[EM] Re: Weighted Median Approval

[Chris Benham]

Mike,
On Sun.Apr.11,  I  referred to the earlier version of  WMA:

> Plain  WMA, as  I  have defined it,  is descended from an earlier 
> version (from Joe Weinstein, Forest tells us) in which each ballot
> approves  as many of  the highest-ranked candidates as possible 
> without their combined weight exceeding half the total weight,
> and then only approves the next ranked candidate if the weight of 
> candidates ranked below this (pivot) candidate is greater  than
> the weight of candidates ranked above it. If the two weights are 
> equal, then the ballot half-approves that candidate.
> The problem with this is that  it fails 3-candidate Condorcet. To 
> distinguish it, this earlier version could perhaps be called
> "Above Median Weighted Approval" (AMWA). 

The earlier version ("AMWA"), doesn't have quite as severe a 
 Later-no-harm failure.  I  think a very reasonable attempt  to get the
best of both is to use both methods, and if they produce different 
winners then elect the one which pairwise beats the other.
This could be called something  like "Weighted Median Approval Runoff", 
or maybe something more cute.

Take this example,  recently posted in the introduction to the "River" 
method.

7:A>C>D>B
6:B>A>C>D
4:B>C>D>A
3:D>B>A>C
2:D>A>B>C
2:D>C>A>B
1:D>B>C>A

25 ballots. All candidates in the Smith set. Candidates A and B have the highest median rank (2),
and of those, according to Forest's formula, B has the higher "generalized" median rank. The Borda
winner is A.
Regarding the Condorcet methods: Schulze, Simpson, River pick A; Tideman and LeGrand pick B.

WMA elects D. WMA-STV and AMWA elect A. A>D, so A wins in the WMA Runoff method.

I wrote:

"I  am pretty sure that the method  that always picks the candidate with
the highest  "generalised median rank", is Woodall's "Quota-Limited Trickle-Down" (QLTD)  rule."

After getting some advice, the word "always" may have been an over-statement. But on average it
would certainly pick that candidate much more often than Bucklin.


Chris Benham