[EM] Re: Weighted Median Approval

[Chris Benham]

Mike,
The reason why  Woodall's  "Quota-Limited Trickle-Down"  method's 
description makes mention of  a quota
etc. is  because that is how Woodall likes to conceive of it, and 
because he is  interested in multi-winner PR
versions. Obviously  single-winner QLTD is easier  and quicker to 
hand-count  than Bucklin, because if  a candidate
takes the lead after any given round, it is sufficient to show that that 
candidate gains a majority on the next round to
declare that candidate the winner. There is no need to check if any 
other candidate also gets a majority, and then see
which majority is the biggest.
Weighted  Median Approval (WMA)  can perhaps be thought of  as 
clone-proofed Bucklin. It  should be at least as
easy to hand-count, and seems to me to be in every way better.

Here is a simple example from Marcus Schulze:
2: A>B>C
3: B>C>A
4: C>A>B

The median rank of  all  the candidates is 2.  Using  Forest's formula 
to break this tie, A is the winner of  what  I  will
simply call  the "Highest Median Rank" (HMR) method. There are 9 
ballots, so the middle one is ballot 5 (with 4 below
and 4 above). For candidate A, spreading out the ballots in a row  in 
order from those on which A is ranked highest to
those on which A is ranked lowest, we see that on the middle ballot A is 
ranked second; so that is A's median rank.
Using Forest's formula to break the tie, we subtract the number of 
 ballots on which A is ranked below second from
the number on which A is ranked above second, and then divide this sum 
by the number of ballots on which A is ranked
exactly second. For A this comes to (2-3)/4 =  -.25.  For B this is 
(3-4)/2 = -.5  and for C it is  (4-2)/3 = -.6666.

None of the candidates have a majority of first preferences, but all of 
them are ranked first or second on all the ballots.
C  has the most first preferences and the most  combined first and 
second preferences, so C is both the Bucklin and
QLTD winner. So QLTD is definitely not  (as I once thought it might be) 
equivalent to the HMR(Forest's formula)
method.

In this simple example, both WMA and  WMA-STV  also elect C.
In WMA-STV, B has the lowest WMA score and so is eliminated.Then C is 
highest-ranked on a majority of the ballots,
and so is elected.

Now we  replace C with the clone set  C1, C2, C3.

2: A>B>C2>C1>C3
3: B>C3>C2>C1>A
4: C1>C2>C3>A>B

Now Bucklin, QLTD and HMR all elect B, showing that they are hopelessly 
clone- dependent. This is an example of  failing
Woodall's  "Clone-Winner" criterion.

> "Clone-Winner: cloning a candidate who has a positive probability of 
>  election should not help any other candidate"


Weights      A:2    B:3    C1:4    C2:0   C3:0    

WMA  approvals
2: A, B
3: B, C3, C2, C1
4: C1, C2, C3, A

WMA  final scores --  A: 6,     B: 5,    C1: 7,    C2: 7,    C3: 7

WMA  gives  a  3-way tie between  the clones, which should probably be 
resolved by picking the tied candidate with the
greatest "weight", C1.
WMA-STV  first eliminates B, which raises C3's top-preference score to 3 
, so then eliminate A, which raises C2's
top-preference score to 2. Then we still have no candidate with a 
majority of top-preferences, and the clones have tied
elimination scores, so as in plain WMA we would break the tie in favour 
of  C1.

Chris Benham