[Election-Methods] Baldiwn does not satisfy reversal symmetry

[Ian Fellows]

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

)