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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