[EM] RE : Ranked Preference benefits

[Juho]

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  
and natural. It can also be called the "Least Additional Votes"  
method since it can be shortly defined by saying "elect the candidate  
that needs least number of additional votes to become the Condorcet  
winner". In the example D needs only 2 votes, so it is clearly the  
winner. I think the least additional votes philosophy makes sense in  
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)
> 50 AD 49 (1)
> 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
>
>
>
> ----
> election-methods mailing list - see http://electorama.com/em for  
> list info


		
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