[EM] [CES #4445] Re: Looking at Condorcet

[Jameson Quinn]

2012/2/6 Dave Ketchum <davek at clarityconnect.com>

> How did we get here?  What I see called Condorcet is not really that.
>
> On Feb 6, 2012, at 10:02 PM, Jameson Quinn wrote:
> ...
>
>
> Say people vote rated ballots with 6 levels, and after the election you
> see a histogram of candidate X and Y that looks like this:
>
> (better)
> 6:Y X
> 5:  Y  X
> 4:     YX
> 3:     XY
> 2:  X  Y
> 1:X Y
> (worse)
> N:123456789
>
> That is, 3 people rated X as 6 and only one person rated them as 1, and
> vice versa for Y.
>
> X wins, right?
>
> If it's Condorcet, not necessarily. This is consistent with a 14:12
> victory for Y over X.
>
>
> I count 15 vs 6, being that all you can say in Condorcet is X>Y, X=Y, and
> X<Y.  There being no cycles in this election, I would not expect any
> variation among Condorcet methods.  Perhaps Jameson was thinking of
> something other than Condorcet - consistent with saying "rated" rather than
> "ranked"?
>

This is not standard notation; I was trying to draw a picture of a
histogram, with a distribution for X that is clearly above the distribution
for Y.

In standard notation, the 14:12 scenario is:
3: X6, Y1
5: X5, Y2
4: X4, Y3

1: Y6, X5
3: Y5, X4
6: Y4, X3
3: Y3, X2
1: Y2, X1

Obviously, X and Y are not the only candidates in this race, or people
wouldn't vote like that.

>
>
>  If you present the pairwise total, it's "obvious" to people that Y
> should win. If you present the histogram, it's at least as "obvious" to
> people that X should win. If what people find obvious isn't even consistent
> (which even just pairwise isn't, of course; that's why there is more than
> one Condorcet system), then you can't elevate "obvious" to an unbreakable
> principle.
>
> ...
>
>
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