[EM] calculating margins in condorcet

[Ernest Prabhakar]

On Apr 20, 2004, at 7:05 AM, Stephane Rouillon wrote:

> Nice, and I suppose you could generalize it
> to cycles using the Condorcet ranking to get your normalizing scheme.
> For example:
> A>B: 30/20
> B>C: 35/15
> C>A: 32/28
> Using ranked pair (margin): A>B>C
> Thus: A(30), B(20), C(8,1...)
> Using ranked pair (winning votes): B>C>A
> Thus: B(35), C(15), A(1,8...)
>
> The only problem I can see is a huge drop when sligthly changing the 
> votes or
> the counting method.

Yeah, its extremely model-dependent, but then again so is everything 
when there's cycles involved, since you don't even have a single 
Condorcet winner.    For Curt's case, where we're just reporting the 
results of a Condorcet single-winner election, typically the interest 
will be focused on the top 2 or 3 candidates, and this would certainly 
explain "who won and by how much."   Even for the lower-tier 
candidates, it would give a sense of 'how far they are behind the 
previous candidate.'

Of course, it would be important to publish the full result matrix for 
real analysis, but this sort of "effective vote" does provide a 
convenient shorthand.

Still, its  a bit of a hack, and I'm open to hearing better ideas.  One 
option is simply to renormalize the total to the actual number of 
voters, which doesn't change the percentages but makes the actual vote 
count come out correctly (which otherwise might bother people).  On the 
other hand, that would make the winner's votes come up less than they 
actually were, which might create a political problem by implying a 
lack of majority support (of the type IRV supporters love to point 
out).

-- Ernie P.

>
> Ernest Prabhakar a écrit :
>
>> Hi Curt,
>>
>> On Apr 19, 2004, at 7:19 PM, Curt Siffert wrote:
>>> Hi, I'm trying to figure out a way to calculate margins for multiple
>>> candidates in a Condorcet-counted election, normalized to 100%.
>>
>> I do it via pairwise ratios for each step.   So, if we have an 
>> ordering
>> A > B > C > D, where, say:
>>
>> A/B = 300/200
>> B/C = 250/100
>> C/D = 200/20
>>
>> Then I would normalize that as:
>> A: 300
>> B: 200
>> C: 80
>> D: 8
>>
>> I don't know if it is "fair", but it is well-defined, and has the
>> useful property that the 'normalizd votes' for the first two 
>> candidates
>> equal actual votes.
>>
>> -- Ernie P.
>>
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