[EM] calculating margins in condorcet

[Stephane Rouillon]

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.

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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