[EM] “Monotonic” Binomial STV
Richard Lung
voting at ukscientists.com
Sun Feb 27 11:08:05 PST 2022
Kristofer,
Thank you for correcting me. I do tend to forget the square root, to
finish the average, the geometric mean.
In some formalisms, a zero numerator implies zero, but the geometric
mean, unlike the arithmetic mean, does not work wih zero, so that result
cannot be infered from it.
The trouble with just putting infinity is that there are different
infinities! One could require 1 vote for a candidate, from the
candidates themselves. Then we have a standard of comparison.
Glancing at your example, tho, I am reminded of an example with small
numbers, in which I had to reduce the 1 vote minimum to 0.1. Traditional
STV resorted to this expedient for small numbers elections, with the
Droop quota. They could not add plus one, because that made the quota
too hard for candidates to win. So, they resorted to a final plus 0,001,
I believe. But then ERS Ballot Services Major Frank Britton realised
that the final constant was never needed. And it is true, in this case,
of Binomial STV that a minimum candidate vote is not needed, as I think
you have suggested.
Moreover, if candidates have a minimum vote, they must also have the
same reverse preference minimum, for the sake of symmetrical treatment -
and perhaps as a neutralising factor.
I don't know yet what will turn out to be the most elegant count
instructions, in this and, no doubt, other instances. Your method
designating keep value infinity may be better, because such results are
nowhere in it, anyway. And it cuts out a troublesome added minimum
constant to candidates votes. I guess your result is correct. One has to
be a bit careful, in general, tho. An extremely bad election count may
be somewhat redeemed by a tolerable exclusion count, getting nowhere
near an exclusion quota. In that case, the not popular but also not
unpopular candidate is perhaps entitled to a quantitative tabulation.
This quandry reminds me of the caution, from an stv count expert, to use
floating point arithmetic in computer coding stv. Meek method, in New
Zealand, uses decimal point, but that might create future difficulties.
Anyway, I appreciate how important it is not to under-estimate the
possible ill consequences of casually considered operations.
The New Zealand government has not made its Meek method open source.
They denied access. Dr David Hill made his coding, of Meek method, open
source. He is a direct descendant of Thomas Wright Hill, who published
the first known instance of tranferable voting (barring the Gospel
Incident of the loaves and fishes). To celebrate the 200th anniversaey,
I re-published this and the public domain code by David Hill. Smashwords
does not allow publishing public domain works, so I had to put it on
Amazon, who charge a minimal fee. However, I could perhaps put it on
archive.org who don't charge, and where I put many of my e-books in pdf
format.
Dr David Hill wrote his code, for Meek method, in Pascal, an early
script. But I did not include his code text for eliminated candidates
last past the post, when the quota surpluses run out. Nor did I include
the code for reducing the quota, when preference voting gives way to
abstentions. Binomial STV does not use either of these expedients.
But Meek method does calculate the keep values of elected candidates:
the quota divided by their total transferable vote, which may increase
with further preferences after a quota is already achieved. The lower
the keep value below unity, the greater the popularity.
Binomial STV greatly extends the Meek method use of keep values, to all
candidates, and for their exclusion, as well as their election. However,
there is no difference in principle to these extended operations, of
transferble voting, by keep value.
The hand count version of binomial stv, tho, drops the distinctively
computerised Meek count of post-quota preference counting. And sticks to
the first order binomial count, making it simpler than traditional
counts, as well as Meek stv.
The New Zealand government hired two software coding firms, as back-up,
paying both, but only using one of them. This shows how arduous and
uncertain the results of their making Dr Hill coding texts executable.
Regards,
Richard Lung.
On 27/02/2022 13:41, Kristofer Munsterhjelm wrote:
> On 27.02.2022 14:04, Richard Lung wrote:
>> 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.
> Another option is to just let infinities be worse than any alternative.
> Since not every candidate can have a zero last preference count, at
> least one candidate must have a finite value and so would be considered
> better than every candidate with an infinite value.
>
>> 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.
> Is 0/49 indeterminate? Shouldn't it just be zero? 0/x = 0 for x not
> equal to zero, and the square root of zero is zero.
>
> But let me in any case revise my example. Who wins in this one?
>
> 50: A>B>C
> 47: B>A>C
> 2: B>C>A
> 1: A>C>B
>
> My calculations are as follows:
>
> The quota is 50.
>
> Election keep value is quota/candidate preferences:
>
> A: 50/51
> B: 50/49
> C: infinity
>
> Exclusion keep value equals quota/candidates reversed first preferences:
>
> A: 50/2
> B: 50/1
> C: 50/97
>
> Geometric mean:
>
> A: square root of (50/51 x 2/50) ~ 0.198
> B: square root of (50/49 x 1/50) ~ 0.143
> C: ~= infinity (or very high)
>
> So B wins, having the lowest keep value. Is this correct?
>
> (You seem to have omitted the square root in your calculations, but it
> shouldn't make a difference. Without the square root, A and B's values
> are 0.0392 and 0.0204 respectively.)
>
> -km
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