[EM] Displaying intermediate results in Condorcet-based elections

[Gervase Lam]

> Date: Wed, 29 Oct 2003 19:10:06 -0800
> From: Rob Brown <rob at hypermatch.com>
> Subject: Re: [EM] Displaying intermediate results in Condorcet-based
>   elections (re: Rob Brown's original question)

> My original request was to suggest a way to produce a single scalar
> score per candidate which enhances, but does not conflict with, a simple
> ordered ranking.

I've thought of another way of doing this without using Kemeny-Young.  I 
should have thought of this earlier: 'Plain' Condorcet.

I'll assume that the weakest pairwise win is the pairwise contest where a 
candidate has obtained the highest number of votes and still lost the 
pairwise contest.

(1) Drop the pairwise contest involving the candidate who obtained the 
highest number of votes but still lost the pairwise contest.

(2) Any undefeated candidate that results from dropping the pairwise 
win(s) is the 'winner'.  If there is no undefeated candidate, goto (1).

(3) To continue determining the ranking of the remaining candidates, drop 
all of the pairwise contests involving the 'winner' then goto (2).

The above steps can be made less iterative:

(1) Order all of the pairwise contests by the number of votes the pairwise 
losing candidate obtained.

(2) The 'winner' is the candidate whose lowest number of votes in the list 
is higher than any other candidate's lowest number of votes.

(3) To continue determining the ranking of the remaining candidates, drop 
all of the pairwise contests involving the 'winner' from the list then 
goto (2).

What you get in the end is a ranking with each candidate getting a score 
equaling the lowest number of votes received for any of the candidate's 
defeats.  The scores will decrease as you move from 1st in the rankings to 
last in the rankings.

It is very easy to invert the decreasing scores so that it increases 
instead.  Just subtract this score from the total number of ballots 
received in the vote.

I'm not too sure whether it is necessary to drop all of the pairwise 
contests involving the winner.  One reason to do this is because once 
there is a winner, we want to know who is the best out of the remaining 
candidates.

On the other hand, I can see the possibility of candidates being tied if 
the winner's pairwise contest are dropped.  An advantage of not dropping 
the pairwise is that the second placed candidate is then also compared 
with the winning candidate.  In other words, the second placed candidate 
is the second best overall, not just against those remaining candidates.
I would appreciate other people mentioning what it is best to do here.

There is one "problem" with this.  If there is a Condorcet winner (i.e. 
has no pairwise loses), then the winner would get a perfect score.  Is 
this of any hindrance?

Now that that is off my chest, I can get to sleep!

Thanks,
Gervase.