I read Richard as saying that BSTV is not defined for scenarios with partial rankings.
That suggests Kristopher's scenarios actually don't have solutions, and scenarios
relevant to Later-no-harm can't be constructed for BSTV.

However, we could still hope to see a solution for a four-candidate scenario. I think
any would do. That would help us determine whether BSTV is monotone in general.

Kevin

Le samedi 21 mai 2022, 12:24:34 UTC−5, Andy Dienes andydienes@gmail.com a écrit :

KM

BTW my ipad doesn't keep a lot of emails, including yours.

My apologies for not answering sooner (privilege of old age).

Your calculation criticism is perfectly fair. I think the best answer I
can give is that C is elected on over the quota. And that literally
No-one or Nemo is excluded, on an even larger exclusion quota.

If Binomial STV implementation progresses, there would no doubt be a
good many other second guesses -- it would be astonishing if there weren't.

My main interests are to see things like the effect of exponential decay
of preferences, in realistic elections, to assess  the balance of power
between election and exclusion counts. And to see how much significance
tests are needed on candidate proximity to the election quota, when that
candidate has not been excluded -- which is not the same as being elected.

Regards,

Richard Lung.

On 21/05/2022 08:57, Richard Lung wrote:

Hello Andy,

these are my free 2 booklets and one book on binomial stv, varying from
about 1500 words to 50,000 words. Tho only part two of the book,
describes the (almost) complete working.

RL.

Smashwords – Elect and Exclude. Binomial STV Hand Count Basics – a book
by Richard Lung https://www.smashwords.com/books/view/1111349

Smashwords – The Super-Vote Supercharged: Binomial STV elections Hand
Count – a book by Richard Lung
https://www.smashwords.com/books/view/1106292

Smashwords – FAB STV: Four Averages Binomial Single Transferable Vote. –
a book by Richard Lung https://www.smashwords.com/books/view/806030

On 21/05/2022 18:25, Andy Dienes wrote:

On 21.05.2022 21:29, Kevin Venzke wrote:

That might be the case, but then we can't say that BSTV passes LNHarm,
as it's out of scope for methods that only handle complete rankings.

Perhaps Richard is saying that he's going to construct BSTV using a
framework and that, due to general properties of that framework, the
resulting method must pass LNHarm. That would let him say that BSTV
passes LNHarm without knowing just what the method is going to become.

But his terminology is indeed rather opaque to me and so I have no
chance of verifying such a general proof. About the only thing I can say
for sure is that BSTV can't pass both LNHarm and LNHelp.

I'll see if I can infer anything from the clarification that BSTV elects
C in my example elections. The keep and exclude values would've been
very useful.

-km

BSTV does not handle equal preferences ("ilections") on principle. I
just don't want to go there for the purpose of elections, (choosing-out).

BSTV handles both all preferences filled and missing rankings.

Help and Harm are both made possible to the Election count by the
Exclusion count, but they are based on the preference information, and
not on some assumption or guess about voters wishes. Thus a candidate
who does not land an exclusion quota is helped to consolidate the
election quota. Whereas an exclusion quota (with surplus exclusions)
will harm the candidates election prospects.

That 3-candidate example was rudimentary, as maybe was my assessment of it!

RL

On 22/05/2022 20:42, Kristofer Munsterhjelm wrote:

On 22.05.2022 22:24, Richard Lung wrote:

Later-no-harm is defined like this for single-winner:

Suppose that candidate A wins. Then a voter who provides an incomplete
ballot (e.g. A>B) can not make A lose by filling in the remaining ranks
(below those already provided) in some way.

Later-no-help is analogously defined that if some other candidate B
wins, a voter who provides an incomplete ballot can not make A win by
ranking additional candidates below A.

Ordinary STV passes both criteria. Are you saying that BSTV passes both
as well? Just making sure I read you right!

In any case, neither criterion requires equal-rank to be supported
(that's why I said that example wasn't important, though I was curious
how it would work). They do, however, require truncation to be
supported. So although I'm not sure what you mean by missing rankings, I
take it that you mean truncation.

That's a fair point. The main thing I was going for was trying to
understand how last preferences would be counted for exclusion counts
when there's truncation. Thus I just made up some numbers that would
give an answer to the question (or at least let me distinguish between
common ways of doing it) just by seeing the keep and exclude values. It
was pretty clear that C was going to win, but just how would've let me
understand the method better.

So if I could ask for one more, it would be this more decisive example,
from which I can learn something by just knowing who won:

19600: A
39366: A>C>B
38957: B>C>A

although I would very much like to know the keep and exclude values too.

-km

The snag is that these and other criteria were invented for what amounts to uninomial elections, that is elections that don't have both, or either, a rational election count and a rational exclusion count. Together they make possible the application of the binomial theorem, to higher order counts. My binomial STV hand count is just a first order binomial count of one election count and one exclusion count.
I am not aware of any untoward effects of tactical voting on the bstv system. I am aware of it doing away with residual irrationalities to traditional stv, including Meek method. Tho I accept that traditional stv (zero-order stv in relation to binomial stv) is a robust system, in practise, as the Hare system of at-large stv/pr.

BSTV counts require values for all preference positions, which are equal to the number of candidates. Any preference position may be an abstention. A citizen who never voted but made an exception of their dislike for Donald or Hilary could abstain on their first preference but vote for either on their second preference, effecting an exclusion, because there is only one vacancy.
That is the theory of it. I don't know how well it would work in practise, because there never has been a practise.
But I do know that democracy is minimised, and evidently works badly, based on single vacancies, in the Anglo-American systems.
Fully fledged binomial stv, FAB STV, does not work on less than 4 or 5 member constituencies, the minimum requirement for a democracy of all the people being represented by their choices.

Thank you for your examples. They have helped clarify my thinking -- somewhat!
According to my (accident-prone) working, A wins on a keep value of  38957/58966.
B also has a less than unity keep value of 38957/39366. The difference is that one can say A has been elected on a quota of 48961.5, with 58966 first preferences.
But B has not reached the elective quota. Tho B has not reached the exclusion quota, that only says B has not been excluded.

Regards,
Richard Lung.

On 23 May 2022, at 6:19 pm, Kristofer Munsterhjelm km_elmet@t-online.de wrote:

Later-no-harm is defined like this for single-winner:

Suppose that candidate A wins. Then a voter who provides an incomplete
ballot (e.g. A>B) can not make A lose by filling in the remaining ranks
(below those already provided) in some way.

Later-no-help is analogously defined that if some other candidate B
wins, a voter who provides an incomplete ballot can not make A win by
ranking additional candidates below A.

Ordinary STV passes both criteria. Are you saying that BSTV passes both
as well? Just making sure I read you right!

In any case, neither criterion requires equal-rank to be supported
(that's why I said that example wasn't important, though I was curious
how it would work). They do, however, require truncation to be
supported. So although I'm not sure what you mean by missing rankings, I
take it that you mean truncation.

That's a fair point. The main thing I was going for was trying to
understand how last preferences would be counted for exclusion counts
when there's truncation. Thus I just made up some numbers that would
give an answer to the question (or at least let me distinguish between
common ways of doing it) just by seeing the keep and exclude values. It
was pretty clear that C was going to win, but just how would've let me
understand the method better.

So if I could ask for one more, it would be this more decisive example,
from which I can learn something by just knowing who won:

19600: A
39366: A>C>B
38957: B>C>A

although I would very much like to know the keep and exclude values too.

-km

On 24.05.2022 21:05, Richard Lung wrote:

The criteria are method-agnostic: for any ranked voting method (in this
case, that supports truncation), if someone gives you a failure example,
you can verify if the method passes or fails the criterion without
knowing anything about the internals of the method.

Put differently, suppose that in a scenario perhaps reminiscent of
Roadside Picnic, a mysterious device falls out of the sky. And it turns
out that this mysterious device calls elections: you can input ranked
ballots with a set of buttons and get the results shown as a series of
lights on the other end.

Then as long as it allows for ballots with truncation, it's possible to
check if a particular ballot where A-first voters truncate can be used
to induce a later-no-harm failure.

Whether the strange technology that makes up the device implements
rational election and exclusion counts doesn't matter. As long as it's a
ranked voting method outputting winners and supporting truncation, the
question "does this pass later-no-harm?" makes sense.

The same goes for things like monotonicity, participation, consistency,
Smith, Condorcet, etc. The criteria say something about the desired
behavior of a method. Nothing about the inner workings makes the
criteria inapplicable (apart from some exceptions like the polynomial
runtime criterion).

Without a mathematical description of the method, you couldn't be sure
it actually passes later-no-harm or later-no-help, but as soon as you
found a counterexample, that would settle the question in the negative.

As a ranked method, it must fail IIA, which means that strategy must
sometimes be possible. And as it fails Condorcet, the obvious starting
place to look is for an election where it doesn't pass Condorcet. For
instance, this:

549: A>B>C
366: B>A>C
366: B>C>A
366: C>A>B

A is the Condorcet winner. The first preferences are:
A: 549, B: 732, C: 366
and last preferences:
A: 366, B: 366, C: 915

so the ratios are:
A: 366/549 = 0.67
B: 366/732 = 0.5
C: 915/366 = 2.5

so B wins. Then the C>A>B voters have an incentive to vote A>C>B instead
(compromising), after which the counts are:

A: 366/915 = 0.29
B: 366/732 = 0.5
C: 915/0   = infinity

and A wins. The C>A>B voters prefer A to B, so the strategy is to their
benefit.

It can be shown that, for methods where a majority can always force an
outcome by coordinating how they vote, then modifying the method so that
it elects the Condorcet winner if there is one never increases the
proportion of elections where strategy is useful, and may indeed reduce it.

So by the keep values: A's first preference count is 58966 and last
preference count is 38957, since the keep value is 38957/58966.

You say that B's keep value is 38957/39366, i.e. first preference count
of 39366 and last preference count of 38957. But that seems to be in
reverse order. Indeed, your HTML page shows that it is 39366/38957.

counting last preferences. I was pretty sure that BSTV would fail
later-no-harm because the standard way of counting truncations, as STV
does, is to consider everybody not ranked to be equal-ranked for last;
and if you had done that, then it would be possible to induce later-no-harm.

The good news is that you avoid this particular problem if you count
anything past truncation simply as abstentions. So I guessed wrong,
which was then cleared up by the example, which shows how useful they
are :-)

However, instead it seems that you get later-no-help failure. Consider
this modified election:

18125: A
20035: A>B>C
18722: A>C>B
34488: B>A>C
38634: C>B>A

By my count, the first preferences are: A: 56882, B: 34488, C: 38634
and the last preferences are:          A: 38634, B: 18722, C: 54523
and the last to first ratios are:      A: 0.68,  B: 0.54,  C: 1.41

so B wins. But if now the A voters fill out their ballot by voting
A>C>B, then B's last preference count changes to 36847 and A wins
instead. This is a violation of later-no-help.

Ordinary STV passes both.

I should note that Condorcet methods, that I prefer, fail both. My point
isn't as much that later-no-harm and later-no-help are intrinsically
good, as that it's much easier to check a claim by concrete evidence
than by references to personal terminology (which may be hard to
understand for others or take a lot of time to get acquainted with).

On a final note, I would say that always counting truncation as
abstention could lead to an unknown candidate problem: suppose there's a
candidate who nobody has heard of and thus nobody bothers to rank. But
he has a dedicated following all of whom rank him first. If nobody
obtains a majority, then this candidate could win, e.g. something like:

3300: A>B>C
3300: B>C>A
3200: C>A>B
2: D

I'm also not entirely sure what's going on with the quota transfers. If,
in the single-winner case, someone who exceeds the quota is
automatically elected, then there's no need for any transfers. However,
if passing the quota doesn't guarantee victory, then later-no-harm
failure might actually be possible. Suppose A is just above the quota
and B is just below it (with B closer to the majority line), then if the
A voters only vote for A, A might win; but if they vote A>B, then the
surplus might be transferred to B and make B win. Perhaps. As I said,
I'm not sure how the logic works in that case.

-km

Kristofer noted in passing a very important and under-appreciated advantage
of Condorcet methods:

It can be shown that, for methods where a majority can always force an outcome
by coordinating how they vote, then modifying the method so that it elects
the Condorcet winner if there is one never increases the proportion of
elections where strategy is useful, and may indeed reduce it.

This is a good reason to routinely include in the description of every
Universal Domain single winner method that satisfies the Majority
Criterion, verbiage to the effect ...

"Lacking a candidate that outranks any opponent on more ballots than not
..."

-Forest

El sáb., 28 de may. de 2022 9:43 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió: