[EM] Notes on a few Later-no-harm methods

[Forest Simmons]

Kristofer noted in passing a very important and under-appreciated advantage
of Condorcet methods:

It can be shown that, for methods where a majority can always force an outcome
by coordinating how they vote, then modifying the method so that it elects
the Condorcet winner if there is one never increases the proportion of
elections where strategy is useful, and may indeed reduce it.

This is a good reason to routinely include in the description of every
Universal Domain single winner method that satisfies the Majority
Criterion, verbiage to the effect ...

"Lacking a candidate that outranks any opponent on more ballots than not
..."

-Forest

El sáb., 28 de may. de 2022 9:43 a. m., Kristofer Munsterhjelm <
km_elmet at t-online.de> escribió:

> On 24.05.2022 21:05, Richard Lung wrote:
> >
> > The snag is that these and other criteria were invented for what
> > amounts to uninomial elections, that is elections that don't have both,
> > or either, a rational election count and a rational exclusion count.
> > Together they make possible the application of the binomial theorem, to
> > higher order counts. My binomial STV hand count is just a first order
> > binomial count of one election count and one exclusion count.
>
> The criteria are method-agnostic: for any ranked voting method (in this
> case, that supports truncation), if someone gives you a failure example,
> you can verify if the method passes or fails the criterion without
> knowing anything about the internals of the method.
>
> Put differently, suppose that in a scenario perhaps reminiscent of
> Roadside Picnic, a mysterious device falls out of the sky. And it turns
> out that this mysterious device calls elections: you can input ranked
> ballots with a set of buttons and get the results shown as a series of
> lights on the other end.
>
> Then as long as it allows for ballots with truncation, it's possible to
> check if a particular ballot where A-first voters truncate can be used
> to induce a later-no-harm failure.
>
> Whether the strange technology that makes up the device implements
> rational election and exclusion counts doesn't matter. As long as it's a
> ranked voting method outputting winners and supporting truncation, the
> question "does this pass later-no-harm?" makes sense.
>
> The same goes for things like monotonicity, participation, consistency,
> Smith, Condorcet, etc. The criteria say something about the desired
> behavior of a method. Nothing about the inner workings makes the
> criteria inapplicable (apart from some exceptions like the polynomial
> runtime criterion).
>
> Without a mathematical description of the method, you couldn't be sure
> it actually passes later-no-harm or later-no-help, but as soon as you
> found a counterexample, that would settle the question in the negative.
>
> > I am not aware of any untoward effects of tactical voting on the bstv
> > system. I am aware of it doing away with residual irrationalities to
> > traditional stv, including Meek method. Tho I accept that traditional
> > stv (zero-order stv in relation to binomial stv) is a robust system,
> > in practise, as the Hare system of at-large stv/pr.
>
> As a ranked method, it must fail IIA, which means that strategy must
> sometimes be possible. And as it fails Condorcet, the obvious starting
> place to look is for an election where it doesn't pass Condorcet. For
> instance, this:
>
> 549: A>B>C
> 366: B>A>C
> 366: B>C>A
> 366: C>A>B
>
> A is the Condorcet winner. The first preferences are:
>         A: 549, B: 732, C: 366
> and last preferences:
>         A: 366, B: 366, C: 915
>
> so the ratios are:
>         A: 366/549 = 0.67
>         B: 366/732 = 0.5
>         C: 915/366 = 2.5
>
> so B wins. Then the C>A>B voters have an incentive to vote A>C>B instead
> (compromising), after which the counts are:
>
>         A: 366/915 = 0.29
>         B: 366/732 = 0.5
>         C: 915/0   = infinity
>
> and A wins. The C>A>B voters prefer A to B, so the strategy is to their
> benefit.
>
> It can be shown that, for methods where a majority can always force an
> outcome by coordinating how they vote, then modifying the method so that
> it elects the Condorcet winner if there is one never increases the
> proportion of elections where strategy is useful, and may indeed reduce it.
>
> > BSTV counts require values for all preference positions, which are
> > equal to the number of candidates. Any preference position may be an
> > abstention. A citizen who never voted but made an exception of their
> > dislike for Donald or Hilary could abstain on their first preference
> > but vote for either on their second preference, effecting an
> > exclusion, because there is only one vacancy.
>
> > That is the theory of it. I don't know how well it would work in
> > practise, because there never has been a practise. But I do know that
> > democracy is minimised, and evidently works badly, based on single
> > vacancies, in the Anglo-American systems.
>
> > Fully fledged binomial stv, FAB STV, does not work on less than 4 or
> > 5 member constituencies, the minimum requirement for a democracy of
> > all the people being represented by their choices.
>
> > Thank you for your examples. They have helped clarify my thinking --
> somewhat!
> > According to my (accident-prone) working, A wins on a keep value of
> 38957/58966.
> > B also has a less than unity keep value of 38957/39366. The
> > difference is that one can say A has been elected on a quota of
> > 48961.5, with 58966 first preferences.
> > But B has not reached the elective quota. Tho B has not reached the
> > exclusion quota, that only says B has not been excluded.
>
> So by the keep values: A's first preference count is 58966 and last
> preference count is 38957, since the keep value is 38957/58966.
>
> You say that B's keep value is 38957/39366, i.e. first preference count
> of 39366 and last preference count of 38957. But that seems to be in
> reverse order. Indeed, your HTML page shows that it is 39366/38957.
>
> From the keep values, it seems that truncations are not included when
> counting last preferences. I was pretty sure that BSTV would fail
> later-no-harm because the standard way of counting truncations, as STV
> does, is to consider everybody not ranked to be equal-ranked for last;
> and if you had done that, then it would be possible to induce
> later-no-harm.
>
> The good news is that you avoid this particular problem if you count
> anything past truncation simply as abstentions. So I guessed wrong,
> which was then cleared up by the example, which shows how useful they
> are :-)
>
> However, instead it seems that you get later-no-*help* failure. Consider
> this modified election:
>
> 18125: A
> 20035: A>B>C
> 18722: A>C>B
> 34488: B>A>C
> 38634: C>B>A
>
> By my count, the first preferences are: A: 56882, B: 34488, C: 38634
> and the last preferences are:           A: 38634, B: 18722, C: 54523
> and the last to first ratios are:       A: 0.68,  B: 0.54,  C: 1.41
>
> so B wins. But if now the A voters fill out their ballot by voting
> A>C>B, then B's last preference count changes to 36847 and A wins
> instead. This is a violation of later-no-help.
>
> Ordinary STV passes both.
>
> I should note that Condorcet methods, that I prefer, fail both. My point
> isn't as much that later-no-harm and later-no-help are intrinsically
> good, as that it's much easier to check a claim by concrete evidence
> than by references to personal terminology (which may be hard to
> understand for others or take a lot of time to get acquainted with).
>
>
> On a final note, I would say that always counting truncation as
> abstention could lead to an unknown candidate problem: suppose there's a
> candidate who nobody has heard of and thus nobody bothers to rank. But
> he has a dedicated following all of whom rank him first. If nobody
> obtains a majority, then this candidate could win, e.g. something like:
>
> 3300: A>B>C
> 3300: B>C>A
> 3200: C>A>B
>    2: D
>
> I'm also not entirely sure what's going on with the quota transfers. If,
> in the single-winner case, someone who exceeds the quota is
> automatically elected, then there's no need for any transfers. However,
> if passing the quota doesn't guarantee victory, then later-no-harm
> failure might actually be possible. Suppose A is just above the quota
> and B is just below it (with B closer to the majority line), then if the
> A voters only vote for A, A might win; but if they vote A>B, then the
> surplus might be transferred to B and make B win. Perhaps. As I said,
> I'm not sure how the logic works in that case.
>
> -km
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