate a "pairwise support count" for each candidate. Each ballot
contributes to that count the number of remaining(!) candidates who are
ranked lower than the specified candidate.

I would of course suggest moving the tiebreaker up here to replace the
current elimination condition :-)

"Eliminate the candidate who has the smallest one-on-one pairwise count
against any other remaining candidate."

You need the pairwise matrix to determine the Condorcet loser anyway, so
there's not much of a gain in complexity.

-km

On 7/21/2021 4:41 AM, Kristofer Munsterhjelm wrote: > I would of course suggest moving the tiebreaker up here to replace the > current elimination condition :-) > > "Eliminate the candidate who has the smallest one-on-one pairwise count > against any other remaining candidate." I tried placing a higher priority on this kind of elimination because it avoids eliminating the Condorcet loser. However it causes odd and undesirable behavior regarding Clone Independence and IIA failures. Using IRV as the tie breaker protects against clone independence failures -- which means it protects against money-based vote splitting tactics. > You need the pairwise matrix to determine the Condorcet loser anyway, so > there's not much of a gain in complexity. A major goal of this redesign is to make the method easier to demonstrate using an animated video. Introducing the idea of a pairwise matrix would not work well in an animated video. It's much too difficult for most voters to understand. Instead, this method can be animated by having upward-pointing arrows emerge from the marks on a ranked-choice ballot. For each candidate, the number of those arrows equals the number of marks to the right of the mark for that candidate. This assumes that unmarked candidates are marked by default at the far-right, lowest ranking level. I'm intending to use candidate names that are readily associated with a single letter, such as: Jay (J), Kay (K), Cici (C), Gigi (G) Each arrow will have the pairwise winner's initial at the top of the arrow and the pairwise loser's initial at the bottom of the arrow. That will allow: * Removing arrows that are associated with the candidate who was just eliminated * Using the same arrows both for checking for Condorcet losers -- when the arrows are grouped into subgroups (one group per candidate, one subgroup per arrow letter pair) -- and checking for the smallest pairwise support count -- when the arrows are not subgrouped. I'm learning that what's easy to describe in words is not necessarily easy to explain using graphics. With fewer and fewer people reading, and more and more people expecting to learn from videos (which are now easy to access, unlike a few decades ago), I'm trying to make that shift in what I create. As always, thank you Kristofer for your wise feedback! Richard Fobes The VoteFair guy On 7/21/2021 4:41 AM, Kristofer Munsterhjelm wrote: > On 21.07.2021 03:19, VoteFair wrote: >> Version 2 of the Ranked Choice Including Pairwise Elimination (RCIPE) >> voting method is now coded. >> >> Here's a link to the scatter plot that shows its Clone Independence and >> IIA (Independence of Irrelevant Alternatives) success rates compared to >> other single-winner methods: >> >> https://www.rankedchoiceoregon.org/img/clone_iia_success_rates.jpg >> >> Based on Kristofer's feedback on the scatter plot, I've moved the IRV >> (instant-runoff voting) data points to the top to show that IRV has a >> zero Clone Independence failure rate. (The CI measurements yield a >> non-zero CI failure rate because IRV does not specify a tie breaker >> method and there are lots of lower-level ties in these "stress test" >> measurements.) I've removed the IRV-BTR data because it's so similar to >> IRV. >> >> Here is how RCIPE version 2 is counted: >> >> * Eliminate the "pairwise losing candidate" (Condorcet loser) when there >> is one. >> >> * Calculate a "pairwise support count" for each candidate. Each ballot >> contributes to that count the number of remaining(!) candidates who are >> ranked lower than the specified candidate. > > I would of course suggest moving the tiebreaker up here to replace the > current elimination condition :-) > > "Eliminate the candidate who has the smallest one-on-one pairwise count > against any other remaining candidate." > > You need the pairwise matrix to determine the Condorcet loser anyway, so > there's not much of a gain in complexity. > > -km >

KM

Kristofer Munsterhjelm

Thu, Jul 22, 2021 1:04 PM

On 7/21/21 5:25 PM, VoteFair wrote:

On 7/21/2021 4:41 AM, Kristofer Munsterhjelm wrote:

I would of course suggest moving the tiebreaker up here to replace the
current elimination condition :-)

"Eliminate the candidate who has the smallest one-on-one pairwise count
against any other remaining candidate."

I tried placing a higher priority on this kind of elimination because it
avoids eliminating the Condorcet loser. However it causes odd and
undesirable behavior regarding Clone Independence and IIA failures.

Using IRV as the tie breaker protects against clone independence
failures -- which means it protects against money-based vote splitting
tactics.

I meant as a replacement for the pairwise support count step, which is
not cloneproof either. Since it's a sum of the number of votes, you
would be very likely to get a unique result, so the IRV and minmax steps
would very rarely come into play.

You need the pairwise matrix to determine the Condorcet loser anyway, so
there's not much of a gain in complexity.

A major goal of this redesign is to make the method easier to
demonstrate using an animated video.

Introducing the idea of a pairwise matrix would not work well in an
animated video. It's much too difficult for most voters to understand.

Instead, this method can be animated by having upward-pointing arrows
emerge from the marks on a ranked-choice ballot.

For each candidate, the number of those arrows equals the number of
marks to the right of the mark for that candidate. This assumes that
unmarked candidates are marked by default at the far-right, lowest
ranking level.

I'm intending to use candidate names that are readily associated with a
single letter, such as:

Jay (J), Kay (K), Cici (C), Gigi (G)

Each arrow will have the pairwise winner's initial at the top of the
arrow and the pairwise loser's initial at the bottom of the arrow.

That will allow:

Removing arrows that are associated with the candidate who was just
eliminated
Using the same arrows both for checking for Condorcet losers -- when
the arrows are grouped into subgroups (one group per candidate, one
subgroup per arrow letter pair) -- and checking for the smallest
pairwise support count -- when the arrows are not subgrouped. >
I'm learning that what's easy to describe in words is not necessarily
easy to explain using graphics.

I see, so you are in effect restricted to the sign-values of the
Condorcet matrix entries.

How about this?

Eliminate the candidate with the least number of winning subgroups.
If there is a tie, break that tie by IRV.

That's simpler and always eliminates the Condorcet loser (because such a
loser obviously has no winning subgroups at all). Because the winning
subgroups depend only on the number of candidates and not on the number
of voters, you should get ties more often, which means that you have a
greater chance of using the clone-independent component (IRV). And you
don't have to introduce un-subgrouped arrows at all.

Alternatively you could do:

Eliminate the Condorcet loser
If there is none, eliminate the candidate with the least number of
winning subgroups,
If there is a tie, break it by ungrouped arrows,
If there's still a tie, break it by IRV,
If there's still a tie, break it by minmax.

But again, the ungrouped mechanic is not cloneproof.

-km

On 7/21/21 5:25 PM, VoteFair wrote: > On 7/21/2021 4:41 AM, Kristofer Munsterhjelm wrote: > > I would of course suggest moving the tiebreaker up here to replace the > > current elimination condition :-) > > > > "Eliminate the candidate who has the smallest one-on-one pairwise count > > against any other remaining candidate." > > I tried placing a higher priority on this kind of elimination because it > avoids eliminating the Condorcet loser. However it causes odd and > undesirable behavior regarding Clone Independence and IIA failures. > > Using IRV as the tie breaker protects against clone independence > failures -- which means it protects against money-based vote splitting > tactics. I meant as a replacement for the pairwise support count step, which is not cloneproof either. Since it's a sum of the number of votes, you would be very likely to get a unique result, so the IRV and minmax steps would very rarely come into play. > > You need the pairwise matrix to determine the Condorcet loser anyway, so > > there's not much of a gain in complexity. > > A major goal of this redesign is to make the method easier to > demonstrate using an animated video. > > Introducing the idea of a pairwise matrix would not work well in an > animated video. It's much too difficult for most voters to understand. > > Instead, this method can be animated by having upward-pointing arrows > emerge from the marks on a ranked-choice ballot. > > For each candidate, the number of those arrows equals the number of > marks to the right of the mark for that candidate. This assumes that > unmarked candidates are marked by default at the far-right, lowest > ranking level. > > I'm intending to use candidate names that are readily associated with a > single letter, such as: > > Jay (J), Kay (K), Cici (C), Gigi (G) > > Each arrow will have the pairwise winner's initial at the top of the > arrow and the pairwise loser's initial at the bottom of the arrow. > > That will allow: > > * Removing arrows that are associated with the candidate who was just > eliminated > > * Using the same arrows both for checking for Condorcet losers -- when > the arrows are grouped into subgroups (one group per candidate, one > subgroup per arrow letter pair) -- and checking for the smallest > pairwise support count -- when the arrows are not subgrouped. > > I'm learning that what's easy to describe in words is not necessarily > easy to explain using graphics. I see, so you are in effect restricted to the sign-values of the Condorcet matrix entries. How about this? - Eliminate the candidate with the least number of winning subgroups. - If there is a tie, break that tie by IRV. That's simpler and always eliminates the Condorcet loser (because such a loser obviously has no winning subgroups at all). Because the winning subgroups depend only on the number of candidates and not on the number of voters, you should get ties more often, which means that you have a greater chance of using the clone-independent component (IRV). And you don't have to introduce un-subgrouped arrows at all. Alternatively you could do: - Eliminate the Condorcet loser - If there is none, eliminate the candidate with the least number of winning subgroups, - If there is a tie, break it by ungrouped arrows, - If there's still a tie, break it by IRV, - If there's still a tie, break it by minmax. But again, the ungrouped mechanic is not cloneproof. -km

V

VoteFair

Thu, Jul 22, 2021 3:31 PM

On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote:

How about this?

Eliminate the candidate with the least number of winning subgroups.
If there is a tie, break that tie by IRV.
...

Isn't the first step basically Copeland's method?

That's an ugly "method" that fails to look beneath the surface.

IRV also fails to look beneath the surface, which is why it too is an
"ugly" method.

Alternatively you could do:

Eliminate the Condorcet loser
If there is none, eliminate the candidate with the least number of
winning subgroups,
If there is a tie, break it by ungrouped arrows,
If there's still a tie, break it by IRV,
If there's still a tie, break it by minmax.

That still uses a very crude/ugly win/loss count approach.

But again, the ungrouped mechanic is not cloneproof.

Being cloneproof is not a goal. The goal is to have a very small failure
rate for clone independence.

Also, electing the Condorcet winner is not a goal. The goal is to have a
very small Condorcet criteria failure rate.

To repeat my concern, attempting to get a zero failure rate will cause
other kinds of failure rates to increase.

I'm still willing to consider improvements, but it needs to find a
balance between what voters can understand -- both through an animated
video and through words -- and what yields low failure rates.

Again, thank you Kristofer for applying your clear understanding to this
revision from RCIPE 1 to RCIPE 2.

Richard Fobes
The VoteFair guy

On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote:

On 7/21/21 5:25 PM, VoteFair wrote:

On 7/21/2021 4:41 AM, Kristofer Munsterhjelm wrote:

I would of course suggest moving the tiebreaker up here to replace the
current elimination condition :-)

"Eliminate the candidate who has the smallest one-on-one pairwise

count

against any other remaining candidate."

I tried placing a higher priority on this kind of elimination because
it avoids eliminating the Condorcet loser. However it causes odd and
undesirable behavior regarding Clone Independence and IIA failures.

Using IRV as the tie breaker protects against clone independence
failures -- which means it protects against money-based vote splitting
tactics.

I meant as a replacement for the pairwise support count step, which is
not cloneproof either. Since it's a sum of the number of votes, you
would be very likely to get a unique result, so the IRV and minmax steps
would very rarely come into play.

You need the pairwise matrix to determine the Condorcet loser

anyway, so

there's not much of a gain in complexity.

A major goal of this redesign is to make the method easier to
demonstrate using an animated video.

Introducing the idea of a pairwise matrix would not work well in an
animated video. It's much too difficult for most voters to understand.

Instead, this method can be animated by having upward-pointing arrows
emerge from the marks on a ranked-choice ballot.

For each candidate, the number of those arrows equals the number of
marks to the right of the mark for that candidate. This assumes that
unmarked candidates are marked by default at the far-right, lowest
ranking level.

I'm intending to use candidate names that are readily associated with
a single letter, such as:

Jay (J), Kay (K), Cici (C), Gigi (G)

Each arrow will have the pairwise winner's initial at the top of the
arrow and the pairwise loser's initial at the bottom of the arrow.

That will allow:

Removing arrows that are associated with the candidate who was just
eliminated
Using the same arrows both for checking for Condorcet losers -- when
the arrows are grouped into subgroups (one group per candidate, one
subgroup per arrow letter pair) -- and checking for the smallest
pairwise support count -- when the arrows are not subgrouped. >
I'm learning that what's easy to describe in words is not necessarily
easy to explain using graphics.

I see, so you are in effect restricted to the sign-values of the
Condorcet matrix entries.

How about this?

Eliminate the candidate with the least number of winning subgroups.
If there is a tie, break that tie by IRV.

That's simpler and always eliminates the Condorcet loser (because such a
loser obviously has no winning subgroups at all). Because the winning
subgroups depend only on the number of candidates and not on the number
of voters, you should get ties more often, which means that you have a
greater chance of using the clone-independent component (IRV). And you
don't have to introduce un-subgrouped arrows at all.

Alternatively you could do:

Eliminate the Condorcet loser
If there is none, eliminate the candidate with the least number of
winning subgroups,
If there is a tie, break it by ungrouped arrows,
If there's still a tie, break it by IRV,
If there's still a tie, break it by minmax.

But again, the ungrouped mechanic is not cloneproof.

-km

On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote: > How about this? > > - Eliminate the candidate with the least number of winning subgroups. > - If there is a tie, break that tie by IRV. > ... Isn't the first step basically Copeland's method? That's an ugly "method" that fails to look beneath the surface. IRV also fails to look beneath the surface, which is why it too is an "ugly" method. > Alternatively you could do: > > - Eliminate the Condorcet loser > - If there is none, eliminate the candidate with the least number of > winning subgroups, > - If there is a tie, break it by ungrouped arrows, > - If there's still a tie, break it by IRV, > - If there's still a tie, break it by minmax. That still uses a very crude/ugly win/loss count approach. > But again, the ungrouped mechanic is not cloneproof. Being cloneproof is not a goal. The goal is to have a very small failure rate for clone independence. Also, electing the Condorcet winner is not a goal. The goal is to have a very small Condorcet criteria failure rate. To repeat my concern, attempting to get a zero failure rate will cause other kinds of failure rates to increase. I'm still willing to consider improvements, but it needs to find a balance between what voters can understand -- both through an animated video and through words -- and what yields low failure rates. Again, thank you Kristofer for applying your clear understanding to this revision from RCIPE 1 to RCIPE 2. Richard Fobes The VoteFair guy On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote: > On 7/21/21 5:25 PM, VoteFair wrote: >> On 7/21/2021 4:41 AM, Kristofer Munsterhjelm wrote: >> > I would of course suggest moving the tiebreaker up here to replace the >> > current elimination condition :-) >> > >> > "Eliminate the candidate who has the smallest one-on-one pairwise >> count >> > against any other remaining candidate." >> >> I tried placing a higher priority on this kind of elimination because >> it avoids eliminating the Condorcet loser. However it causes odd and >> undesirable behavior regarding Clone Independence and IIA failures. >> >> Using IRV as the tie breaker protects against clone independence >> failures -- which means it protects against money-based vote splitting >> tactics. > > I meant as a replacement for the pairwise support count step, which is > not cloneproof either. Since it's a sum of the number of votes, you > would be very likely to get a unique result, so the IRV and minmax steps > would very rarely come into play. > >> > You need the pairwise matrix to determine the Condorcet loser >> anyway, so >> > there's not much of a gain in complexity. >> >> A major goal of this redesign is to make the method easier to >> demonstrate using an animated video. >> >> Introducing the idea of a pairwise matrix would not work well in an >> animated video. It's much too difficult for most voters to understand. >> >> Instead, this method can be animated by having upward-pointing arrows >> emerge from the marks on a ranked-choice ballot. >> >> For each candidate, the number of those arrows equals the number of >> marks to the right of the mark for that candidate. This assumes that >> unmarked candidates are marked by default at the far-right, lowest >> ranking level. >> >> I'm intending to use candidate names that are readily associated with >> a single letter, such as: >> >> Jay (J), Kay (K), Cici (C), Gigi (G) >> >> Each arrow will have the pairwise winner's initial at the top of the >> arrow and the pairwise loser's initial at the bottom of the arrow. >> >> That will allow: >> >> * Removing arrows that are associated with the candidate who was just >> eliminated >> >> * Using the same arrows both for checking for Condorcet losers -- when >> the arrows are grouped into subgroups (one group per candidate, one >> subgroup per arrow letter pair) -- and checking for the smallest >> pairwise support count -- when the arrows are not subgrouped. > >> I'm learning that what's easy to describe in words is not necessarily >> easy to explain using graphics. > > I see, so you are in effect restricted to the sign-values of the > Condorcet matrix entries. > > How about this? > > - Eliminate the candidate with the least number of winning subgroups. > - If there is a tie, break that tie by IRV. > > That's simpler and always eliminates the Condorcet loser (because such a > loser obviously has no winning subgroups at all). Because the winning > subgroups depend only on the number of candidates and not on the number > of voters, you should get ties more often, which means that you have a > greater chance of using the clone-independent component (IRV). And you > don't have to introduce un-subgrouped arrows at all. > > Alternatively you could do: > > - Eliminate the Condorcet loser > - If there is none, eliminate the candidate with the least number of > winning subgroups, > - If there is a tie, break it by ungrouped arrows, > - If there's still a tie, break it by IRV, > - If there's still a tie, break it by minmax. > > But again, the ungrouped mechanic is not cloneproof. > > -km

KM

Kristofer Munsterhjelm

Sat, Jul 24, 2021 9:19 PM

On 7/22/21 5:31 PM, VoteFair wrote:

On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote:

How about this?

Eliminate the candidate with the least number of winning subgroups.
If there is a tie, break that tie by IRV.
...

Isn't the first step basically Copeland's method?

No, because there's no elimination in Copeland (and it doesn't pass
LIIA). It would just elect the candidate/s with the most winning subgroups.

That's an ugly "method" that fails to look beneath the surface.

IRV also fails to look beneath the surface, which is why it too is an
"ugly" method.

That leads me to wonder which is the case.

You said you couldn't replace the IRV tiebreaker with minmax elimination
because IRV is cloneproof and minmax is not -- that clone independence
was important because it "protects against money-based vote splitting
tactics". So I found something that invokes IRV's clone independence
more often.

But then clone independence is not important after all because the
methods are ugly. I can't quite determine whether clone independence is
important or not.

But again, the ungrouped mechanic is not cloneproof.

Being cloneproof is not a goal. The goal is to have a very small failure
rate for clone independence.

Then you could check the alternatives by that metric. A method seeming
ugly may not necessarily have any bearing on the rates of failure.

Also, electing the Condorcet winner is not a goal. The goal is to have a
very small Condorcet criteria failure rate.

To repeat my concern, attempting to get a zero failure rate will cause
other kinds of failure rates to increase.

That's true. You implicitly need some kind of valuation of the different
failure rates. For instance, if you want LNHarm and LNHelp, you have to
give up either monotonicity or mutual majority. Which it's going to be
depends on what values you place on the different criteria.

The same would hold for rates. Say you want to find the method that
minimizes w * x, where x is the rates of each failure type
(monotonicity, vote splitting, teaming, crowding, favorite betrayal...).
Then the weights of the w vector provide a measure of indifference: how
much of failure type 1 is an acceptable trade for one unit of failure
type 2?

Or to put it differently: if the method insists on a zero failure rate
for Condorcet loser, why shouldn't it insist on a zero failure rate for
Condorcet winner, say? And, equivalently, if "merely a low rate of
failure" is good enough for the Condorcet criterion (or say, clone
independence), why is it not good enough for Condorcet loser?

I'm still willing to consider improvements, but it needs to find a
balance between what voters can understand -- both through an animated
video and through words -- and what yields low failure rates.

Again, thank you Kristofer for applying your clear understanding to this
revision from RCIPE 1 to RCIPE 2.

You're welcome :-)

-km

On 7/22/21 5:31 PM, VoteFair wrote: > On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote: > > How about this? > > > > - Eliminate the candidate with the least number of winning subgroups. > > - If there is a tie, break that tie by IRV. > > ... > > Isn't the first step basically Copeland's method? No, because there's no elimination in Copeland (and it doesn't pass LIIA). It would just elect the candidate/s with the most winning subgroups. > That's an ugly "method" that fails to look beneath the surface. > > IRV also fails to look beneath the surface, which is why it too is an > "ugly" method. That leads me to wonder which is the case. You said you couldn't replace the IRV tiebreaker with minmax elimination because IRV is cloneproof and minmax is not -- that clone independence was important because it "protects against money-based vote splitting tactics". So I found something that invokes IRV's clone independence more often. But then clone independence is not important after all because the methods are ugly. I can't quite determine whether clone independence is important or not. > > But again, the ungrouped mechanic is not cloneproof. > > Being cloneproof is not a goal. The goal is to have a very small failure > rate for clone independence. Then you could check the alternatives by that metric. A method seeming ugly may not necessarily have any bearing on the rates of failure. > Also, electing the Condorcet winner is not a goal. The goal is to have a > very small Condorcet criteria failure rate. > > To repeat my concern, attempting to get a zero failure rate will cause > other kinds of failure rates to increase. That's true. You implicitly need some kind of valuation of the different failure rates. For instance, if you want LNHarm and LNHelp, you have to give up either monotonicity or mutual majority. Which it's going to be depends on what values you place on the different criteria. The same would hold for rates. Say you want to find the method that minimizes w * x, where x is the rates of each failure type (monotonicity, vote splitting, teaming, crowding, favorite betrayal...). Then the weights of the w vector provide a measure of indifference: how much of failure type 1 is an acceptable trade for one unit of failure type 2? Or to put it differently: if the method insists on a zero failure rate for Condorcet loser, why shouldn't it insist on a zero failure rate for Condorcet winner, say? And, equivalently, if "merely a low rate of failure" is good enough for the Condorcet criterion (or say, clone independence), why is it not good enough for Condorcet loser? > I'm still willing to consider improvements, but it needs to find a > balance between what voters can understand -- both through an animated > video and through words -- and what yields low failure rates. > > Again, thank you Kristofer for applying your clear understanding to this > revision from RCIPE 1 to RCIPE 2. You're welcome :-) -km

RT

Richard, the VoteFair guy

Sun, Jul 25, 2021 3:30 PM

On 7/24/2021 2:19 PM, Kristofer Munsterhjelm wrote:

On 7/22/21 5:31 PM, VoteFair wrote:

On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote:

How about this?

Eliminate the candidate with the least number of winning subgroups.
If there is a tie, break that tie by IRV.
...

Isn't the first step basically Copeland's method?

No, because there's no elimination in Copeland (and it doesn't pass
LIIA). It would just elect the candidate/s with the most
winning subgroups.

I see you're right, of course.

I admit your suggestion is clever because it includes Condorcet loser
elimination.

Yet I'm sure lots of non-math-savvy voters will not trust that the
candidate with the least number of wins is not always the least popular.
I too share that lack of trust.

Keep in mind that lots of voter don't trust the idea that the winner of
all the pairwise contests is always the most popular.

But then clone independence is not important after all because the
methods are ugly. I can't quite determine whether clone independence is
important or not.

It's important that the failure rate is small. But it doesn't need to be
zero.

That's true. You implicitly need some kind of valuation of the different
failure rates. For instance, if you want LNHarm and LNHelp, you have to
give up either monotonicity or mutual majority. Which it's going to be
depends on what values you place on the different criteria.

I'm not concerned about monotonicity, LNHarm, LNHelp or any other
failures that are difficult to exploit. I'm much more concerned about
exploitable failures.

Admittedly, as a fan of Condorcet-Kemeny, I favor looking deep into the
ballots, and I favor ways of "sorting" that basically move the biggest
pairwise counts into one half of the usual matrix while moving the
smallest pairwise counts into the other half, where the dividing line is
the diagonal where candidates are paired with themselves.

Or to put it differently: if the method insists on a zero failure rate
for Condorcet loser, why shouldn't it insist on a zero failure rate for
Condorcet winner, say? And, equivalently, if "merely a low rate of
failure" is good enough for the Condorcet criterion (or say, clone
independence), why is it not good enough for Condorcet loser?

I admit I'm intentionally avoiding a zero failure rate for Condorcet
winner because that makes the method into a Condorcet method, and those
have been vilified (portrayed as evil) by the FairVote organization, and
to some extent by STAR fans.

Plus, just as a voter is not likely to trust that the candidate with the
fewest wins is least popular, they aren't likely to trust that the
candidate who wins all the pairwise matches is most popular.

So at this point I'm still happy with eliminating the Condorcet loser as
the top priority and otherwise eliminating the candidate who has the
smallest pairwise support count (which basically counts how many
remaining candidates are ranked below the candidate being scored).

At this point I continue to be open to suggestions for something better,
but that window of time is closing very soon.

Again, thank you Kristofer for your wise feedback!

Richard Fobes
The VoteFair guy

On 7/24/2021 2:19 PM, Kristofer Munsterhjelm wrote:

On 7/22/21 5:31 PM, VoteFair wrote:

On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote:

How about this?

Eliminate the candidate with the least number of winning subgroups.
If there is a tie, break that tie by IRV.
...

Isn't the first step basically Copeland's method?

No, because there's no elimination in Copeland (and it doesn't pass
LIIA). It would just elect the candidate/s with the most winning subgroups.

That's an ugly "method" that fails to look beneath the surface.

IRV also fails to look beneath the surface, which is why it too is an
"ugly" method.

That leads me to wonder which is the case.

You said you couldn't replace the IRV tiebreaker with minmax elimination
because IRV is cloneproof and minmax is not -- that clone independence
was important because it "protects against money-based vote splitting
tactics". So I found something that invokes IRV's clone independence
more often.

But then clone independence is not important after all because the
methods are ugly. I can't quite determine whether clone independence is
important or not.

But again, the ungrouped mechanic is not cloneproof.

Being cloneproof is not a goal. The goal is to have a very small
failure rate for clone independence.

Then you could check the alternatives by that metric. A method seeming
ugly may not necessarily have any bearing on the rates of failure.

Also, electing the Condorcet winner is not a goal. The goal is to have
a very small Condorcet criteria failure rate.

To repeat my concern, attempting to get a zero failure rate will cause
other kinds of failure rates to increase.

That's true. You implicitly need some kind of valuation of the different
failure rates. For instance, if you want LNHarm and LNHelp, you have to
give up either monotonicity or mutual majority. Which it's going to be
depends on what values you place on the different criteria.

The same would hold for rates. Say you want to find the method that
minimizes w * x, where x is the rates of each failure type
(monotonicity, vote splitting, teaming, crowding, favorite betrayal...).
Then the weights of the w vector provide a measure of indifference: how
much of failure type 1 is an acceptable trade for one unit of failure
type 2?

Or to put it differently: if the method insists on a zero failure rate
for Condorcet loser, why shouldn't it insist on a zero failure rate for
Condorcet winner, say? And, equivalently, if "merely a low rate of
failure" is good enough for the Condorcet criterion (or say, clone
independence), why is it not good enough for Condorcet loser?

I'm still willing to consider improvements, but it needs to find a
balance between what voters can understand -- both through an animated
video and through words -- and what yields low failure rates.

Again, thank you Kristofer for applying your clear understanding to
this revision from RCIPE 1 to RCIPE 2.

You're welcome :-)

-km

On 7/24/2021 2:19 PM, Kristofer Munsterhjelm wrote: > On 7/22/21 5:31 PM, VoteFair wrote: >> On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote: >> > How about this? >> > >> > - Eliminate the candidate with the least number of winning subgroups. >> > - If there is a tie, break that tie by IRV. >> > ... >> >> Isn't the first step basically Copeland's method? > > No, because there's no elimination in Copeland (and it doesn't pass > LIIA). It would just elect the candidate/s with the most > winning subgroups. I see you're right, of course. I admit your suggestion is clever because it includes Condorcet loser elimination. Yet I'm sure lots of non-math-savvy voters will not trust that the candidate with the least number of wins is not always the least popular. I too share that lack of trust. Keep in mind that lots of voter don't trust the idea that the winner of all the pairwise contests is always the most popular. > But then clone independence is not important after all because the > methods are ugly. I can't quite determine whether clone independence is > important or not. It's important that the failure rate is small. But it doesn't need to be zero. > That's true. You implicitly need some kind of valuation of the different > failure rates. For instance, if you want LNHarm and LNHelp, you have to > give up either monotonicity or mutual majority. Which it's going to be > depends on what values you place on the different criteria. I'm not concerned about monotonicity, LNHarm, LNHelp or any other failures that are difficult to exploit. I'm much more concerned about exploitable failures. Admittedly, as a fan of Condorcet-Kemeny, I favor looking deep into the ballots, and I favor ways of "sorting" that basically move the biggest pairwise counts into one half of the usual matrix while moving the smallest pairwise counts into the other half, where the dividing line is the diagonal where candidates are paired with themselves. > Or to put it differently: if the method insists on a zero failure rate > for Condorcet loser, why shouldn't it insist on a zero failure rate for > Condorcet winner, say? And, equivalently, if "merely a low rate of > failure" is good enough for the Condorcet criterion (or say, clone > independence), why is it not good enough for Condorcet loser? I admit I'm intentionally avoiding a zero failure rate for Condorcet winner because that makes the method into a Condorcet method, and those have been vilified (portrayed as evil) by the FairVote organization, and to some extent by STAR fans. Plus, just as a voter is not likely to trust that the candidate with the fewest wins is least popular, they aren't likely to trust that the candidate who wins all the pairwise matches is most popular. So at this point I'm still happy with eliminating the Condorcet loser as the top priority and otherwise eliminating the candidate who has the smallest pairwise support count (which basically counts how many remaining candidates are ranked below the candidate being scored). At this point I continue to be open to suggestions for something better, but that window of time is closing very soon. Again, thank you Kristofer for your wise feedback! Richard Fobes The VoteFair guy On 7/24/2021 2:19 PM, Kristofer Munsterhjelm wrote: > On 7/22/21 5:31 PM, VoteFair wrote: >> On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote: >> > How about this? >> > >> > - Eliminate the candidate with the least number of winning subgroups. >> > - If there is a tie, break that tie by IRV. >> > ... >> >> Isn't the first step basically Copeland's method? > > No, because there's no elimination in Copeland (and it doesn't pass > LIIA). It would just elect the candidate/s with the most winning subgroups. > >> That's an ugly "method" that fails to look beneath the surface. >> >> IRV also fails to look beneath the surface, which is why it too is an >> "ugly" method. > > That leads me to wonder which is the case. > > You said you couldn't replace the IRV tiebreaker with minmax elimination > because IRV is cloneproof and minmax is not -- that clone independence > was important because it "protects against money-based vote splitting > tactics". So I found something that invokes IRV's clone independence > more often. > > But then clone independence is not important after all because the > methods are ugly. I can't quite determine whether clone independence is > important or not. > >> > But again, the ungrouped mechanic is not cloneproof. >> >> Being cloneproof is not a goal. The goal is to have a very small >> failure rate for clone independence. > > Then you could check the alternatives by that metric. A method seeming > ugly may not necessarily have any bearing on the rates of failure. > >> Also, electing the Condorcet winner is not a goal. The goal is to have >> a very small Condorcet criteria failure rate. >> >> To repeat my concern, attempting to get a zero failure rate will cause >> other kinds of failure rates to increase. > > That's true. You implicitly need some kind of valuation of the different > failure rates. For instance, if you want LNHarm and LNHelp, you have to > give up either monotonicity or mutual majority. Which it's going to be > depends on what values you place on the different criteria. > > The same would hold for rates. Say you want to find the method that > minimizes w * x, where x is the rates of each failure type > (monotonicity, vote splitting, teaming, crowding, favorite betrayal...). > Then the weights of the w vector provide a measure of indifference: how > much of failure type 1 is an acceptable trade for one unit of failure > type 2? > > Or to put it differently: if the method insists on a zero failure rate > for Condorcet loser, why shouldn't it insist on a zero failure rate for > Condorcet winner, say? And, equivalently, if "merely a low rate of > failure" is good enough for the Condorcet criterion (or say, clone > independence), why is it not good enough for Condorcet loser? > >> I'm still willing to consider improvements, but it needs to find a >> balance between what voters can understand -- both through an animated >> video and through words -- and what yields low failure rates. >> >> Again, thank you Kristofer for applying your clear understanding to >> this revision from RCIPE 1 to RCIPE 2. > > You're welcome :-) > > -km

RL

Richard Lung

Sun, Jul 25, 2021 9:00 PM

A few comments from Richard Lung (not the VoteFair guy, (who is not to be confused, if I remember rightly, with Santucci, the vote guy!).

As pointed out to Susan Simmons, which she acknowledged, eliminating candidates, during the count, loses voting information, before the count is over. It is not necessary with a binomial count, unlike all existing methods (uninomial counts) which employ elimination as an afterthought to an essentially uninomial election count.
I repeat, in case you missed it, science demands one truth (to aspire to) not two. Therefore an election count and an exclusion/elimination count must be symmetrical. Call it symmetrical count requirement. But that is a binomial count.
And a binomial count indeed does imply higher order counts, governed by the binomial theorem. But a simple coherent first order binomial count should be sufficient for democratic representation.
Whereas, FAB STV is the whole caboodle perhaps relevant to data mining. This system is monotonic, not vulnerable to strategic shuffling the preference orders. It avoids premature exclusion, and indeed premature election! It entirely avoids later harm, not just for transfer of surplus preferences.
It meets the Laplace condition of weighting a whole range of preferences in order of importance, unlike Condorcet pairing, whether or not the pairs are weighted in relative importance.
FAB STV recognises elections as statistical estimates of representation, and employs up to four averages to maximise accuracy. It accepts the "Impossibility" of a deterministic election result, and moves on. I beseech you all to do the same!

Yours sincerely,
Richard Lung.

On 25 Jul 2021, at 4:30 pm, Richard, the VoteFair guy electionmethods@votefair.org wrote:

On 7/24/2021 2:19 PM, Kristofer Munsterhjelm wrote:

On 7/22/21 5:31 PM, VoteFair wrote:

On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote:

How about this?

Eliminate the candidate with the least number of winning subgroups.
If there is a tie, break that tie by IRV.
...

Isn't the first step basically Copeland's method?

No, because there's no elimination in Copeland (and it doesn't pass
LIIA). It would just elect the candidate/s with the most
winning subgroups.

I see you're right, of course.

I admit your suggestion is clever because it includes Condorcet loser elimination.

Yet I'm sure lots of non-math-savvy voters will not trust that the candidate with the least number of wins is not always the least popular. I too share that lack of trust.

Keep in mind that lots of voter don't trust the idea that the winner of all the pairwise contests is always the most popular.

But then clone independence is not important after all because the
methods are ugly. I can't quite determine whether clone independence is
important or not.

It's important that the failure rate is small. But it doesn't need to be zero.

That's true. You implicitly need some kind of valuation of the different
failure rates. For instance, if you want LNHarm and LNHelp, you have to
give up either monotonicity or mutual majority. Which it's going to be
depends on what values you place on the different criteria.

I'm not concerned about monotonicity, LNHarm, LNHelp or any other On 7/24/2021 2:19 PM, Kristofer Munsterhjelm wrote:

On 7/22/21 5:31 PM, VoteFair wrote:

On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote:

How about this?

Eliminate the candidate with the least number of winning subgroups.
If there is a tie, break that tie by IRV.
...

Isn't the first step basically Copeland's method?

No, because there's no elimination in Copeland (and it doesn't pass
LIIA). It would just elect the candidate/s with the most
winning subgroups.

I see you're right, of course.

I admit your suggestion is clever because it includes Condorcet loser elimination.

Yet I'm sure lots of non-math-savvy voters will not trust that the candidate with the least number of wins is not always the least popular. I too share that lack of trust.

Keep in mind that lots of voter don't trust the idea that the winner of all the pairwise contests is always the most popular.

But then clone independence is not important after all because the
methods are ugly. I can't quite determine whether clone independence is
important or not.

It's important that the failure rate is small. But it doesn't need to be zero.

That's true. You implicitly need some kind of valuation of the different
failure rates. For instance, if you want LNHarm and LNHelp, you have to
give up either monotonicity or mutual majority. Which it's going to be
depends on what values you place on the different criteria.

I'm not concerned about monotonicity, LNHarm, LNHelp or any other failures that are difficult to exploit. I'm much more concerned about exploitable failures.

Admittedly, as a fan of Condorcet-Kemeny, I favor looking deep into the ballots, and I favor ways of "sorting" that basically move the biggest pairwise counts into one half of the usual matrix while moving the smallest pairwise counts into the other half, where the dividing line is the diagonal where candidates are paired with themselves.

Or to put it differently: if the method insists on a zero failure rate
for Condorcet loser, why shouldn't it insist on a zero failure rate for
Condorcet winner, say? And, equivalently, if "merely a low rate of
failure" is good enough for the Condorcet criterion (or say, clone
independence), why is it not good enough for Condorcet loser?

I admit I'm intentionally avoiding a zero failure rate for Condorcet winner because that makes the method into a Condorcet method, and those have been vilified (portrayed as evil) by the FairVote organization, and to some extent by STAR fans.

Plus, just as a voter is not likely to trust that the candidate with the fewest wins is least popular, they aren't likely to trust that the candidate who wins all the pairwise matches is most popular.

So at this point I'm still happy with eliminating the Condorcet loser as the top priority and otherwise eliminating the candidate who has the smallest pairwise support count (which basically counts how many remaining candidates are ranked below the candidate being scored).

At this point I continue to be open to suggestions for something better, but that window of time is closing very soon.

Again, thank you Kristofer for your wise feedback!

Richard Fobes
The VoteFair guy

On 7/24/2021 2:19 PM, Kristofer Munsterhjelm wrote:

On 7/22/21 5:31 PM, VoteFair wrote:
On 7/22/2021 6:04 AM, Kristofer Munsterhjelm wrote:

How about this?

Eliminate the candidate with the least number of winning subgroups.
If there is a tie, break that tie by IRV.
...