[EM] RE : Ranked Preference benefits

Juho juho4880 at yahoo.co.uk
Sun Nov 5 15:28:39 PST 2006

```I believe with Simpson you refer to the method that is also called
minmax. Minmax(margins) is one of my favourites. It is very simple
method since it can be shortly defined by saying "elect the candidate
winner". In the example D needs only 2 votes, so it is clearly the
many settings. I.e. the utility value of that candidate is in many
settings the best.

I tend to think that the strong dislike of Condorcet loser is not
well justified. Many people set the Smith set as a key requirement to
any Condorcet completion method. I just sent another mail in the EM
list where I tried to explain why the "real world 3D aesthetics" are
not the right way to handle cyclic preferences in the voting methods.
If the candidates would form a queue where best candidates take the
first places and the worst ones take the last places, then it would
make sense not to pick the winner from the end of the queue. But with
cyclic preferences there are no such queues. Also the visual image
that the Smith set candidates would form a group that has to be put
at the beginning of the queue is not well justified since that
thinking totally ignores the fact that the cyclic defeats within that
group should give some "negative points" to the members of that group.

Juho Laatu

On Nov 5, 2006, at 16:27 , <mrouse1 at mrouse.com> <mrouse1 at mrouse.com>
wrote:

> Juho wrote:
>> 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.
>
> I ran the example through the Rob LeGrand's election calculator at
> http://cec.wustl.edu/~rhl1/rbvote/calc.html and got the following:
>
> Winner: (A,B,C, requires tiebreaker)
> Baldwin*
> Black*
> Borda*
> Copeland*
> Nanson*
> Raynaud*
> Schulze*
> Small*
> Tideman*
>
> Winnder: D
> Dodgson
> Simpson
>
> Which seemed like an odd result to me. Any idea why Dodgson and
> Simpson
> gave the Condorcet loser?
>
> On a tangential note, as a test I tried it with IFNOP to see if it
> would
> resolve there, and got the following:
>
> 25:A>B>D>C
> 25:B>C>D>A
> 25:C>A>D>B
> 24:D>A=B=C
>
> Left hand side is normal comparison (>), right hand side is reverse
> (<),
> number in parenthesis is the difference.
> 25 AB 25 (0)
> 25 AC 25 (0)
> 25 BC 25 (0)
> 50 BD 49 (1)
> 50 CD 49 (1)
>
> Since D is the Condorcet loser, it is dropped.
>
> AB 0+0+0+25+0+0
> AC 0+0+0+0+0+25
> BA 0+0+0+0+0+25
> BC 0+0+0+25+0+0
> CA 0+0+0+25+0+0
> CB 0+0+0+0+0+25
>
> This gives A=B=C -- a tiebreaker would be needed with this method
> as well.
> (It's D's voters' fault for not giving a complete preference, and an
> election giving D the win would seem to open such methods for
> manipulation.) A runoff election between A, B, and C would probably
> be the
> fairest method.
>
> Michael Rouse
> mrouse1 at mrouse.com
>
>
>
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