[EM]Ranked-Pairs (wv) can lose a Cond. Winner

[Elisabeth Varin/Stephane Rouillon]

Matt --

Adam says adding 1/2 votes leads winning votes to become margins.
I do not know, I am trying to check. But the result seems to corroborate
what he says. Using your matrix, I do not find that A still wins.
I find back the result I had with margins (X=1):

>By the way,
>with x=0, margins finds a triple equality.
>With x>0 it locks B>C, then gets stuck between the two last pairwise
>comparison.
>Winning votes used as tie-breaker would give C>A. Thus B wins.
>Relative margins used as tie-breaker would give A>B. Thus A wins.

Check by yourself.

Steph.

matt at tidalwave.net a écrit :

> On 24 Nov 2002 at 14:41, Elisabeth Varin/Stephane Rouillon wrote:
>
> > 3: A
> > 2: A > B > C
> > 2: B > A > C
> > 2: B > C > A
> > 4: C
> > Ranked pairs with winning votes produces:
> > A (7) > C (6) , B (6) > C (4) and A (5) > B (4).
> > A is the Condorcet winner and wins.
> > Margins and relative margins produce of course the same result.
> > If I am one of the two B > A > C voter, my 2nd (A)
> > choice harms my favorite 1st choice (B).
> > The proof is, if I and my co-thinker vote B only:
> > 3: A
> > 2: A > B > C
> > 2: B   (truncated !)
> > 2: B > C > A
> > 4: C
> > Ranked pairs with winning votes produces:
> > B (6) > C (4), C (6) > A (5) and A (5) > B (4) can't lock.
> > B wins now.
>
> This appears to be an example that illustrates a more stable outcome is achievable
> by counting equal ranked options 1/2 vote each.  With an additional 1/2 vote for
> each non-voted pair the option pair tally matrix after > 2: B (truncated !) looks like:
> #       A       B       C
>         7.0     7.0     6.0
>         6.0     6.0     7.5
>         7.0     5.5     8.0
> and A still wins (RP and SSD).  You guys are (mis?)computing the tally matrix like:
> #       A       B       C
>         7.0     5.0     5.0
>         4.0     6.0     6.0
>         6.0     4.0     8.0
> (diagonal represents "approval count").
>
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