[EM] “Monotonic” Binomial STV

Richard Lung voting at ukscientists.com
Sun Feb 27 05:04:40 PST 2022


Thank you, Kristofer,


for first example.

The quota is 100/(1+1) = 50.

Election keep value is quota/(candidates preference votes)

for A: 50/51

B: 50/49

C: 50/0  Which, of course is infinite. It may be convenient, for tidy 
book-keeping, that small elections require that each candidate votes for 
themself. Then the keep value maximum simply equals the quota. 
Generally, it is not necessary to make this stipulation, for large scale 
elections, because no candidate, however miserable, ever gets no votes.

Exclusion keep value equals quota/(candidates reverse preference vote):

A: 50/1

B: 50/0

C: 50/99

Geometric mean keep value ( election keep value multiplied by inverse 
exclusion keep value):

A:  50/51 x 1/50  ~ 0,0196

B: 50/49 x 0/50 = 0/49 is indeterminate. The closest determinate 
approximation gives 1/49, not quite as low a keep value as 1/51 for A, 
who is therefore the winner.

C: 50/0 x 99/50 = 99/0 is indeterminate. The closest determinate 
approximation is 99/1. This keep value properly signifies huge 
unpopularity, since an elective keep value is unity.

However, this example does illustrate the problem of the relative 
importance of the election and exclusion counts. My guess is that the 
problem would resolve itself in the entropy of preference voting, which 
falls off exponentially with lower preferences.

Laws of physics are time-reversible in principle. But in practise, on 
the classical or macroscopic scale, entropy intervenes, to give time a 
one-way direction. (Binomial STV, in small scale elections, below the 
level of much or any significance for parametric statistics, might be 
likened to time-reversible quantum scale inter-actions, with regard to 
the relative importance of election and exclusion, the direction of 
choice, positive or negative.)

This falling off, of non-abstentions, would be countered, to some 
extent, by the power of binomial stv to exclude, as well as elect, 
candidates. And there is no reason in principle why a binomial stv 
election might actually be more of an exclusion of generally disliked 
candidates (just as often is FPTP). Still, it should be borne in mind, 
that this very exclusive power of binomial stv should deter adversarial 
candidate line-ups.

Regards,

Richard Lung.



On 26/02/2022 22:28, Kristofer Munsterhjelm wrote:
> On 26.02.2022 15:02, Kristofer Munsterhjelm wrote:
>> On 26.02.2022 13:21, Richard Lung wrote:
>>> Thank you, Forest,
>>>
>>> Your example is the kind of example that Riker gave.
>> Thank you for providing information about how to calculate who wins
>> according to Binomial STV.
>>
>> I have a few examples of my own. Could you tell me who wins, and how the
>> wins are calculated, for these single-winner elections?
>>
>> First:
>>
>> 51: A>B>C
>> 48: B>A>C
>>   1: B>C>A
>>
>> And second:
>>
>> 36: A>B>C
>> 34: B>C>A
>> 30: C>A>B
> Oops, I probably should've noticed that Forest's example is a variant of
> my second election, so I don't think I need the outcome and calculations
> for that one to verify whether the way I think the method works (for
> three candidates) is right.
>
> The first one would still be useful, though :-)
>
> -km


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