[EM] Web interfaces and Condorcet scores

[Rob Brown]

At 11:54 AM 10/27/2003, Rob Speer wrote:
>This interface is *very nice*. I applaud you.

Thanks!

(and is it just me, or are there entirely too many Rob's around here? ;) )

>However, now that I've seen your interface, I find it amazingly
>intuitive, and I'd like to ask your permission to use it in my project.

Cool....sounds very interesting, I'd be thrilled to have my interface used 
in your project...maybe you could email me offline and we can talk about 
it.  I definitely don't consider the interface "finished" yet.

>One display of information that I think could be useful is to have the
>list of rankings, and show the vote margins between adjacent candidates
>in the rankings. This is, of course, seriously glossing over the
>details, but it does give a general idea of "how close" the election
>was. "Bob is the winner; Bob defeated Sam by 17 votes; Sam defeated Jane
>by 5 votes..."
>
>The only worry would be if a Condorcet method could produce a ranking
>where someone loses to the candidate directly below him, which would
>confuse and alarm people by displaying a negative margin of defeat.
>However, it seems that generally cycles are unrolled in such a way that
>this does not happen.

Yeah, I tried that, and wasn't happy with that behaviour, where a higher 
ranking candidate could have a lower score (which would also be true if you 
simply showed the pairwise score vs. the #1 candidate).  However, after 
sleeping on the problem....these are my thoughts:

I first tried to distill this problem down to "what should a score really 
represent?" and then started asking a slightly different question, "what 
should the difference between two adjacent candidates' scores represent?"

For instance, going back to a plurality vote, if candidate A has a vote 
count of 200, and the candidate one above him, B, has 204, that means that 
4 additional votes for A could allow A to tie B.  That is pretty simple to 
understand, and is meaningful in saying how "close" it was between the two 
candidates.

So I'm thinking, what I need to concentrate on is finding the "delta score" 
between a candidate and the one above him.  This value should always be 
positive, of course, and should represent the number of votes necessary to 
tie that candidate, but that is *not* necessarily the pairwise margin 
between those two candidates (since as you mention, that could be negative, 
which would confuse people).

As an example:

         A       B       C       D
A       X       81      30      64
B       19      X       29      22
C       70      71      X       69
D       36      78      31      X

Pairwise victories:
A -> C  (70 30)
B -> A  (81 19)
B -> C  (71 29)
B -> D  (78 22)
D -> A  (64 36)
D -> C  (69 31)

Ranking:

B:      number of wins: 3
D:      number of wins: 2
A:      number of wins: 1
C:      number of wins: 0

So:

For D to tie B in the final ranking, D would simply need to tie B in a 
pairwise ranking (which would need 56 votes). So we'll say that B beats D 
by 56 votes.

For A to tie D in the final ranking, A would either have to tie D in a 
pairwise ranking (which would need 38 votes) or tie B pairwise (which would 
take 62 votes).  The smaller of these is 38, so we'll say that that D beats 
A by 38 votes.

For C to tie A in the final ranking, C would either have to tie A in a 
pairwise ranking (which would need 40 votes) tie B (which would take 42 
votes), or tie D (which would take 38 votes).  The smallest of these is 38, 
so we'll say that that A beats C by 38 votes (even though the 38 represents 
NOT the pairwise margin for C vs. A, but for C vs. D).

So now our scoring looks like this:

B: x + 38 + 38 + 56
D: x + 38 + 38
A: x + 38
C: x

We still need to find a reasonable value to give to x, but other than that 
we are done.

Admittedly, this is a simpler election than many, but I think its safe to 
say that it is *always* possible to determine the minimum number of votes a 
candidate would need to overtake (i.e. tie) the candidate above him.

Aside from the actual algorithm used to compute it (which will get complex 
when you get into Condorcet ties), does this seem like a reasonable way to 
score them?

-rob