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 ---- Election-Methods mailing list - see http://electorama.com/em for list info