[EM] Multiple Same Choices

[Rob Lanphier]

On Mon, 26 Feb 1996 DEMOREP1 at aol.com wrote:
> Answer. In a single winner case, a voter may (a) equally approve 2 or more
> candidates for the office at a given choice level (first, second, etc.)
> and/or (b) not want his/her second (or later) choice vote(s) to help elect a
> person who he/she regards as an extremist (and may only "strategically" vote
> for such an extremist to offset some other voter's vote for some even worse
> extremist).

A voter who considers someone an extremist won't rank that candidate, or 
rank them very low, only above someone who is *even more* extreme.  Their 
vote will never help that candidate get elected, except when the 
alternative is even more extreme, in which case it is exactly what that 
voter wanted.

Reason (a) is a good reason.  Reason (b) is not.

I'm trying to understand the mathematical implications of this as they 
apply to the algorithm I use to calculate the Condorcet winner.  Right 
now I count the votes against a candidate in determining their worst 
defeat.  Identical rankings throws a bit of a monkey wrench into the 
business:

Example Ballot:
Joe Left      1
Sally Middle  1
Martha Right  2

What should the pairwise set of results be?

A)
Left-0.5    Middle-0.5
Left-1      Right-0
Middle-1    Right-0

B)
Left-0      Middle-0
Left-1      Right-0
Middle-1    Right-0

I'm inclined to say B, because the implication is that this voter would
not have voted in an election between Left and Middle. Therefore, neither
Left nor Middle should have even half of a vote counted against them in
this particular election. 

Since unranked candidates are on the "same level", my program already has 
to deal with this issue.  When tabulating a specific pair of candidates, 
any ballot where neither candidate is ranked isn't counted.  So, in the 
name of consistancy, ballots where both candidates are ranked identically 
should also not be counted (in this pairwise calculation).

My brain hurts.

Rob Lanphier
robla at eskimo.com
http://www.eskimo.com/~robla