Could you give an example of that max B(x)S(x) calculation?

And how do you guarantee that said candidate is in the Smith Set?

Here's an alternative approval top two/three runoff for you:

Find the approval winner and approval runner up.

Count all ballots that approve of either AW or ARU. Find the approval
winner on ballots that don't approve either, call that the Exclusive
Approval candidate, XA.

If the either/or ballots amount to more that a threshold value, say 75%,
then have a two person general election. Otherwise, include XA as a third
candidate for the general.

On Mon, Jun 6, 2022, 20:49 Forest Simmons forest.simmons21@gmail.com
wrote:

On 07.06.2022 05:49, Forest Simmons wrote:

I'd say there are two ideas of generalizing STAR, call them cardinal and
ordinal. The cardinal one should preserve the property that in an LCR
election where the centrist is highly rated, he wins, but where he's
ranked poorly, he loses. (Making this cloneproof without just degrading
to plain Range is hard!)

The ordinal one, on the other hand, should find the CW whenever he
exists, and probably also elect from some nice set (Smith, Landau or
Banks). STAR itself doesn't do this.

I haven't thought about it in detail, but I think Range's de jure IIA
compliance implies that Benham-Range and BTR-Range are both Smith. They
might even be equivalent to Smith,Range. So if you want to look for a
simpler way of explaining Smith,Range, those may be it.

My thinking is like this, for Benham-Range:

Benham-Range eliminates the current Range loser until there's a CW.
Since Range passes IIA, eliminating without normalization doesn't change
the order of the other candidates. Hence all but one Smith set member
will be eliminated, after which the highest scoring Smith set member
becomes the CW and wins. So Benham-Range is just Smith,Range and may be
easier to explain.

Benham-BTR I'm much less sure about because it protects the loser who
pairwise defeats the other. This is more like STAR. It might be that a
prospective loser is protected all the way up the Smith set, and then
beats the highest rated Smith set member pairwise -- unless such a
scenario is impossible.

For runoffs, I think you could argue in two ways. Either the runoff is
about telling two quite good candidates apart, or it's about giving each
faction of the electorate "their" candidate to vote for. The former
suggests (if it's a manual runoff) just picking the two Smith set
members with the highest rating; the latter, something more like "the
candidate rated highest by voters who rated the first candidate low".

If I had to choose, I'd probably go for the former, though it could
affect the turnout in the general: if both candidates are good, a lot of
voters may simply decide not to vote in the general.

Automating the former, STAR would probably do a good job because it's
unusual for a candidate to be the CW yet not in the top two by highest
rating. At least it gets the bang for the buck award, even if it's not
perfect :-)

-km

Very good observations ...and I like your method suggestions. Benham-Range
(without the renormaliization),unlike ordinal Benham, should also be
monotone and precinct summable if I understand it correctly.

El mar., 7 de jun. de 2022 5:12 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:

That's the idea of my suggestion ... pit the Martin Harper Lottery top
probability candidate against the score winner ... or I could have
suggested a runoff between the two top probability lottery candidates.

Better yet ... do Sequential Pairwise Elimination on the support of the MH
Lottery, where the agenda order is from least to greatest probability.

The MH Lottery is based on an idea for
Range-->Plurality DSV: given all of the Range ballots, how should you
decide which candidate ballot B should support (from the superstitious
one-person-one-vote point of view)?

Answer: the candidate X should have a high score B(X) on ballot B, and
should have high support  from the other ballots as well. The support of X
from the other candidates can be gauged by its total score S(X). So X
should be the candidate that maximizes the product B(X)S(X).

Which is why the Martin Harper Lottery probability for candidate X is the
percentage of the ballots B for which X maximizes the product B(X)S(X).

[In 2001 Martin Harper posted to the EM list a definitive explanation for
diehard IRV supporters of how Approval utterly satisfies "one person one
vote" by way of a memorable Maxwell's Demon vote shuttling mechanism. So it
is only fitting that Harper get the credit for this lottery idea!]

Benham-Range is ISDA (department of agriculture?). In fact, it elects the
lowest score candidate that defeats all of the higher score candidates.

So it is a simple proposal with lots of heuristic rationales and criteria
compliances going for it.

Another similar ISDA proposal that I call RSTAR for reasons soon apparent,
always elects the highest score Smith candidate in the three candidate case:

RSTAR like STAR can be formulated as a procedure that takes a set of score
ballots as input, and selects as output a winner from among the scored
candidates on the input ballots.

For comparison, here's a formulation of STAR in that format:

Let beta be a set of Score ballots.

Then the STAR output STAR(beta) is the winner of the two candidate runoff
between X=Score(beta) and Y=Score(beta'), where beta' is the set of ballots
with candidate X temporarily removed.

The RSTAR output RSTAR(beta) is the winner of the two candidate runoff
between X=Score(beta) and Y=RSTAR(beta'), where beta' is the set of ballots
with candidate X temporarily removed.

See the comparison (and tiny contrast)? If you see it and understand it,
then you know that RSTAR must stand for Recursive STAR.

This is the most succinct formulation of SPE(Score), Sequential Pairwise
Elimination with a Score based agenda.

Since SPE is a respectable, classical method with a hoary history of use in
deliberative assemblies, as prescribed by Robert's Rules of Order, this
would be an impeccable public proposal.

The Recursive formulation should serve only as a bridge between STAR and
the less exotic SPE formulation; we don't want any heads to explode!

What say ye?

-Forest

["Ye" is an archaic form of "ya'll" or "youse guys" ;>)]

El mar., 7 de jun. de 2022 10:51 a. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:

Hi Forest,

If I am understanding RSTAR correctly, it should be the same as electing
"the lowest score candidate that defeats all of the higher score
candidates," right?

It certainly sounds intuitive. How would it compare to, e.g. Score Chain
Climbing?

-Andy

On Tue, Jun 7, 2022 at 6:36 PM Forest Simmons forest.simmons21@gmail.com
wrote:

Actually, you have just described the winner of Benham,Score, which
successively eliminates the lowest score candidates until there remains one
candidate that defeats all of the other remaining candidates.

RSTAR also starts at the bottom, but eliminates losers of bottom score
pairwise contests until there remains a candidate that defeats all of the
other remaining candidates.

It is easy to check that in the case of three Smith members, (1) Score
Chain Climbing, elects the highest score Smith member that defeats the
lowest score Smith member ... (2) Benham-Score elects the lowest score
Smith member that defeats every higher score member, and (3) RSTAR elects
the highest score Smith member.

Since public elections rarely have more than three Smith members, that
summary pretty much tells us what to expect from these three methods.

The best one to propose/adopt depends on simulation results and which
method has the simplest, most appealing counting procedure(s), since each
of them can be tallied by several different procedures to check the result
of the 'official' procedure.

If burial is a perceived threat, I would recommend Chain Climbing.
Otherwise, I would recommend the Single Pairwise Elimination description of
RSTAR ... unless it turns out that some description of Benham,Score turns
out to be more appealing to the lay voter.

I am assuming that all three of these will have roughly the same VSE, which
we already know to be better than plain Score from published simulations in
the case of Smith,Score and plain STAR. If one does stand out, we should
see if the voters care about the difference.

The beauty of the above three methods is that they give us Independence
from Smith Dominated Alternatives (ISDA) without having to construct the
Smith set, or even mention it. In fact the simplest procedures have no need
to even mention the word Condorcet, which some people have been sternly
warned against ... like countries with state religions warning their
citizens against reading any literature from non-state religions, or
Christian priests prohibiting Bible reading by their parishioners. [In
sixteenth century England, William Tyndale was burned at the stake for
making an English translation of the Bible so that common English speakers
could read it for themselves.]

-Forest

El mar., 7 de jun. de 2022 4:29 p. m., Andy Dienes andydienes@gmail.com
escribió:

Both Benham-Score and Score based Sequential Pairwise Elimination can be
cast as Pairwise Sorted Score methods.

In both cases you start with the candidates listed in Score order, high to
low, top to bottom, and swap adjacent candidates that are out of order
pairwise until the final list is a pairwise "beat path" finish order. All
of the out-of-order adjacent pairs have been "rectified."

The difference in the two methods is the rectification priority. For
Benham, as long as there remains any adjacent pair out of order, rectify
the pair closest to the top of the list. For SPE, as long as there is any
adjacent pair out of order,  rectify the pair closest to the bottom of the
list. Thus we make a distinction between bubble sort and sink sort.

In either case the Smith set will end up at the top of the list.

However, if we think of rectification as correction of a tentative order,
based on error prone estimated merit, why not start with the pair for which
the score order is most tentative ... least confident?

Suppose the tentative order looks like
A***************BC, with B and C close in estimated score. Which
adjacent pair appears to be more secure, i.e. less tentative in its order?
Thinking in terms of hypothesis testing, which hypothesis would yield
higher confidence A>B or B>C?

That is the rationale for "Score Sorted Margins": prioritize rectifications
with least confidence in their score order ... i.e. where their positions
are only marginally distinguished by score.

This sorting rule makes use of information that is ignored by the bubble
and sink sort rules, and so gives it an edge (no matter how slight) on
accuracy.

It also satisfies a strong Reverse Symmetry property not enjoyed by other
sorts of sorting: if every ballot score x is replaced by 100-x, the finish
order will be precisely reversed.

For these reasons, all else being equal, I currently recommend Score Sorted
Margins over the other potential score/pairwise hybrids.

Here's an analogy that might help explain it:

Suppose that we want to have a Round Robin basketball tournament to rank
the teams from best to least.

It is agreed that if one team defeats all of the other teams, it will be
the champion team. But what if no team beats all of the others? Then the
teams tied for the most wins will have a free-throw contest to resolve the
tie.

But that just decides which team is ranked first. How about second, third
etc?

Somebody comes up with the idea of having the free throw contest moved up
front to the beginning of the tornament, to serve not only its tie breaking
purpose (should the need arise) but also to effectively and fairly seed the
match order.

Each team has each of their 12 players make as many free throws in a row as
possible before missing. The teams are listed in order of total team
successful free-throws, their team "seed score".

We start by interspersing question marks between the adjacent teams on the
list.

Then as long as there remains one or more question marks, find the question
mark separating the two teams closest in their team seed scores. Have these
teams play each other. If the one with the higher seed score beats (or
ties) the one with the lower seed score, simply remove the question mark.

Otherwise, swap the positions of the two teams in the list, remove the
question mark between them, and place question marks between them and their
other neighbors (if not already in place) to reflect the fact that their
relative rank order has not yet been confirmed.

When all of the question marks are gone, the final team finish order has
been reached.

Why is contest priority given to the teams closest in seed score?

Because if two teams are far apart in seed score, it is less likely that a
game between them would result in a swap, i.e. it is less likely that they
are out of order, i.e. more likely that just leaving them alone would not
be a mistake ... so not as urgent to test them against each other.

Thanks!

-Forest

P.S. IRV is like having a tournament consisting of successive free for all
games, where the team making the fewest points is eliminated from the next
free for all. until there are only two teams left, at which point a
regulation game by standard rules determines the tournament winner ... in
all more like a demolition derby than a sports tournament.

El mar., 7 de jun. de 2022 4:29 p. m., Andy Dienes andydienes@gmail.com
escribió: