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<div class="moz-forward-container">postscript.</div>
<div class="moz-forward-container"><br>
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<div class="moz-forward-container">Actually that example of
splitting B into Bo and Be wouldnt be possible with STV, of any
stripe, which is only concerned with individual candidates. They
may form into parties, making the analysis of internal and cross
party support possible. But STV is not of parties splitting into
smaller parties. There are good philosophical reasons for this.
But the afore-mentioned analysis is perhaps the most pertinent.<br>
</div>
<div class="moz-forward-container">Never the less, a main interest
of seeing real examples of binomial stv, (with enough voters to
form a binomial distribution) is to see whether best keep values
sometimes have to be over-ruled by the distance of their total
candidates vote from the quota, in standard deviations.</div>
<div class="moz-forward-container"><br>
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<div class="moz-forward-container">Thank you, Kristofer, also for
the helpful pointers.</div>
<div class="moz-forward-container">I have uploaded a pdf format of
Dr David Hill code of Meek method (with previous qualifications)
free on archive.org</div>
<div class="moz-forward-container">Your suggestion to stay with keep
values as ratios makes sense, rather than divide into something
with approximate points, whatever the number base, decimal point
or floating point. <br>
</div>
<div class="moz-forward-container">By the way, with regard to FAB
STV: Four Averages Binomial STV, all the arithmetic operations
involved, the four averages and the binomial theorem, should
already be coded in software libraries.</div>
<div class="moz-forward-container"><br>
</div>
<div class="moz-forward-container">Regards,</div>
<div class="moz-forward-container">Richard Lung.<br>
</div>
<div class="moz-forward-container"><br>
</div>
<div class="moz-forward-container"><br>
-------- Forwarded Message --------
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<th valign="BASELINE" nowrap="nowrap" align="RIGHT">Subject:
</th>
<td>Re: [EM] “Monotonic” Binomial STV</td>
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<th valign="BASELINE" nowrap="nowrap" align="RIGHT">Date: </th>
<td>Mon, 28 Feb 2022 01:59:28 +0000</td>
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<th valign="BASELINE" nowrap="nowrap" align="RIGHT">From: </th>
<td>Richard Lung <a class="moz-txt-link-rfc2396E" href="mailto:voting@ukscientists.com"><voting@ukscientists.com></a></td>
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<th valign="BASELINE" nowrap="nowrap" align="RIGHT">To: </th>
<td>Kristofer Munsterhjelm <a class="moz-txt-link-rfc2396E" href="mailto:km_elmet@t-online.de"><km_elmet@t-online.de></a>,
Forest Simmons <a class="moz-txt-link-rfc2396E" href="mailto:forest.simmons21@gmail.com"><forest.simmons21@gmail.com></a></td>
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<th valign="BASELINE" nowrap="nowrap" align="RIGHT">CC: </th>
<td>EM <a class="moz-txt-link-rfc2396E" href="mailto:election-methods@lists.electorama.com"><election-methods@lists.electorama.com></a></td>
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<br>
<br>
<br>
Kristofer,<br>
<br>
Your reasoning is perfectly right. But the point is that neither
of the B candidates nearly reach the Droop quota. In fact
splitting the B votes further removes the B party from the quota.
it is true I haven't said that here. But in your first example,
several weeks ago, the winner A missed the quota by half a vote,
which is statistically insignificant. More-over, in that former
example, candidates B and C, technically with marginally better
keep values, had votes that completely failed the standard
deviation test for distance from the quota.<br>
<br>
Here the same must apply if the sample size of the total vote is
big enough. Take the square root of 100 votes times probability of
success and corresponding probability of failure or 1/2 in both
cases out of a unit total probability. So sq rt of 100 x 1/2 x 1/2
gives one standard deviation of 5 votes. Two standard deviations
are significant to a 95% probability that the B vote is not in
the region of the quota as a chance fluctuation from the quota. in
fact, Bo is over 3 SDs adrift from the quota, which yields about
one chance in a thousand that the Bo vote is is just a random
departure from the quota, of no significance.<br>
<br>
This example is even more helpful than your previous one, because
it makes clear to me that statistical rules of representation by
average and significant dispersion must be observed, in
conjunction with elevating exclusion to the same rational status
as election.<br>
<br>
Regards,<br>
<br>
Richard Lung.<br>
<br>
<br>
On 27/02/2022 21:43, Kristofer Munsterhjelm wrote:<br>
<blockquote type="cite">On 27.02.2022 20:08, Richard Lung wrote:<br>
<blockquote type="cite">Kristofer,<br>
<br>
Thank you for correcting me. I do tend to forget the square
root, to<br>
finish the average, the geometric mean.<br>
<br>
In some formalisms, a zero numerator implies zero, but the
geometric<br>
mean, unlike the arithmetic mean, does not work wih zero, so
that result<br>
cannot be infered from it.<br>
<br>
The trouble with just putting infinity is that there are
different<br>
infinities! One could require 1 vote for a candidate, from the<br>
candidates themselves. Then we have a standard of comparison.<br>
<br>
Glancing at your example, tho, I am reminded of an example
with small<br>
numbers, in which I had to reduce the 1 vote minimum to 0.1.
Traditional<br>
STV resorted to this expedient for small numbers elections,
with the<br>
Droop quota. They could not add plus one, because that made
the quota<br>
too hard for candidates to win. So, they resorted to a final
plus 0,001,<br>
I believe. But then ERS Ballot Services Major Frank Britton
realised<br>
that the final constant was never needed. And it is true, in
this case,<br>
of Binomial STV that a minimum candidate vote is not needed,
as I think<br>
you have suggested.<br>
</blockquote>
Mathematically, you could take the limit of adding a very small
number<br>
of first preferences, as that number goes to zero. That's
probably the<br>
spirit of the 0.001 addition.<br>
<br>
<blockquote type="cite">Moreover, if candidates have a minimum
vote, they must also have the<br>
same reverse preference minimum, for the sake of symmetrical
treatment -<br>
and perhaps as a neutralising factor.<br>
<br>
I don't know yet what will turn out to be the most elegant
count<br>
instructions, in this and, no doubt, other instances. Your
method<br>
designating keep value infinity may be better, because such
results are<br>
nowhere in it, anyway. And it cuts out a troublesome added
minimum<br>
constant to candidates votes. I guess your result is correct.
One has to<br>
be a bit careful, in general, tho. An extremely bad election
count may<br>
be somewhat redeemed by a tolerable exclusion count, getting
nowhere<br>
near an exclusion quota. In that case, the not popular but
also not<br>
unpopular candidate is perhaps entitled to a quantitative
tabulation.<br>
</blockquote>
If my result is correct, that does lead to a problem. In the
example I<br>
provided, A is the majority winner, but B wins. So while the
method is<br>
at least monotone in the single-winner case with three
candidates, it<br>
fails both majority and mutual majority.<br>
<br>
This means that seen as a whole, Binomial STV fails Droop<br>
proportionality (since Droop proportionality implies mutual
majority for<br>
the single-winner case). It does not pass the proportionality
criteria<br>
that I would associate twith the term "STV".<br>
<br>
(When I said that I would find it very surprising for an STV
method to<br>
be proven monotone, I was expecting that the method would pass
Droop<br>
proportionality but fail monotonicity. But it turns out I was
wrong:<br>
instead it seems to be the other way around.)<br>
<br>
It's true that the majority criterion failure is somewhat
contrived, but<br>
ordinary STV always passes, even in artificial elections like
the one I<br>
provided. It can probably also be done for more realistic
scenarios.<br>
Here's an attempt to do that.<br>
<br>
Suppose we have two parties, A and B, each of which fields a
single<br>
candidate, and one of which has a 20 pp lead on the other:<br>
<br>
60: A>B<br>
40: B>A<br>
<br>
A is the majority winner and should win (if the method is
majority<br>
rule). But now suppose a more extreme faction of the B party
runs its<br>
own candidate. Most of the A-voters and B-voters prefer the
original<br>
candidate (Bo) to the extremist (Be). The B voters in particular
prefer<br>
the original, but they consider party unity most important of
all and<br>
won't vote A second. So we have:<br>
<br>
40: A>Bo>Be<br>
20: A>Be>Bo<br>
33: Bo>Be>A<br>
7: Be>Bo>A<br>
<br>
The quota is 100.<br>
<br>
A's keep value is sqrt(50/60 x 40/50) = 0.815<br>
Bo's keep value is sqrt(50/33 x 20/50) = 0.778<br>
Be's keep value is sqrt(50/7 x 40/50) = 2.39<br>
<br>
So the B party wins by running additional candidates -- enough
to cancel<br>
out a 20 pp lead. This is a teaming incentive - and it's strong
enough<br>
to counteract a majority win.<br>
<br>
It may be unrealistic to suppose that any of the A party voters
would<br>
prefer Be to Bo: if A is left-wing, Bo is center-right and Be is
far<br>
right, that would seem very strange. But (if my calculations are
right)<br>
the more voters prefer A>Bo>Be, the worse the effect gets.<br>
<br>
Every voting method supported by Rob LeGrand's voting calculator
at<br>
<a class="moz-txt-link-freetext" href="https://web.archive.org/web/20200813191652/http://www.cs.angelo.edu/~rlegrand/rbvote/calc.html">https://web.archive.org/web/20200813191652/http://www.cs.angelo.edu/~rlegrand/rbvote/calc.html</a><br>
gives the win to the majority candidate, both for the election
of my<br>
prior post and the one I just provided. That includes Borda,
which is<br>
otherwise known for its serious teaming incentive.<br>
<br>
<blockquote type="cite">This quandry reminds me of the caution,
from an stv count expert, to use<br>
floating point arithmetic in computer coding stv. Meek method,
in New<br>
Zealand, uses decimal point, but that might create future
difficulties.<br>
Anyway, I appreciate how important it is not to under-estimate
the<br>
possible ill consequences of casually considered operations.<br>
<br>
The New Zealand government has not made its Meek method open
source.<br>
They denied access. Dr David Hill made his coding, of Meek
method, open<br>
source. He is a direct descendant of Thomas Wright Hill, who
published<br>
the first known instance of tranferable voting (barring the
Gospel<br>
Incident of the loaves and fishes). To celebrate the 200th
anniversaey,<br>
I re-published this and the public domain code by David Hill.
Smashwords<br>
does not allow publishing public domain works, so I had to put
it on<br>
Amazon, who charge a minimal fee. However, I could perhaps put
it on<br>
archive.org who don't charge, and where I put many of my
e-books in pdf<br>
format.<br>
</blockquote>
Using archive.org would have the benefit that you could upload
your<br>
source code separately in machine-readable format; it's not
limited to<br>
ebooks.<br>
<br>
I would say that for election methods that use iterative methods
on<br>
rationals, like Meek does, it would be preferable to use exact
rational<br>
arithmetic instead of either decimals or floating point.<br>
<br>
Even though rational arithmetic considerably slows down the
method, it<br>
fully eliminates numerical precision and round-off problems.<br>
<br>
-km<br>
</blockquote>
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