Mike, In response to this part of my last post:
http://groups.yahoo.com/group/election-methods-list/files/wood1996.pdfThis is "Election 2" from the same paper (page 11): 12: A>B>C>D>E>F 11: C>A>B>D>E>F 10: B>C>A>D>E>F 27: D>E>F 60 ballots. Smith set comprises ABC.
Bucklin, QLTD, HMR agree with WMA (and RP and BP) in electing A. WMA-STV elects C.
you wrote:Now we add 6 A>D ballots: 12: A>B>C>D>E>F 06: A>D 11: C>A>B>D>E>F 10: B>C>A>D>E>F 27: D>E>F 66 ballots. Smith set is ABCD, Schwartz set is ABC.
Now Bucklin, HMR, and QLTD all elect D. Adding ballots all with A ranked first, causing A to lose, demonstrates that those methods fail Mono-add-top.
In Chris's example, A is the winner with Bucklin. Wouldn't most agree that half of the voters isn't a majority? So when D gets 33 our of 66 votes, D doesn't win? In a subsequent round A gets a majority, a vote total greater than half of the number of voters.I reply:
OK, I failed to spot that a small difference between Bucklin and QLTD, that Bucklin is about "more than half",
whereas QLTD is about a "quota" of exactly half makes a difference in this example of Woodall's.
However, this problem disappears if instead of adding six, we instead add seven A>D ballots.
So Bucklin (and QLTD and HMR) fails Mono-add-top.
Another little mistake in my last post has come to my attention:
C doesn't have any second preferences, so we say that C needs all of them, and needs to use some third preferences and soQLTD winner is B.40: A>B>C 25: B>A>C 35: C>B>A 100 ballots. B is the CW.
The "quota" is 50 (%). To reach this, A needs (10/25 = .4) of his second preferences, whereas B needs (25/75 = .33333)
of his second preferences to reach the quota. C needs (15/40 = .375). B's fraction is the smallest, so B wins.
loses to any candidate who can make a quota using only first and second preferences.
Chris Benham
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