[EM] Comparing ranked versus unranked methods

[Forest Simmons]

On Wed, 6 Feb 2002, Adam Tarr wrote:

> 
> I _think_ I follow what you are saying.  What I am arguing (perhaps 
> incorrectly!) is that my method of iteratively electing and removing candidates 
> is exactly equivalent to trying to maximize the PAV score of an entire set of 
> candidates.  That is, repeat the following process k times for a k-seat council:
> 
> 1) Tally the ballots.
> 2) Select the Approval winner.  This candidate is elected.
> 3) For each ballot that contained the Approval winner, decay all of the 
> remaining votes on that ballot by n/(n+1) (d'Hondt) or (2n-1)/(2n+1) (Webster), 
> where n is the number of candidates that ballot has helped elect so far. (For a 
> manual count, simply mark the ballot each time it elects a candidate.)
> 4) Remove the elected candidate from the ballot.

This is equivalent to sequential PAV (a form of conditional PAV) but not
to PAV itself as can be seen by the following two winner example:

# of 
ballots   approved candidates

n         A and C
n         B and C
51-n      A only
49-n      B only

As long as 2*n is greater than 51, the winning combination is {A,C} in
sequential PAV.

But when 2*n is not too much greater than 51, the winning combination is
still {A,B} in regular PAV.

If we assume the d'Hondt rule, then for 2*n between 51 and 65 the two
methods differ.

Note that when 2*n is in the low end of this range (say 2*n<58) the
combination {A,B} definitely gives better coverage, but if one of the
candidates is to be designated President and the other Vice President,
then the {C,A} combination (from sequential PAV) is slightly more
appropriate. 

Forest