STV^1
Thank you, Filip,
The first order Binomial STV is one election count and one exclusion
count, exactly like it (being symmetrical; an iteration). For the
election count I use Meek method of surplus transfers. The distinction,
of that computer count, over the traditional hand counts, is that
preferences, for an already elected candidate, with a quota, are still
recorded. Meek did that by updating the candidate keep value (the quota
divided by a candidates total transferable vote).
Unlike Meek method, I do keep values for every candidate, losers as well
as winners. Candidates in deficit of a quota have keep values of more
than unity, signifying they are excluded. The exclusion count is run
exactly like the election count but with the preferences reversed, so a
quota now becomes an exclusion quota. The rule is simple: an election
count elects candidates reaching the quota. An exclusion count excludes
candidates reaching a quota. One voters preferences is another voters
unpreferences. There is no difference in principle between them.
Binomial STV (symbolised as STV^; first order Binomial STV would be
STV^1. Any order bimomial STV would be STV^n. Preceding forms of STV,
including Meek method, are STV^0. The ballot paper looks just like any
Ranked Choice Vote. But the instructions are different. Every voting
method has voters instructions.
The instructions are, in the case of your example: There are four seats
available and ten candidates to choose from. Your first four preferences
would more or less help to elect candidates. Your next 6 preferences (if
you choose to make them) would less or more help to exclude those
candidates. So, a tenth preference counts as much against a candidate,
as your first preference would count for a candidate. But you don't have
to give any order of preference. A carte blanche is equivalent to NOTA.
If a quota of abstentions is reached, one of the seats is left empty.
This election also gives voters the rational power to exclude candidates.
Some candidates may be both popular and unpopular enough to gain both
election and exclusion quotas. They are both "alive" and "dead" to the
electorate. (A case of "Schrodingers candidate" according to Forest
Simmons.) Whether they are elected or excluded is determined by a
Quotient of the exclusion quota divided by the election quota. If the
ratio is one or less, they are elected; if not, excluded. (The Quotient
is the square of a geometric mean.)
When inverted, the exclusion count is like a second-opinion election.
The geometric means of the candidates election keep values and inverse
exclusion keep value establish the over-all order of popularity of the
candidates (from lowest to highest over-all keep values.
All the voters abstentions have to be counted, to establish whether they
care more to elect or exclude candidates. This also means there is no
reduction of the quota with abstentions, as in Meek method. Counting
abstentions observes the conservation of (preferential) information.
I hired a programmer for first order Binomial STV, which, unlike the
higher orders, should be much simpler than Meek method, and simpler in
conception than the hand counts. However I have always supported them
all my adult life, and am now an old man.
Kristofer found the GitHub link to the programmers coding, which he sent me:
https://github.com/Esrot-Clients/STV_CSV/tree/master
The programmer also sent me a "frontend" for the use of voters:
And he sent me two manuals, which I cannot attach, in case useful to a
technical person, unlike myself, because the moderator doen't allow
messages over a certain size. I think it was a different reason last time.
Regards,
Richard Lung.
If a voter's lower ranks exclude candidates as much as higher ranks count in their favour, then I can't see how the result will end up looking anything like proportional representation, which I believe is still your goal. Take a simple case with two parties (or factions as parties don't need to be explicit), the larger of which has 2/3 of the support and the smaller 1/3. The candidates from the smaller party will be at the top on 1/3 of the ballots, and in a normal PR method would get about a third of the seats. But with this binomial count, they would be excluded and more besides by the 2/3 ranking them at the bottom.
Toby
On Friday, 5 April 2024 at 21:33:52 BST, Richard Lung voting@ukscientists.com wrote:
Thank you, Filip,
The first order Binomial STV is one election count and one exclusion
count, exactly like it (being symmetrical; an iteration). For the
election count I use Meek method of surplus transfers. The distinction,
of that computer count, over the traditional hand counts, is that
preferences, for an already elected candidate, with a quota, are still
recorded. Meek did that by updating the candidate keep value (the quota
divided by a candidates total transferable vote).
Unlike Meek method, I do keep values for every candidate, losers as well
as winners. Candidates in deficit of a quota have keep values of more
than unity, signifying they are excluded. The exclusion count is run
exactly like the election count but with the preferences reversed, so a
quota now becomes an exclusion quota. The rule is simple: an election
count elects candidates reaching the quota. An exclusion count excludes
candidates reaching a quota. One voters preferences is another voters
unpreferences. There is no difference in principle between them.
Binomial STV (symbolised as STV^; first order Binomial STV would be
STV^1. Any order bimomial STV would be STV^n. Preceding forms of STV,
including Meek method, are STV^0. The ballot paper looks just like any
Ranked Choice Vote. But the instructions are different. Every voting
method has voters instructions.
The instructions are, in the case of your example: There are four seats
available and ten candidates to choose from. Your first four preferences
would more or less help to elect candidates. Your next 6 preferences (if
you choose to make them) would less or more help to exclude those
candidates. So, a tenth preference counts as much against a candidate,
as your first preference would count for a candidate. But you don't have
to give any order of preference. A carte blanche is equivalent to NOTA.
If a quota of abstentions is reached, one of the seats is left empty.
This election also gives voters the rational power to exclude candidates.
Some candidates may be both popular and unpopular enough to gain both
election and exclusion quotas. They are both "alive" and "dead" to the
electorate. (A case of "Schrodingers candidate" according to Forest
Simmons.) Whether they are elected or excluded is determined by a
Quotient of the exclusion quota divided by the election quota. If the
ratio is one or less, they are elected; if not, excluded. (The Quotient
is the square of a geometric mean.)
When inverted, the exclusion count is like a second-opinion election.
The geometric means of the candidates election keep values and inverse
exclusion keep value establish the over-all order of popularity of the
candidates (from lowest to highest over-all keep values.
All the voters abstentions have to be counted, to establish whether they
care more to elect or exclude candidates. This also means there is no
reduction of the quota with abstentions, as in Meek method. Counting
abstentions observes the conservation of (preferential) information.
I hired a programmer for first order Binomial STV, which, unlike the
higher orders, should be much simpler than Meek method, and simpler in
conception than the hand counts. However I have always supported them
all my adult life, and am now an old man.
Kristofer found the GitHub link to the programmers coding, which he sent me:
https://github.com/Esrot-Clients/STV_CSV/tree/master
The programmer also sent me a "frontend" for the use of voters:
And he sent me two manuals, which I cannot attach, in case useful to a
technical person, unlike myself, because the moderator doen't allow
messages over a certain size. I think it was a different reason last time.
Regards,
Richard Lung.
Election-Methods mailing list - see https://electorama.com/em for list info
Thank you Toby
As far as I understand your argument, I think the answer is that the
lower preferences do not preclude the higher preferences. The lower
preferences do not count against the higher preferences for other
candidates. Higher preferences make candidates more likely to be elected
(up to the number of seats). Lower preferences make other candidates
more likely to be excluded.
The converse is the case as we move from the election count to an
exclusion count. Here the higher preferences make certain candidates
more likely to be excluded. The lower preferences (up to the total
number of candidates) make the candidates more likiely to be elected.
Thus an exclusion count may actually help a candidate to be elected,
under certain conditions. Suppose a candidate wins something like an
election quota. Suppose they also do not have many exclusion votes (with
the preferences reversed); much less than an exclusion quota. The
exclusion quota is inverted to give a second opinion election quota.
Then it is used to give an over-all keep value for the candidate.
Multiplying the election keep value by the inverted exclusion keep value
can improve the candidates over-all keep value, to unity or less than
unity, signifying election.
My programmer (of whose work I do not understand the first thing) said
it works fine. Your list manager recently released the link. I added a
manual, as an attachment to that e-mail.
If I have not fully answered your quetion or you have more questions,
please let me know.
Regards,
Richard Lung.
On 18/04/2024 22:18, Toby Pereira wrote:
If a voter's lower ranks exclude candidates as much as higher ranks
count in their favour, then I can't see how the result will end up
looking anything like proportional representation, which I believe is
still your goal. Take a simple case with two parties (or factions as
parties don't need to be explicit), the larger of which has 2/3 of the
support and the smaller 1/3. The candidates from the smaller party
will be at the top on 1/3 of the ballots, and in a normal PR method
would get about a third of the seats. But with this binomial count,
they would be excluded and more besides by the 2/3 ranking them at the
bottom.
Toby
On Friday, 5 April 2024 at 21:33:52 BST, Richard Lung
voting@ukscientists.com wrote:
Thank you, Filip,
The first order Binomial STV is one election count and one exclusion
count, exactly like it (being symmetrical; an iteration). For the
election count I use Meek method of surplus transfers. The distinction,
of that computer count, over the traditional hand counts, is that
preferences, for an already elected candidate, with a quota, are still
recorded. Meek did that by updating the candidate keep value (the quota
divided by a candidates total transferable vote).
Unlike Meek method, I do keep values for every candidate, losers as well
as winners. Candidates in deficit of a quota have keep values of more
than unity, signifying they are excluded. The exclusion count is run
exactly like the election count but with the preferences reversed, so a
quota now becomes an exclusion quota. The rule is simple: an election
count elects candidates reaching the quota. An exclusion count excludes
candidates reaching a quota. One voters preferences is another voters
unpreferences. There is no difference in principle between them.
Binomial STV (symbolised as STV^; first order Binomial STV would be
STV^1. Any order bimomial STV would be STV^n. Preceding forms of STV,
including Meek method, are STV^0. The ballot paper looks just like any
Ranked Choice Vote. But the instructions are different. Every voting
method has voters instructions.
The instructions are, in the case of your example: There are four seats
available and ten candidates to choose from. Your first four preferences
would more or less help to elect candidates. Your next 6 preferences (if
you choose to make them) would less or more help to exclude those
candidates. So, a tenth preference counts as much against a candidate,
as your first preference would count for a candidate. But you don't have
to give any order of preference. A carte blanche is equivalent to NOTA.
If a quota of abstentions is reached, one of the seats is left empty.
This election also gives voters the rational power to exclude candidates.
Some candidates may be both popular and unpopular enough to gain both
election and exclusion quotas. They are both "alive" and "dead" to the
electorate. (A case of "Schrodingers candidate" according to Forest
Simmons.) Whether they are elected or excluded is determined by a
Quotient of the exclusion quota divided by the election quota. If the
ratio is one or less, they are elected; if not, excluded. (The Quotient
is the square of a geometric mean.)
When inverted, the exclusion count is like a second-opinion election.
The geometric means of the candidates election keep values and inverse
exclusion keep value establish the over-all order of popularity of the
candidates (from lowest to highest over-all keep values.
All the voters abstentions have to be counted, to establish whether they
care more to elect or exclude candidates. This also means there is no
reduction of the quota with abstentions, as in Meek method. Counting
abstentions observes the conservation of (preferential) information.
I hired a programmer for first order Binomial STV, which, unlike the
higher orders, should be much simpler than Meek method, and simpler in
conception than the hand counts. However I have always supported them
all my adult life, and am now an old man.
Kristofer found the GitHub link to the programmers coding, which he
sent me:
https://github.com/Esrot-Clients/STV_CSV/tree/master
The programmer also sent me a "frontend" for the use of voters:
And he sent me two manuals, which I cannot attach, in case useful to a
technical person, unlike myself, because the moderator doen't allow
messages over a certain size. I think it was a different reason last time.
Regards,
Richard Lung.
Election-Methods mailing list - see https://electorama.com/em for list
info