On Jan 8, 2011, at 12:55 PM, Stephen Turner wrote:
Dear EM fans!
I was wondering if anyone can think of a source
for the following simple observation, as it might make
a nice little paper. It amounts to seeing how the Borda
count can be made Condorcet-compliant by
replacing the mean by the median as detailed below.
All ballots are total rankings of n candidates, and
X, Y are two distinct candidates. For any ballot,
write s(X,Y) for (rank of Y) - (rank of X) which of
course is the difference of their Borda scores.
The s(X,Y) are all non-zero integers, at most
n-1 in absolute value.
We will suppose that there are no pairwise (i.e.
Condorcet) ties.
Given an election profile, write b(X,Y) for the mean
of all the s(X,Y), and write c(X,Y) for the
median of the same data set.
Now b(X,Y)>0 iff X beats Y in the Borda count,
and b(X,Y)>0 for all Y iff X is the Borda winner.
Also c(X,Y)>0 iff X beats Y pairwise, and
c(X,Y)>0 for all Y iff X is the Condorcet winner.
What makes all this true is the fact that the mean of
the differences (of the Borda scores for X and Y) equals
the difference of their means; c(X,Y) is the median of the
differences, which is not the same as the difference
of the medians.
If there is no Condorcet winner, then you
could resort to one of the established methods
to break the tie (RP, Schulze, minimax, ...)
so, if this is a method to identify the Condorcet winner when there is
one, how is this preferable to just identifying the CW the old-
fashioned way: 1. total up all pairwise vote subtotals from the
precincts, 2. start out with each candidate a "potential winner", 3.
for every pair, clear the "potential winner" bit for the candidate who
loses in the "runoff" of that pairing, 4. after every pair is
runoffed, the only candidate standing with the "potential winner" flag
set is the CW.
if you are going to explicitly resort to another cycle-breaking method
if there is no CW, then i don't see why there should be any other
method than the simple one to identify the CW if there is a CW.
If there can be pairwise ties then c(X,Y)>0
only implies that Y does not beat X pairwise. There is a
range 1<=c(X,Y)<=(n/2) - 1 in which either X beats
Y or they tie - both are possible.
As calculating the median is relatively expensive, the
above probably is not useful as an algorithm.
Any thoughts?
mine are only wonder for what the method gains you.
it *is* interesting to use this to compare the difference between
Borda and Condorcet. in Borda A beats B if the mean of the ranking
distance of A over B is greater than zero. in Condorcet, it's the
same if you replace "mean" with "median".
--
r b-j r...@audioimagination.com
"Imagination is more important than knowledge."
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