[EM] Displaying intermediate results in Condorcet-based elections (re: Rob Brown's original question)

[Adam Tarr]

Rob Brown wrote:

>Well, I have nothing against using beatpaths if the scores really do 
>represent the way the results are tabulated

They do represent the way the results are tabulated, but I don't think 
that, in general, the "scores" that beatpaths produce will be 
satisfying.  For example, if you take

49% Bush
12% Gore>Bush
12% Gore>Nader
27% Nader>Gore

A beatpath-score approach would give both Bush AND Nader a score of 27, 
since this is the strength of the only defeat Gore suffers.  While this 
seems right to me for Nader, I think giving Bush a score of 49 seems more 
right.

>(except that I must admit, I've yet to quite understand how to work that 
>method into my code .. I will look into it further unless I first hit upon 
>something that is really really easy to code and works really well :) ).

I would again direct you to Rob LeGrand's website.  The description of 
"Schulze" at http://cec.wustl.edu/~rhl1/rbvote/desc.html is pretty easy to 
follow for someone who understands Condorcet (I think).  You can also 
e-mail either Rob or Eric Gorr, as both have implemented beatpath.

>  I have heard your suggestion of instead using the pairwise score against 
> the winner, but don't feel it meets my criteria of being consistant with 
> the logic used for determining ranking order.

I respectfully disagree.  Think about it this way.  When there is a 
Condorcet winner (a "beats all" candidate), then every candidate lost by 
their margin against the winner.  Where they finished relative to everyone 
else is pretty much irrelevant, since they couldn't beat the 
winner.  Condorcet is clone-independent after all.

(When there is a circular tie, there simply isn't any 1-D set of scores 
that can accurately represent what is going on.  So the performance of a 
scalar measure in this situation will never really be "perfect".)

>I prefer not have something that is "use this method, unless a certain 
>case happens that causes a noticable conflict and then fall back on a 
>completely different system".

Personally, I would be happy with a scoring system that always used every 
candidate's pairwise count against the winner, and allowed the winner to 
have a lower score than someone else in the case of circular ties.  After 
all, when there's a circular tie there's really no set of 1-D scores that 
can represent that cycle.  I suspect that you (and more to the point, your 
client) would not be satisfied with this.  So I proposed the beatpath 
approach as a fallback in this case.

Here's another potential technique that avoids beatpaths.

--------------------------

1)  Calculate the winner and the Smith set using ranked pairs.

2)  For all alternatives outside the Smith set, their score is the number 
of votes they received pairwise against the winner.

3a)  If the Smith set contains only one alternative (i.e. the Condorcet 
winner), then the winner's score is the winning alternative's pairwise vote 
total against the losing alternative with the highest score (as defined in 
step 2).

3b)  If the Smith set contains more than one alternative (i.e there is a 
cyclic tie) then find the alternative within the Smith set that the winner 
has its strongest victory against (in terms of winning votes).  Score every 
alternative in the Smith set based on their pairwise vote totals against 
this alternative.  Assign this alternative a score equal to its pairwise 
votes against the winner.

-----------------------------

3b basically amounts to breaking the cyclic tie in the fashion that 
guarantees the winner its best possible score.  A simple approach, and one 
that I think gives very intuitive results.

-Adam














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