[EM] Re: Chris's example, Bucklin, A wins

[Chris Benham]

Mike,
In response to this part of my last post:

>This 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.
>
http://groups.yahoo.com/group/election-methods-list/files/wood1996.pdf

>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.
>
>  
>
you wrote:

>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:

>> 40: A>B>C
>> 25: B>A>C
>> 35: C>B>A
>> 100 ballots. B is the CW.
>>
>  QLTD winner is B.
> 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.
>
C doesn't have any second preferences, so we say that C needs  all of 
 them, and needs to use some third preferences and so
loses to any candidate who can make a quota using only first and second 
preferences.

Chris Benham