It has more to do with strategy than equity.

I prefer minimizing opposition to the defeat, because that relieves
tension, resentment, and polarization. Perhaps it is possible to win
without alienating the opposition, i.e. without all of the bad feelings.

Notice in the three examples, there was zero opposition against the victory
of A over B, so no hard feelings if A beats B.

On the other hand, the C beats D defeat is highly polarized ... 4999 angry
losers.

But my question which of the three claims is most defensible statistically
... i.e. which one is least likely to be reversed by a random fluctuation
in the weather on election day in a parallel universe.

It seems to me that the margins result (F defeating G) is the most stable
in this case. Is that always true?

On Sat, Sep 3, 2022, 5:33 PM robert bristow-johnson <
rbj@audioimagination.com> wrote:

Actually, max likelihood analysis would say that the A>B defeat was less
likely to be reversed because max likelihood estimation would estimate
Prob(B>A)=0.

But Baysian estimation would start with Prob(B>A)=Prob(A>B)>0<Prob(A=B)
prior probabilities, and then adjust using Bayes' law to posterior
probabilities in the order Prob(A=B)>Prob(A>B)>Prob(B>A)>0

and Prob(B>A) > Prob(G>F), so according to Bayes, the F>G defeat is
stronger (less likely to be reversed in a parallel universe) than the A>B
defeat.

Somebody should do the precise Bayesian calculations to verify (or refute)
my statistical intuition.

On Sat, Sep 3, 2022, 6:30 PM Forest Simmons forest.simmons21@gmail.com
wrote:

Forest – this isn’t an answer, but a dissertation on a related topic.
   I recently read Condorcet's Essai, and then commentaries by
Todhunter, Black and Young. I concluded that Condorcet's probability
theory was all at sea, and that his commentators were too kind to him.
In my view you cannot make sense of voting using pure probability theory
(even turning a blind eye to the faults of a jury model); you need to
fall back on statistics. This makes it possible to get away from
Condorcet's untenable independence assumption and also to correct
another fatal error in his method. He assumes that when a voter votes
A>B>C, he is no likelier to be right in placing A above C than in
placing A above B. I believe that this is impossible if erroneous
rankings are treated as random errors. But if you try to find the
relation between the likelihoods of one-place errors and two-place
errors using pure probability theory, you just can't get started. You
have to use a statistical model - eg. assume that true candidate merits
are Gaussianly distributed, estimated merits are contaminated by
Gaussian noise, and that a voter ranks candidates in decreasing order of
estimated merit. If you do this, you can find the relative chances of
one-place and two-place errors (subject to suitable distributional
assumptions).
   Having done this, you can say: "I will give candidate A one point
more than candidate B whenever he comes one place higher in a ballot,
and x points more whenever he comes two places higher"; x=1 gives the
Condorcet criterion, x=2 gives the Borda count. By my calculation, the
optimal x is almost exactly 2 (and almost independent of distributional
assumptions), and the Borda count is therefore almost optimal under a
jury model with 3 candidates. Condorcet thought his jury theorem showed
his own criterion to be optimal in the same case.
   So I would say that you have to recast your question: let x and y be
the candidate merits, distributed as N(0,1); let the noise be
distributed as N(0,sigmasquared); let sigmasquared be governed by some
fairly diffuse prior (maybe 1/sigmasquared). What is the probability
that x>y when each of 100 noisy estimates of x is larger than the
corresponding noisy estimate of y?
   If I have time, I might attempt the calculation.
      CJC

On 04/09/2022 04:43, Forest Simmons wrote:

I've not been receiving messages for a month or so.

JFS Ross has a chapter, in Elections and Electors, on Borda method, including some of the names you mention. His main point is that Laplace adjudicated on Borda vs Condorcet, and chose Borda. The reason given is that greater and lesser preferences should not be treated as of equal importance, as Condorcet pairing does.
Laplace gave a proof, in Analytic Theory of Probabilities, which my computer refused to access from Internet Archive.
A Laplace proof is a big deal. He would often say, in Celestial Mechanics, this is too obvious to need proof. When his American translator saw that, he said, he knew he was in for a hard nights work. Laplace was rated one of historys half dozen greatest mathematicians.

It does not really matter, tho. The Laplace position was illuminated by SS Stevens, on the scales of measurement, in Science, in early 1940s. The advance of Borda method on Condorcet pairing is the progression to an interval scale of measurement.
Standard statistical practise has a name for Borda count. It is weighting in arithmetic progression. Statisticians use this when they assume an arithmetic series adequately weights the successive importance of a series of data classes, to find a representative average. It is thus an assumed interval scale.
Transferable voting Gregory method is a real interval scale, achieved by the standard technique, known as weighting in arithmetic proportion, when the real values of the class intervals in the data collection are known. -- a preferable more accurate calculation. (The interval scale is next to the ratio scale in accuracy, where election methods use both assumed and real ratio scales, by way of election quotas.)
Condorcet pairing stages binary counts, associated with the weakest scale, the categorical scale. Tho, in both instances, Condorcet and Borda, the ranked choice is on the ordinal scale.

Regards,
Richard Lung.

On 4 Sep 2022, at 4:22 pm, Colin Champion colin.champion@routemaster.app wrote:

Forest – this isn’t an answer, but a dissertation on a related topic.
I recently read Condorcet's Essai, and then commentaries by Todhunter, Black and Young. I concluded that Condorcet's probability theory was all at sea, and that his commentators were too kind to him. In my view you cannot make sense of voting using pure probability theory (even turning a blind eye to the faults of a jury model); you need to fall back on statistics. This makes it possible to get away from Condorcet's untenable independence assumption and also to correct another fatal error in his method. He assumes that when a voter votes A>B>C, he is no likelier to be right in placing A above C than in placing A above B. I believe that this is impossible if erroneous rankings are treated as random errors. But if you try to find the relation between the likelihoods of one-place errors and two-place errors using pure probability theory, you just can't get started. You have to use a statistical model - eg. assume that true candidate merits are Gaussianly distributed, estimated merits are contaminated by Gaussian noise, and that a voter ranks candidates in decreasing order of estimated merit. If you do this, you can find the relative chances of one-place and two-place errors (subject to suitable distributional assumptions).
Having done this, you can say: "I will give candidate A one point more than candidate B whenever he comes one place higher in a ballot, and x points more whenever he comes two places higher"; x=1 gives the Condorcet criterion, x=2 gives the Borda count. By my calculation, the optimal x is almost exactly 2 (and almost independent of distributional assumptions), and the Borda count is therefore almost optimal under a jury model with 3 candidates. Condorcet thought his jury theorem showed his own criterion to be optimal in the same case.
So I would say that you have to recast your question: let x and y be the candidate merits, distributed as N(0,1); let the noise be distributed as N(0,sigmasquared); let sigmasquared be governed by some fairly diffuse prior (maybe 1/sigmasquared). What is the probability that x>y when each of 100 noisy estimates of x is larger than the corresponding noisy estimate of y?
If I have time, I might attempt the calculation.
CJC

Election-Methods mailing list - see https://electorama.com/em for list info

Hi,

I'm afraid that defeat strength is such a tangible concept that it inclines us to view it as
a finished product that we can now simply customize as we please. Or worse: That we are
duty-bound to measure it "accurately." We maximize some desired property, readily identified
within a single pairwise contest, and then apply the minmax algorithm to this ranking, and
expect it to result in a method whose properties have some kind of relationship to the
property we were maximizing, or properties that will at least be as good as the defeat
strength metric sounded.

I couldn't count how many times I had a nice idea, tried to use minmax or River with it,
and discovered some horrible basic flaw. Or just mediocre results.

I see Forest wants to lock a 1%:0% victory first because there is no one around to object
to that particular point. Only the local question is considered, not whether "hard feelings"
might come from some other factor than this. That's the key issue. What can we say about the
overall method that results from applying this logic? We have no real idea, without
checking.

The question then of why anything would "represent defeat strength better than margins" is
strange to me. Defeat strength is just a possible tool to get to the winner. There's no
need for it to represent anything with fidelity to any principle. It might be a marketing
concern, I admit. But when evaluating method properties you only need to know who wins when;
you won't talk about the underlying algorithm and indeed you don't even need to know what
it is.

I don't know what to make of the term "equity" below. Can reducing strategic incentives
cause inequity? How can I tell whether a method is equitable? (Of course then I mean its
results being equitable, rather than its method of calculation.)

Usually there's a simple approach to reducing strategy: If a lot of people want something,
such that they could lie on the ballot in order to get it, then just let them have it. Then
that's a complaint of inequity you can avoid, if they were to not get what they want. Maybe
there are other types of inequity, but I'm not sure what, staying within a majoritarian
perspective.

Kevin
votingmethods.net

Le samedi 3 septembre 2022 à 20:31:14 UTC−5, Forest Simmons forest.simmons21@gmail.com a écrit :

I'm happy that this thread has excited so much interest ... and astute
comments ... obviously the result of careful and thoughtful study and
experimentation.

"Equity" was not the apt word. There are many kinds of fairness, as the
subject of fair division makes clear , what I try to aim for is the kind
of fairness that minimizes reasonable people thinking that the rules were
stacked against them from the start ... an admittedly  nebulous concept.

What is the best tweak of Borda that restores some semblance of clone
independence?

I once suggested something called "ranked rankings" that (like Borda)
removes the burden of numerical quantification from the voter .. who is
only required to supply a rough relative strength preference between
alternatives:

A=B>C>D>>>E>>F=G, for example.

A and B would both get 100% approval whileF and G would each get zero.
The separation between B and C, and C and D would both be X, while the
separation between D and E would be 3x, and between E and F would be 2x.

The total number of chevrons (the ">" symbols) is seven, so we must have
7X=100, which means x is 14 + 2/7 or approximately 14.3

The voter doesn't have to do any of this arithmetic, which allows the
"ranked rankings" to be turned into ratings.

If all of the rankings are of equal strength (one chevron each, say), then
the resulting ratings will agree with the normalized Borda scores. In other
words, this way of counting "ranked rankings" style ballots is a precise
generalization of Borda that satisfies clone independence (for all
practical purposes).

In particular, there is no need to indicate strength of preference if they
are all roughly equal in strength.

Does any version of this idea have any potential as a basis for a public
proposal?

-Forest

On Sun, Sep 4, 2022, 1:01 PM Kevin Venzke stepjak@yahoo.fr wrote:

I think there are some democratic principles that ain't so nebulous.

How is this not a form of Score Voting?

This is where Condorcet and I differ from Borda (or Warren Smith).  I consider this a principle, not just a "desirable property" of elections in a democracy:

1. “One person, one vote”. Every enfranchised voter has an equal influence on government in elections because of our inherent equality as citizens and this is independent of any utilitarian notion of personal investment in the outcome. If I enthusiastically prefer Candidate A and you prefer Candidate B only tepidly, your vote for Candidate B counts no less (nor more) than my vote for A. The effectiveness of one’s vote – how much their vote counts – is not proportional to their degree of preference but is determined only by their franchise. A citizen with franchise has a vote that counts equally as much as any other citizen with franchise. For any ranked ballot, this means that if Candidate A is ranked higher than Candidate B then that is a vote for A, if only candidates A and B are contending (such as in the RCV final round). It doesn’t matter how many levels A is ranked higher than B, it counts as exactly one vote for A.

How is it that the thoughtful voter, that considers the effect of their vote on the count of the votes that ultimately elects a candidate, can avoid this arithmetic?  I don't get it.

Marks on a ballot don't have human rights (I liked how the North Dakota Supreme Court put it 111 years ago).  Enfranchised voters, as persons having equality under the law, must be the things that have equal strength.

I have a problem with Score Voting (or any cardinal voting) in public elections (and I know that Fargo ND, of all places, are doing Approval Voting).  I look at Borda as quasi-cardinal or "Score-Lite".

--

r b-j . _ . _ . _ . _ rbj@audioimagination.com

"Imagination is more important than knowledge."

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