[EM] (P1) tweak up of AV meth. using random walk may fail

Craig Carey research at ijs.co.nz
Wed Jan 26 03:28:31 PST 2000


Contents:
- AV1 is proven in a 4 candidate example to not satisfy (P1). So it is
 is not being advocated.
- An example 


At 19:50 26.01.00 , Blake Cretney wrote:
>Craig Carey wrote:
...

The plausibility of AV1 has been removed already.

>> ----------------------------------------------------------------------
>>                       AV1 preferential voting method
>> 
>> AV1 is AV0 except that the quotas are defined.
>> 
>>  Quota =
>>       1 / (number of remaining candidates)
>> 
>> Hence, if there are n candidates, then the quota at the 1st stage is
>>  1/n (e.g. 1/10 if there are 10 candidates). The quota for the 2nd
>>  stage =
>>       1 / (n - 'number of candidates eliminated in 1st stage')
>> 
>> ----------------------------------------------------------------------
>
>37 A C B
>26 B A
>24 C B A
>13 D C
>
>With 4 candidates, quota is 1/4, or in this case 25 votes.  C and D
>are eliminated.  B wins.
>
>But, if we change 2 votes from A C B to C B A, we get
Make it 2.1
>
>35 A C B
>26 B A
>26 C B A
>13 D C
>
 34.9  A C B
 26.0  B A
 26.1  C B A
Tot = 87, 2nd Quota = 29  (not 33)

>D is eliminated.  Then, with a quota of 1/3=33 votes, B and C are

The B.C. "AVQ" Quota = 1/3 of the initial number of votes.

>eliminated.  A wins.
>
>Note that in regular AV, A wins both.

The Alteration AV1 is therefore: (ACB+)--(CBA+) which is failed by (P1)
 because of what happens to candidate A. (the (ACB) is altered at and
 after the preference for A, perhaps with loss of preferences for A (or
 votes since A has the 1st preference), but A changes from a loser into
 a winner).

So that seems to be good enough : AV1 is failed by (P1) people can be
 free to ignore AV1. This is no time to develop a thoery to say how
 AV1 is better.

...
>It's almost always easier to prove that a method fails a criterion
>than that it passes one.  It makes sense to try to find a
>counter-example before trying to find a proof.

I could be given an estimate of a metric of the complexity of the program
 that numerically found the result, if any.

>
>> 
>> The quota can't be above 1/3 because that could reject all
>>  candidates, for example, when candidates have the 1st preference
>>  counts of 1/3 after being perturbed by (3/v, -1/v, -2/v), and v
>>  is large enough to take all three under the quota.
>
>You are using the quota in a different way that I was.  In the method
>I will call AVQ, all candidates must get more than 1/3 of the 1st
>preference vote, or be eliminated right away.  If this would eliminate

This is not able to be understood, is it?. Suppose the are 12 candidates
 to start with and at least 3 stages would eliinate 2 candidates, where
 is the 1/3 test applied?.


>all candidates, the plurality winner is chosen instead.  This
>elimination is only done at the beginning, not at every step, and is
>not based on the number of candidates.  Here is an example:
>
>40 A
>32 B
>28 C B
>
>Since B and C fall beneath the quota, they are both eliminated.  A
>wins.  It is my contention, backed up by the evidence I gave
>previously, that AVQ meets P1.

I am sure that would be wrong if AVQ was defined in a way that was
 an actual definition. A method won't pass (P1) if all the surfaces
 do not meet properly and form a gentle piecewise-flat surface.

Any method with an "if then else" in it, comes under more suspicion of
 failing (P1). Your AVQ has this feature: "if all are rejected at any
 stage then one will be picked".

>
>> >Here's a quote from Lord Alexander (speaking about AV)
>> >
>> >> I find this approach wholly illogical. Why should the second 
>> >> preferences of those voters who favoured the two stronger
>> >> candidates on the first vote be totally ignored and only those
>> >> who support the lower placed and less popular candidates get a
>> >> second bite of the cherry?

That seems to me, to be considering not less than 4 candidates.

Maybe the sentence is saying that this is possible: if there are 10
 candidates and 1 winner is to be found, and also if in the first stage
 their 1st preference counts are similar, then even the candidate with
 the 2nd or 3rd or 4th largest 1st preference count, could be eliminated.

 A method satisfying STV's SPC (or (P1)) is allowed to peek at subsequent
 preferences, but it is not allowed to use that information if that leads
 to a violation of SPC. I presume this is a quite empty idea when there
 is only 3 candidates.

 Another way to disregard the discovered 1st preference ordering at any
 stage [eliminate candidates with major superficial support at the start
 which prevents papers preferring lesser candidates getting multiple
 bites at the cherry], is to use information from previous stages of the
 count.
 That could allow an elimination of the 2nd best candidate instead the
 last which AV would do. SPC could be still held. [For example, removing
 minor parties in sequence, an election in North UK could have the Tory
 candidate removed in the early stages of the counting.]

...
>
>That's why I favour Condorcet-type methods.  Let me point out that
>both AV1 and AVQ suffer from vote-splitting, and therefore may be
>considered to waste votes.

What's the reason why. Condorcet is a method that returns the wrong
 number of winners. As far as it goes (which isn't far) , it satisfies
 (P1).
Vote splitting (of some sort) is an ineradicable feature of the very best
 methods, to the approx detriment of the populace that has to vote in
 such methods.

>
>40 A
>31 B C A
>29 C B A
            Tot = 100
>
>Here, B and C have split the vote.

So what?, people need to strategically vote in ways that do not correspond
 to their intent. The alternative is have the method provide alterations
 where a 1st preference loser can become a winner (a (P1) violation). That
 is avoided, and then a problem turns up with 2nd preferences.

...
>Are you aware of Arrow's theorem, which says that in any realistic
>method, you are always going to have the possibility of spoiler
>candidates, who do not win, but still alter the election result?

Numerical examples that can be followed may be better, and there is less
 chance of total error than tends to occurs when tailing behind Social
 Decision Theorists.

>
>That doesn't mean, however, that the election has to be determined by
>a "string of dead losers". AV tries to avoid this, but not necessarily
>in the best possible way.

Let Mr B.C. comment on "dead losers".

>
>In particular, in the example

 45 A
 12 B A
 43 C B A

>AV [and IFPP] eliminates B, then elects A.  In this example, C is
>clearly acting as a spoiler, since, if C was not running

 45 A
 55 B A

>B would win.
>
>Since the C B A voters prefer B to A, we can say that these voters
>are ill-served by the fact that their support is locked behind the
>dead loser C, until it is too late to help the viable option B.  In
>fact, SPC mandates this locking away.

The mandated by SPC argument can be this:
A wins the first (45:A, 12:BA, 43:CB) if B loses this:

 45  A
 12  B
 43  C B
 q=33.3.., A wins

and C loses this:

 45  A
 12  B A
 43  C
 q=33.3.., A wins

So: either preferences harm candidates of subsequent preferences, or
 preferences after preferences for a candidate can harm that candidate
 [an SPC violation], or there is a dispute over the winners of the
 2 small elections above [e.g. FPP/FPTP is preferred].

>
>As a criterion for excluding spoilers, I prefer the Local
>Independence from Irrelevant Alternatives Criterion (LIIAC).  I have
>no idea what they mean by "Local".  But in any case, this criterion
>says that you should find the smallest non-empty set of candidates
>such that no candidate outside the set is majority preferred to any

This is pairwise comparing idea, an idea of much reduced credibility
 (ref. Condorcet) due to the failure to remain plausible in any idiot's
 opinion when applied to extremely large problems.

>candidate in the set.  Then, the candidates outside the set should not
>have any effect on the election outcome.

So it finds the wrong number of winners?. The constraint that the right
 number of winners be found dramatically reduces the complexity of
 arguments so long as STV's SPC rule hasn't been sacrificed.

Is there a dispute over whether is it essential to make it an axiom, that
 the method return the right number of winners?.

I regard it as vital that the method get the right number of winners.
 That simplifies derivations from rules.

Mr BC wrote "In fact, SPC mandates this locking away." I gave a proof
 and the proof assumed that the number of winners was exactly one rather
 than anything from 0 (otherwise 1) to 3. Making any sort of real
 progress in deriving solutions would become much harder if ideas like
 LIIAC are introduced. Worse, LIIAC makes the method too impractical to
 use, and it about impossible to see how it is plausible given that it
 seems awfully suspect in very large elections. B.C. himself stopped
 giving examples as soon as LIAAC was introduced


>However we define "dead loser", we would not expect that a dead loser
>would be preferred by a majority to a viable candidate.  LIIAC
>guarantees that if you have a bunch of "dead" candidates, none of whom
>are preferred to any of the viable candidates, none of the dead
>candidates can affect the election outcome.

What are the winners?. The price of getting any sized winner set the
 method feels in the mood for, is just too high. A 2nd election could be
 done, but it is, is it not, that a LIIAC method could fail to find even
 one loser?.

STV's SPC should not be lost, since it is a tremendous simplifier of the
 mathematics. Once SPC is imposed, STV ends up being the only known
 choice ?, for 20 or more candidates, and when there are less.

---------------------------------------------------------------------------
To Mr David Catchpole:

I'll skip over the possibly alleged bad Japanese voting methods. I failed
 to realize your comments about ("D E M O R E P" & " " & "1") [an achetypal
 'adult' democracy advocate] were such a light and real attempt at humour.
 Missed that. On the 1 vs. 2 candidate problem: a 1 candidate example is good
 enough because D at AOL's algorithm has a loop and there is no way to exit it,
 even when the only candidate is found to be a winner.

At 09:16 25.01.00 , DEMOREP1 at aol.com wrote:
>Attention juvenile idiots on this list-
>
>Demorep1 happens to be a male.
>
>Your idiot postings at least give the readers a few laughs (but zero lasting 
>election method math info).

You send in many messages: what about an admission of having programmed
 a bug into your pr algorithm?. If you adapted to every criticism, the
 method could end up being almost STV?. Do you consider STV to be too
 complex?.



Mr. G. A. Craig Carey, Auckland



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