[Election-Methods] Baldiwn does not satisfy reversal symmetry
Ian Fellows
ifellows at ucsd.edu
Wed Jan 2 16:21:24 PST 2008
Thank you for the correction Dr. Schulze. I have corrected my page
(http://thefell.googlepages.com/statisticalsnipstprelections), and see that
you have corrected the wikipedia article
(http://en.wikipedia.org/wiki/Nanson's_method) as well.
Cheers,
Ian
Hallo,
the following example demonstrates that the Baldwin
method violates reversal symmetry.
Situation #1:
5 ACB
4 BAC
2 CBA
The initial Borda scores are 14 for candidate A,
10 for candidate B, and 9 for candidate C.
Candidate C is eliminated, because candidate C
has the lowest Borda score.
The new Borda scores are 5 for candidate A
and 6 for candidate B.
Candidate A is eliminated, because candidate A
has the lower Borda score.
Thus, candidate B is the Baldwin winner.
Situation #2:
The individual rankings are inverted.
5 BCA
4 CAB
2 ABC
The initial Borda scores are 8 for candidate A,
12 for candidate B, and 13 for candidate C.
Candidate A is eliminated, because candidate A
has the lowest Borda score.
The new Borda scores are 7 for candidate B
and 4 for candidate C.
Candidate C is eliminated, because candidate C
has the lower Borda score.
Thus, candidate B is the Baldwin winner.
Markus Schulze
)
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