[EM] STV^1

Richard Lung voting at ukscientists.com
Fri Apr 5 13:16:12 PDT 2024


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:

https://votingstv.cloud/

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.



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