# [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
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

)

```