[EM] Voter Space Borda (VSB)

[Forest Simmons]

If you just want to know the method, skip the next three paragraphs, and
begin at "METHOD."

Working with Joe Weinstein's weighted median idea has brought me around
again to Richard's issue space idea.

Richard says that if candidates were uniformly distributed among the
voters, then Copeland and Borda would give excellent answers. He then
suggests filling in the gaps between real candidates with virtual
candidates, calculating the Copeland or Borda scores, and then taking the
real candidate with the highest score.

Playing around with Joe's idea has made me realize that the best way to
get the virtual candidates uniformly distributed among the voters is to
make all of the voters into virtual candidates.

METHOD:

Step 1.  Use each ballot to rank the voters; the more their ballot agrees
with yours, the higher you rank them. There are various ways to do this,
but I suggest incorporating the following condition:

Condition A. If your ballot shows a preference of candidate X over
cadidate Y, then you should rank all voters that profess X as a unique
favorite above all voters whose ballots show Y as a unique favorite.

Step 2.  The ballot of the voter with the highest average rank is used to
pick the winner of the election.

[End of Method Description.]

Remarks:

If issue space is one dimensional, and the voter rankings reflect this,
then it is very likely that the "voter with the highest average rank" in
step 2 above will be located at the "voter median" position in issue
space, and therefore the method winner will be the "voter median
candidate," i.e. the Condorcet Winner.

Some readers may detect similarities with Voter Space Borda and Kemeny's
method. Instead of the Kemeny distance between my suggested social order
and yours, we use your rank as determined by my ballot.  Instead of
entertaining all possible social orders, we stick with the orders on
actual ballots, which gives us a computationally feasible method, unlike
Kemeny.

If Condition A is adhered to, then our ranking distance removes the clone
problem associated with the Kemeny distance.


Let's see how Voter Space Borda fares on an example in which plain Borda
would violate the Majority Criterion:

6 A>B>C
4 B>C>A

Each of the six voters in the first faction would order all of the A
voters before all of the B voters, and each of the four voters in the
second faction would rank all of the B voters before all of the A voters,
so we would have:

6 a1=a2=a3=a4=a5=a6>b1=b2=b3=b4
4 b1=b2=b3=b4>a1=a2=a3=a4=a5=a6

Each A voter has 6*4=24 wins and 4*4=16 losses, which gives a Borda score
of 24-16=8.

Each B voter has 4*6=24 wins and 6*6=36 losses, which yields a Borda score
of 24-36=-12.

The A voters are all tied for first place.

But this is no problem since they all agree that A should be the winner!

Have fun!

Forest