[EM] “Monotonic” Binomial STV
Kristofer Munsterhjelm
km_elmet at t-online.de
Sun Feb 27 13:43:11 PST 2022
On 27.02.2022 20:08, Richard Lung wrote:
> 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.
Mathematically, you could take the limit of adding a very small number
of first preferences, as that number goes to zero. That's probably the
spirit of the 0.001 addition.
> 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.
If my result is correct, that does lead to a problem. In the example I
provided, A is the majority winner, but B wins. So while the method is
at least monotone in the single-winner case with three candidates, it
fails both majority and mutual majority.
This means that seen as a whole, Binomial STV fails Droop
proportionality (since Droop proportionality implies mutual majority for
the single-winner case). It does not pass the proportionality criteria
that I would associate twith the term "STV".
(When I said that I would find it very surprising for an STV method to
be proven monotone, I was expecting that the method would pass Droop
proportionality but fail monotonicity. But it turns out I was wrong:
instead it seems to be the other way around.)
It's true that the majority criterion failure is somewhat contrived, but
ordinary STV always passes, even in artificial elections like the one I
provided. It can probably also be done for more realistic scenarios.
Here's an attempt to do that.
Suppose we have two parties, A and B, each of which fields a single
candidate, and one of which has a 20 pp lead on the other:
60: A>B
40: B>A
A is the majority winner and should win (if the method is majority
rule). But now suppose a more extreme faction of the B party runs its
own candidate. Most of the A-voters and B-voters prefer the original
candidate (Bo) to the extremist (Be). The B voters in particular prefer
the original, but they consider party unity most important of all and
won't vote A second. So we have:
40: A>Bo>Be
20: A>Be>Bo
33: Bo>Be>A
7: Be>Bo>A
The quota is 100.
A's keep value is sqrt(50/60 x 40/50) = 0.815
Bo's keep value is sqrt(50/33 x 20/50) = 0.778
Be's keep value is sqrt(50/7 x 40/50) = 2.39
So the B party wins by running additional candidates -- enough to cancel
out a 20 pp lead. This is a teaming incentive - and it's strong enough
to counteract a majority win.
It may be unrealistic to suppose that any of the A party voters would
prefer Be to Bo: if A is left-wing, Bo is center-right and Be is far
right, that would seem very strange. But (if my calculations are right)
the more voters prefer A>Bo>Be, the worse the effect gets.
Every voting method supported by Rob LeGrand's voting calculator at
https://web.archive.org/web/20200813191652/http://www.cs.angelo.edu/~rlegrand/rbvote/calc.html
gives the win to the majority candidate, both for the election of my
prior post and the one I just provided. That includes Borda, which is
otherwise known for its serious teaming incentive.
> 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.
Using archive.org would have the benefit that you could upload your
source code separately in machine-readable format; it's not limited to
ebooks.
I would say that for election methods that use iterative methods on
rationals, like Meek does, it would be preferable to use exact rational
arithmetic instead of either decimals or floating point.
Even though rational arithmetic considerably slows down the method, it
fully eliminates numerical precision and round-off problems.
-km
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