“Monotonic” Binomial STV
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
On 27.02.2022 20:30, Kevin Venzke wrote:
Hi Kristofer/Richard,
I wonder not just about the square root, but also if the quota has some additional
role in the method, perhaps when there are 4+ candidates.
Because this expression:
( quota / keep ) * ( exclude / quota )
Appears to simplify to:
( exclude / keep )
This creates the appearance that the quota has no effect on the outcome.
Richard stated that final values below unity are electable. It looks like there
will always be an electable candidate, unless it's a complete tie, or perhaps if
there is some other rule not yet stated here.
It probably is a special case of a more general method.
I think this special case can be further simplified by removing the
square root because the square root is monotone for nonnegative
integers: sqrt(x) > sqrt(y) iff x > y. And for the sake of simplicity,
the rule "elect the least keep value" can be replaced with "elect the
candidate with the greatest score" by flipping numerator and denominator.
So we have: keep value of A is sqrt( (quota/first prefs) * (last prefs /
quota))
which is proportional to sqrt(last prefs / first prefs)
which is monotonely related to (last prefs)/(first prefs)
and so in score terms, score(A) = fpA/lpA
where fpA is the number of first preferences for A, and lpA is the
number of last preferences for A, and the candidate with the highest
score is elected,
will elect the same candidates as (this special case of) Binomial STV does.
(You could try to add this method to the simulator we've been
discussing, I guess, and see where it ends up :-) Though it has the same
ambiguity as Antiplurality about what to do with truncation and equal-rank.)
It seems to me that the 3-candidate 1-winner case of this method is monotone.
It would help to see a four-candidate election resolved, too.
Yes; or a three-seat election with two winners and three candidates.
-km
Hi Kristofer,
Le dimanche 27 février 2022, 15:53:46 UTC−6, Kristofer Munsterhjelm km_elmet@t-online.de a écrit :
(You could try to add this method to the simulator we've been
discussing, I guess, and see where it ends up :-) Though it has the same
ambiguity as Antiplurality about what to do with truncation and equal-rank.)
I can do a plot with no truncation allowed. One may have to go to:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2022-February/date.html
in order to see this in a fixed-width font:
......................IRV....................................................
.............................................................................
..............................................BPW............................
.............................................................................
............IFPP....................CIRV.....................................
.....................ChainRO.................................................
...................................fpAfpC....................................
.............................................................................
.............................................................................
..................................MinMax.....................................
..................................CFPP.......................................
..............................................Bucklin........................
.............................................................................
FPP............DAC/DSC.......................................................
.............................................................................
.............................................................................
.........................................Borda...............................
.............................................................................
.............................................................................
.............................................................................
.........................................................................Anti
.........................................................................plur
......................................................BinSTV.............alit
I have 3-candidate Binomial STV as slightly closer to Borda than to Antiplurality,
but still pretty distant from either.
Kevin
hello again Kevin,
I consider equal preferences to be "illections" defined by the
multi-nomial theorem, rather than "elections"; maybe a suitable task for
future generations, but not this old man.
"Truncation" I seem to recall has something to do with iteration
methods, not my subject.
Regards,
Richard Lung.
On 28/02/2022 00:38, Kevin Venzke wrote:
Hi Kristofer,
Le dimanche 27 février 2022, 15:53:46 UTC−6, Kristofer Munsterhjelm km_elmet@t-online.de a écrit :
(You could try to add this method to the simulator we've been
discussing, I guess, and see where it ends up :-) Though it has the same
ambiguity as Antiplurality about what to do with truncation and equal-rank.)
I can do a plot with no truncation allowed. One may have to go to:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2022-February/date.html
in order to see this in a fixed-width font:
......................IRV....................................................
.............................................................................
..............................................BPW............................
.............................................................................
............IFPP....................CIRV.....................................
.....................ChainRO.................................................
...................................fpAfpC....................................
.............................................................................
.............................................................................
..................................MinMax.....................................
..................................CFPP.......................................
..............................................Bucklin........................
.............................................................................
FPP............DAC/DSC.......................................................
.............................................................................
.............................................................................
.........................................Borda...............................
.............................................................................
.............................................................................
.............................................................................
.........................................................................Anti
.........................................................................plur
......................................................BinSTV.............alit
I have 3-candidate Binomial STV as slightly closer to Borda than to Antiplurality,
but still pretty distant from either.
Kevin
Election-Methods mailing list - see https://electorama.com/em for list info
Hi Richard,
hello again Kevin,
I consider equal preferences to be "illections" defined by the
multi-nomial theorem, rather than "elections"; maybe a suitable task for
future generations, but not this old man.
"Truncation" I seem to recall has something to do with iteration
methods, not my subject.
Truncation is when the voter doesn't rank all the candidates. They list some
preferred choice(s) and then stop. It seemed likely that your method doesn't allow
voters to do that, so I had to prohibit it for all methods in order to make a plot.
Equal ranking is e.g. when the voter has more than one favorite with no ordering
between them. It can be useful to support this because, if done right, it may be
possible to guarantee to the voter that they can safely rank their favorite
candidate no lower than tied for first.
Equal ranking wasn't allowed for my plot, either.
Kevin
Kristofer,
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.
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.
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.
Regards,
Richard Lung.
On 27/02/2022 21:43, Kristofer Munsterhjelm wrote:
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
Hello Kevin,
My method does allow voters to abstain. In fact it's an essential part
of it, for the relative importance of election and exclusion counts. So,
all the preferences are counted, in a complete dimension of choice.
For an e-lection or choosing out, I don't have equal preferences or
choosing-in.
Regards,
Richard Lung.
On 28/02/2022 01:23, Kevin Venzke wrote:
Hi Richard,
hello again Kevin,
I consider equal preferences to be "illections" defined by the
multi-nomial theorem, rather than "elections"; maybe a suitable task for
future generations, but not this old man.
"Truncation" I seem to recall has something to do with iteration
methods, not my subject.
Truncation is when the voter doesn't rank all the candidates. They list some
preferred choice(s) and then stop. It seemed likely that your method doesn't allow
voters to do that, so I had to prohibit it for all methods in order to make a plot.
Equal ranking is e.g. when the voter has more than one favorite with no ordering
between them. It can be useful to support this because, if done right, it may be
possible to guarantee to the voter that they can safely rank their favorite
candidate no lower than tied for first.
Equal ranking wasn't allowed for my plot, either.
Kevin
On 28.02.2022 02:59, Richard Lung wrote:
Kristofer,
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.
This is pretty hard to parse - you're using your personal terminology
that I'm not familiar with. But I would imagine that the candidates who
are close to the quota would merit election; and, if you're doing a
statistical test, it should elect the candidates who are above the quota
rather than below it. In the example I provided, the Bs are below the
quota and A is above it, yet A is not elected. The distance that B is
below the quota just acts as further evidence that B should not be elected.
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.
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.
I don't understand your reasoning. Are you saying that flipping a 60-40
outcome merely by fielding additional candidates is okay? Or is the
method missing a rule that disqualifies candidates far below the quota?
If it's the latter, could you do my example again, demonstrating that rule?
-km
On 28.02.2022 01:38, Kevin Venzke wrote:
Hi Kristofer,
Le dimanche 27 février 2022, 15:53:46 UTC−6, Kristofer Munsterhjelm km_elmet@t-online.de a écrit :
(You could try to add this method to the simulator we've been
discussing, I guess, and see where it ends up :-) Though it has the same
ambiguity as Antiplurality about what to do with truncation and equal-rank.)
I can do a plot with no truncation allowed. One may have to go to:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2022-February/date.html
in order to see this in a fixed-width font:
I'm using a dedicated mail program, so no problem there :-)
......................IRV....................................................
.............................................................................
..............................................BPW............................
.............................................................................
............IFPP....................CIRV.....................................
.....................ChainRO.................................................
...................................fpAfpC....................................
.............................................................................
.............................................................................
..................................MinMax.....................................
..................................CFPP.......................................
..............................................Bucklin........................
.............................................................................
FPP............DAC/DSC.......................................................
.............................................................................
.............................................................................
.........................................Borda...............................
.............................................................................
.............................................................................
.............................................................................
.........................................................................Anti
.........................................................................plur
......................................................BinSTV.............alit
I have 3-candidate Binomial STV as slightly closer to Borda than to Antiplurality,
but still pretty distant from either.
That makes sense. As I remarked to Forest, although the vast majority of
election methods that have been proposed have win regions that are
unions of convex polytopes, I think this one might be an exception.
Using, in effect log(fpA)-log(lpA) > max over other candidates X:
log(fpX)-log(lpX) might well induce a curved win region.
It's an interesting question if there's some (property based) reason
that most methods are linear in this sense. Perhaps there's some
property that can only be passed by such methods. But if there is, I
don't know what it is.
I did mention some nonlinear methods of my own a long time ago, and how
they might be even more strategy resistant than the linear ones.
However, I'm not sure I was right back then, because I thought I found a
method that was much better than fpA-fpC, yet the optimal method finding
program returns a method that's very close to fpA-fpC. So something
could easily be off.
-km
postscript.
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.
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.
Thank you, Kristofer, also for the helpful pointers.
I have uploaded a pdf format of Dr David Hill code of Meek method (with
previous qualifications) free on archive.org
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.
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.
Regards,
Richard Lung.
-------- Forwarded Message --------
Subject: Re: [EM] “Monotonic” Binomial STV
Date: Mon, 28 Feb 2022 01:59:28 +0000
From: Richard Lung voting@ukscientists.com
To: Kristofer Munsterhjelm km_elmet@t-online.de, Forest Simmons
forest.simmons21@gmail.com
CC: EM election-methods@lists.electorama.com
Kristofer,
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.
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.
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.
Regards,
Richard Lung.
On 27/02/2022 21:43, Kristofer Munsterhjelm wrote:
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