On 12/05/2022 20:12, Richard Lung wrote:

On 12.05.2022 21:27, Richard Lung wrote:

That's no problem, as it's not really relevant to my question :-) Let me
rephrase.

Some time ago, you gave examples of how to call a single-winner Binomial
STV election. In those examples, everybody voted according to their full
preferences. What I'm wondering is how the calculations (of exclude and
keep values) are done when the ballots are truncated early, or when
voters equal-rank some of the candidates.

So, taking for instance, this single-winner election:

How would Binomial STV determine the winner? And what are the
candidates' keep and exclude values?

And similarly, for the election

What are the candidates' keep and exclude values, and who wins in
single-winner Binomial STV?

-km

Just a quick reply. I'm not much familiar with notation.

Binomial stv is a statistical count that doesn't apply for very small
numbers. For that, there is non-parametric statistics. There is no hard
and fast rule. I'd say about 32 votes minimum. but that's just a
minimum. There is a law of large numbers for better approximations.

I forget the meaning of truncated, kindly explained to me. If you mean
what happens with abstentions, they are counted towards the quota for a
vacancy.

With regard to equal preferences, my position is one of both principle
and expediency. They are counted by the multinomial theorem. But to me
that is an "illection" count, a choosing-in of candidates, not a
choosing-out or election.

I recognise there is probably a place for illection, but not necessariy
in an election. That is for future studies -- not by me.

If a voter is really stuck between two candidates, he or she can toss a
coin for it, and there will be no over-all bias.

At least in elections, I think bringing in a multinomial multitude of
possibilities into an election is an astronomical distraction from the
purpose of electing.

I would like to see the consequences in practise of large electorates
limited use of their preferences.

Regards,

Richard Lung.

On 12/05/2022 22:48, Kristofer Munsterhjelm wrote:

On 14.05.2022 10:33, Richard Lung wrote:

Alright, that's not important to my examples either, so I can easily
just multiply the numbers.

Later-no-harm is a criterion that restricts what happens when people who
leave some candidates unranked, later go on to rank them. For a method
to pass later-no-harm, it needs to support ballots where not all
candidates are ranked. Such ballots are usually called "truncated".

A truncated ballot is the kind of ballot that's allowed in optional STV
such as in New Zealand but disallowed in non-optional STV and AV such as
the ones used in Australia.

So suppose that we have three candidates running for election: A, B, and
C. A truncated ballot is one where, for instance, the voter ranks A and
B and leaves C off the ballot. This is represented in EM notation by

1: A>B

for a single voter, or

1000: A>B

if a thousand voters voted that way. Similarly, a voter who votes:

1: A

is simply expressing a preference for A, being indifferent between B and
C but considering both to be lower ranked than A. And a voter who votes:

1: A=B>C

is expressing indifference between A and B, but consider both to be
better than C.

Could you give an example? I find concrete examples much easier to work
with.

In the election:

3000: A=B>C
5000: B=C>A
13000: C=A>B

who wins single-winner Binomial STV, and what are the keep and exclude
values?

Note that the previous election isn't really relevant to my question, so
if you'd prefer not to answer that one, that's okay. But I would very
much like to know the keep and exclude values, and winner, for this
election:

3000: A>B>C
5000: A>C
13000: C

That is: 3000 voters rank A first, B below A, and C below B;
5000 voters rank A first, C below A, and have not expressed any opinion
about B except that he's (implicitly) ranked below C;
and 13000 voters rank C first and have expressed no opinion about
whether A is better or worse than B, only that both are worse than C.

Now, perhaps your practical implementation of Binomial STV hasn't
defined what happens in the case of truncation. But then its
later-no-harm compliance is undefined, not applicable, because
everything that either fails or passes later-no-harm has to produce
outcomes when given truncated ballots.

As a final note, if determining the keep and exclude values requires
some calculation that's very complex in the number of voters, then feel
free to replace 3000 with 3, 5000 with 5, and 13000 with 13. I'm just
interested in the outcome of some concrete elections with truncation in
them, so I can get a handle on Binomial STV's behavior. It doesn't
have to be the ones I listed.

-km

Hi Richard,

Do you say it doesn't apply to very small numbers because you seek to avoid
divisions by zero in the math? I don't think there is any minimum number of voters
that will guarantee that.

It's the omission of a candidate from a voter's ranking.

This does bring to mind to a suggestion which would not satisfy Later-no-harm:
Suppose that BSTV automatically splits the voter into two when they omit the last
two candidates:

480: A      <-- These voters omit B and C
405: B>C>A
115: C>A>B

So BSTV would automatically split them into two blocs to achieve the following
ballots:

240: A>B>C  <-- half of the bloc
240: A>C>B  <-- the other half
405: B>C>A
115: C>A>B

BSTV elects A.

But now, suppose that the 480 voters had actually decided that they felt that B is
better than C, so they instead voted like this:

480: A>B>C  <-- different vote
405: B>C>A
115: C>A>B

Now BSTV elects B. The A>B>C voters do not prefer that outcome, so they were
better off omitting the lower rankings.

That can never happen in ordinary STV, so we say STV satisfies Later-no-harm.

Kevin

Hello Kevin,

Avoiding divisions by zero is a different unrelated issue.
It applies to the geometric mean, the most essential average for binomial stv.
No, it's just that the binomial distribution doesn't become apparent, except with larger numbers of items.
The omission of a candidate preference implies an abstention, which counts towards a vacancy quota. All preferences are counted: conservation of information. (It led to the holographic principle, in physics!)

Tossing a coin just means that, for voters with no preference between 2 candidates, they will leave it to chance, who gets the prior preference, so that neither one will get an over-all undeserved advantage, to any statistically significant degree. This does assume a large enough sample to iron out any chance advantage to one of the unprefered candidates.

Nothing of the sort, you conjecture, takes place, in binomial stv, which does not omit candidates. That is the point of the exclusion count. BSTV is system of keep values: the quota divided by candidates votes, in surplus or deficit of a quota; counting preferences and reverse preferences (election and exclusion counts).

Regards,
Richard Lung.

On 14 May 2022, at 5:36 pm, Kevin Venzke stepjak@yahoo.fr wrote:

Hi Richard,

Do you say it doesn't apply to very small numbers because you seek to avoid
divisions by zero in the math? I don't think there is any minimum number of voters
that will guarantee that.

It's the omission of a candidate from a voter's ranking.

This does bring to mind to a suggestion which would not satisfy Later-no-harm:
Suppose that BSTV automatically splits the voter into two when they omit the last
two candidates:

480: A      <-- These voters omit B and C
405: B>C>A
115: C>A>B

So BSTV would automatically split them into two blocs to achieve the following
ballots:

240: A>B>C  <-- half of the bloc
240: A>C>B  <-- the other half
405: B>C>A
115: C>A>B

BSTV elects A.

But now, suppose that the 480 voters had actually decided that they felt that B is
better than C, so they instead voted like this:

480: A>B>C  <-- different vote
405: B>C>A
115: C>A>B

Now BSTV elects B. The A>B>C voters do not prefer that outcome, so they were
better off omitting the lower rankings.

That can never happen in ordinary STV, so we say STV satisfies Later-no-harm.

Kevin

Binomial STV is later no harm, unlike Borda count, because uses keep
values , equivalent to Gregory method, for both election and exclusion
counts.

Larger elections are more than about just multiplying numbers. Having
thousands of voters for a few preference perms is not realistic. That
never happens in real elections.

Likewise, Condorcet cycles (I thought I saw the 3-candidate case in one
of your examples) are of negligible frequency, especially the more
candidates and the more voters.

There is no problem with regard to truncations if it involves
abstentions. All preferences are counted: conservation of information.
And so have to be given for a binomial stv count. All that happens to
abstentions including total abstention or NOTA is that they go to a
vacancy quota for a seat to remain unfilled. I don't think any of my
examples show this, but the election of "Mr No-one" or Captain Nemo is
no different in principle from any other election to a seat.

Binomial stv would have the advantage for Australia that last
preferencs, beyond the number of seats, more or less count against the
candidate, in an exclusion count, instead of eventually electing some
unwanted party nominee, on the other side of the political fence.

No equal preferences or "ilections" in elections. I don't know of any
elections that are ilections (with equal preferences), except from
theorists.

Regards,

Richard Lung.

On 14/05/2022 10:45, Kristofer Munsterhjelm wrote:

On 16.05.2022 08:20, Richard Lung wrote:

I would like to check that for myself. That's why I've asked (three
times) if you could give me the concrete keep and exclude values, and
the winners, for particular example elections involving truncation.

Could you please do that?

-km

Thank you, Kristofer,

I refer you to the post as a whole, not just the first couple of lines,
for my answer.

It's not just a matter of "shut up and calculate" to quote a famous
grouse of hapless quantum theory students.

(You make me suspect you are an instructor. However, I appreciate your
consideration and competance.)

Regards,

Richard Lung.

On 16/05/2022 10:13, Kristofer Munsterhjelm wrote:

Hi Richard,

I, too, have had a hard time understanding exactly what are the mechanics
of Binomial STV. I have read your full posts, and I think it would clear it
up for me if you would indeed give a "shut up and calculate" worked example
of the type that Kristofer has sent. Please, without the philosophy and
motivation for the method interspersed, just walk through the calculation
of winners for Binomial STV the same way a computer program would.

-Andy

On Sat, May 21, 2022 at 3:57 AM Richard Lung voting@ukscientists.com
wrote: