[EM] Smith//3Score (was MAM vs Schulze)

[Michael Ossipoff]

Yes, the method still allows chicken dilemma defection to succeed.

I tried an example in which the B faction is about as large as possible in
comparison to the C faction.

99: A>B
98: B (sincere is B>C)
100: C

This results in a cycle, so everyone is in the Smith-set.

B has more Borda points than anyone else.

It looks as if it isn't possible to have CD in a strategically good,
reliable, uncriticizable.  rank method.

ICT isn't good as a ranking method. Only as a 3-slot method in which the
middle rating is used only in Chicken-Dilemma situations.

In ICT, a candidate you rank middle doesn't get much protection from you.

But, in a chicken dilemma situation you don't expect B to be a CWs anyway.

3-Slot ICT is still my favorite, to be used as an Approval version rather
than a ranking method, with the middle rating only for chicken dilemma.

Plain MMPO meets Weak CD, FBC, LNHa, and has wv-like strategy.

Though it fails CD's strong original version, if the defection is a
burial--The method's wv burial defensive advice would warn the A voters to
not rank B, if A is likely to be CWs.

That's a unique, impressive & amazing set of advantages.

But there are some strongly-felt criticisms to it. I've answered them, and
it seems to me that only one of them is a genuine problem:

...the possibility of the perpetual burial fiasco.

But I've told here why there's something a bit mutually contradictory about
that fiasco's requirements.

So I suggest that it doesn't rule out MMPO or MAM, though it makes them
just a little questionable & unreliable.

...but still worth a try because of big advantages.

I'd hoped that Bucklin with conditional votes would be a good CD method.
But it's not as good as I'd hoped, because the conditional votes option can
be strategically taken advantage of, resulting in another chicken dilemma,
differently-caused.

I don't know if that rules out the conditional option, but it supports the
conclusion that a CD rank method always costs.

Michael Ossipoff


On Oct 9, 2016 3:19 PM, "Forest Simmons" <fsimmons at pcc.edu> wrote:

>
>
> (Continued below)
>
>
>> Now, how do we adapt this to general rankings? We assume that equal top
>> rankings and equal bottom or multiple truncations are allowed.
>>
>> For each ballot on which a candidate is ranked above bottom but below top
>> that candidate receives one point.  For each ballot on which the candidate
>> is ranked top or equal top that candidate receives two points.
>>
>> The Smith candidate with the greatest number of points wins.
>>
>> [End of definition]
>>
>> Note that the method does satisfy CD unlike Smith//ImplicitApproval.
>> Jameson's idea of three slot scores makes it work.
>>
>> How does it do on burial?
>>
>
> Note that for three candidates and complete rankings the method is just
> Smith//Borda.  But Smith//Borda is extremely clone dependent.  The three
> category (top, middle, bottom) point system is key to making the method
> clone proof.
>
> For public proposal it is important to have a non-cryptic definition of
> Smith.
>
> The smallest set each of whose members pairwise beats every candidate
> outside the set is succinct, but probably (by itself) too cryptic for the
> non-technical reader.
>
> I think the idea of a beat-path is very intuitive.  I remember that ten or
> fifteen years ago Michael said that he had success explaining Schulze to
> various people on the street from the beat-path perspective.
>
> If A beats B beats C beats D, then that chain of defeats is called a
> beat-path from A to D.
>
> The human brain is adept at generalizing this kind of definition to a path
> with more or fewer steps..
>
> For the purpose of defining the Smith set we don't even need to talk about
> the strength of a beat-path.
>
>   Candidate X is a member of Smith iff for each candidate Y (other than X
> itself), there is a beat-path from X to Y.
>
> In typical public elections we expect Smith to be a cycle of three.  In
> that case Smith//3Score is the same as Max Covering 3Score.  In general the
> latter is preferable because no matter how large Smith may be it continues
> to elect an uncovered candidate.
>
> For public consumption it is mainly a question of ease of understanding.
>
> Which is easier to understand a sequence of successive pairwise defeats or
> a covering relation.
>
> Another way to elect the same candidate when the top cycle has only three
> members is to list the candidates by their point totals (highest total at
> the top of the list) and starting with the bottom candidate move up the
> list replacing pairwise losers with winners.
>
> In other words the winner will be the highest listed candidate that beats
> both the champion from below and every (if any) candidate above it in the
> list.
>
> I would be perfectly happy with any of these proposals, though there is an
> obvious soft spot in my heart for the covering version.
>
> Another reason I like the covering version is that various strengthenings
> of the covering relation can be put to good use to refine this method.
> More on that in another post.
>
> Forest
>
>
> ----
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>
>
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