[EM] Comparing ranked versus unranked methods

[Forest Simmons]

On Sun, 9 Mar 2003, Olli Salmi wrote:

> Forest,
>
> I've found this method of yours fascinating.
>
> At 18:45 +0200 6.2.2002, Forest Simmons wrote:
> >>  One option might be to decay a voter's ballot based on the position of the
> >>  elected candidate on the ballot... a sort of Borda-based decay setup.
> >
> >Right.  It turns out that since the sequence 1 + 1/2 + ... + 1/n  is
> >asymptotic to ln(n) that there is a simple way of getting a Borda version
> >of PR:
> >
> >Suppose that there are to be three winners, and that {X,Y,Z} is a set of
> >candidates to be scored by ballot B.  Just take the logarithm of the sum
> >of the Borda scores for the three candidates on ballot B.
> >
> >The three member subset with the highest sum of logarithms over all
> >ballots is the winning subset.
> >
> >Since the sum of the logs is the log of the product we can avoid
> >logarithms altogether in the statement (although computationally we would
> >still want to use them if there are more than a few dozen voters):
> >
> >The three member subset with the highest product of ballot totals is the
> >winning combination.
>
> I tried it for a while and got strangely suitable results, but then I
> found a problem. It can give the majority of the seats to the
> minority in some cases:
> Five to be elected.
> 41 ABC
> 20 DEF
> 20 FDE
> ABDEF wins.
>
> Is there a way round this?

Borda PR inherits this problem from Borda, so Borda PR should be limited
to applications where Borda's defects don't matter, i.e. in sports,
robotics, and as a benchmark of the limits of social utility for more
viable election methods.


>
> I didn't quite know what to do if the Borda score was 0, so I added 1
> to the Borda score. It's the same problem if you use 0 for the
> logarithm when the Borda score is O.
>

In multiwinner elections the Borda score should be greater than zero on
every ballot, unless all truncated candidates get scores of zero. It is
more in the spirit of Borda if two candidates sharing the bottom level
each get (0+1)/2=1/2 point, for example.


> This is how I did my calculations. v=votes, B=Borda score. I'm sorry
> that the columns are not properly aligned
>
> 	ABCFD
> 	v	B	B'	Ln(B+1)		v*Ln(B+1)
> ABC	41	12	14	2,5649493575	105,16292366
> DEF	20	8	11	2,1972245773	43,944491547
> FDE	20	9	12	2,302585093	46,05170186
> Sum						195,15911706
>

I've inserted a column B' for the Borda score that would result from
giving each of the bottom three (0+1+2)/3=1 point when the bottom three
are tied.

The same could be done for the other five combinations, but I don't think
that it would change the over all result.

Forest