[EM] Borda Count to the median

[Blake Cretney]

On Thu, 26 Oct 2000, "Michael A. Rouse" wrote: 
> This is my first post, and I haven't had a chance yet to go through
> 
> the archive to see if this method has been covered, so if it has 
> obvious flaws forgive me :)

--snip--

> Conduct a standard election with BC-style rank ballots. Starting
> with 
> each candidate's highest-ranked ballots, count out a number equal
> to 
> the smallest absolute majority of total votes cast. If 1000 (or
> 1001) 
> votes were cast, you would take the 501 highest-rated votes for the
> 
> candidate. Use these votes in a Borda Count. The candidate with the
> 
> highest total is the winner.

Here are some concerns, assuming I understand the method properly. 
Consider the following example:

45 A B C
25 B C A
30 C B A

A 45*2+6*0=90
B 25*2+26*1=76
C 30*2+21*1=81

So, A wins.  Plurality would also pick A.  Condorcet's method picks
B.  IRV would pick C.

I see a few problems with A being the winner.

First, A is ranked last by a majority.

Assuming sincere rankings, if these same voters were asked to pick
the worst candidate, candidate A would also win for worst candidate
by the same method.  However, no candidate can be both best and
worst.

Note that every voter placed B and C together on their ballot (A
never comes between).  It is therefore possible that B and C
represent some similar ideology or party.  We can therefore consider
this an example of vote splitting.

By the way, I recommend you check out my Electoral Methods Resource. 
It isn't an official FAQ, but it tries to fill that kind of role.

http://www.fortunecity.com/meltingpot/harrow/124

---
Blake Cretney