Kevin, you wrote: >40 A>B>D >35 D>B >25 C>D >B wins, with 75 votes in the second round. > >Now let's raise B on some ballots: >40 A=B>D >35 D>B >25 C>D >Now candidate D wins, with 100 votes in the second round. >So, quite clearly, ER-Bucklin fails monotonicity.
Yes, this is interesting. However, I suggest that ER-Bucklin(whole) should perhaps be tallied in a different way. In the second example, you assume for tally purposes that D is in 2nd place on the A=B>D ballots. However, I suggest that we should still consider D to be in 3rd place, because there are still two candidates who are ranked strictly ahead of D. If we do this, I think that B still wins. Here's another example: a ballot marked A>B=C=D>E>F=G=H=I>J. I think that we should consider A to be in 1st place, B, C, and D to be in 2nd place, E to be in 5th place, F, G, H, and I to be in 6th place, and J to be in 10th place. Thus, the ballot would not count in favor of E until the 5th round, and it would not count in favor of J until the 10th round. If we use this method, does ER-Bucklin pass monotonicity again? If so, I suggest that it is probably better than the version of ER-Bucklin(whole) that you used in your example. I've added a page for ER-Bucklin on the electowiki. Please feel free to add to it, correct it, criticize it, etc. http://wiki.electorama.com/wiki/ER-Bucklin all my best, James Green-Armytage http://fc.antioch.edu/~james_green-armytage/voting.htm ---- Election-methods mailing list - see http://electorama.com/em for list info