[EM] RE : Ranked Preference benefits

[Juho]

I wrote:
> One option would be to limit the selection to Smith Set only.

To be more exact, that could mean e.g. eliminating all candidates  
outside the Smith set or one of them at the elimination round when  
there are candidates outside that set.

Juho Laatu


On Nov 5, 2006, at 10:18 , Juho wrote:

> I reply to myself to give a more complete answer to one of Chris
> Benham's points.
>
> On Nov 4, 2006, at 3:03 , Juho wrote:
>> On Nov 3, 2006, at 19:50 , Chris Benham wrote:
>>> That at least meets  Majority Loser  and is relatively easy to
>>> operate. Also in common with IRV it meets
>>> Dominant Mutual Third, Majority for Solid Coalitions and  Condorcet
>>> Loser.
>>
>> As we discussed earlier I think many criteria like Condorcet loser
>> lose their original idea in some extreme cases like with strong
>> cycles. For many criteria it is also ok if they are met in all the
>> "usual" cases, not necessarily in 100% of the cases. Therefore I
>> think it is best not to treat the criteria as boolean values (filled
>> or not) but as nice exact definitions that should be supported by
>> practical examples (real life failure cases) that demonstrate that
>> the vulnerabilities are real, not just theoretical.
>
> Not respecting the Condorcet loser criterion (unlike the DMC +
> preference strengths variant that Chris Benham talked about) is one
> of the properties of (the described version of) the Ranked
> Preferences Method. The reason I included this feature is however not
> related to use of the preference strengths but comes directly from
> using the basic minmax with margins to pick the winner from the
> matrix (at each cycle). If one wants to fight that problem, limiting
> the winners to the Smith set is maybe the most common approach. But I
> think electing Condorcet loser in some of the extremely cyclic cases
> may actually be the best choice available.
>
> My usual example is this:
> 25:A>B>D>C
> 25:B>C>D>A
> 25:C>A>D>B
> 24:D
>
> There are four parties of about equal size. D is the Condorcet loser
> but on the other hand she is two votes short of being the Condorcet
> winner. The other three candidates are badly cyclic and would each
> need 26 additional votes to become Condorcet winners. In this light D
> is not a bad winner after all. Not electing the Condorcet loser is a
> good property in 99.9% of the elections but in the remaining tricky
> cases things may look different.
>
> If you don't buy this argumentation (that is actually minmax related,
> not ranked preference related) you probably find some slightly
> different variant of the Ranked Preferences Method better. One option
> would be to limit the selection to Smith Set only.
>
> Juho Laatu
>
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