Request for criticism
Rob LeGrand
honky1998 at yahoo.com
Mon Mar 19 12:52:39 PST 2001
Howdy all,
In http://groups.yahoo.com/group/election-methods-list/message/6764 I outlined
my plan for a simulation to test the expected social utility of several
different ranked voting methods. I think I've received all the criticism I can
expect, so here are my answers for Martin (6765) and Craig (6767).
> Which criteria are these, btw? - from the methods you've picked, I'd assume
> the Condorcet Criterion - yet you include Borda and Copeland - odd.
I included methods that could be calculated from a cumulative pairwise matrix,
not all of which could be considered acceptable methods. For example, even if
Borda turns out to be the best of the lot at satifying voters overall, I
wouldn't recommend using Borda in real elections because of its weak resistance
to strategic voting. I very much would have liked to include IRV, but I
haven't found the time to implement it; for the methods I've included I only
need to generate the ranked votes one at a time and update the cumulative
pairwise matrix P, but for IRV I'd have to save every single ranked vote before
beginning the elimination counts. IRV is much more of a mess to implement. I
wish IRVies would realize what a nightmare it would be in practice.
Also note that of the methods I included, neither beatpath(P) nor Condorcet(P)
(in http://groups.yahoo.com/group/election-methods-list/message/4985 Norm
called it all-votes) satisfies the Condorcet criterion. I don't know if that's
been pointed out on the list before or not; I'll post a proof if requested. So
my list of thirteen ranked methods includes only ten Condorcet methods
(including Copeland). The six I'm most interested in comparing are
beatpath(M), beatpath(W), beatpath(T), Tideman(M), Tideman(W) and Tideman(T).
Condorcet(M), Condorcet(W) and Condorcet(T) are almost as good, but they aren't
strictly clone-independent and don't meet criteria like Condorcet Loser.
Copeland is much less acceptable.
> I'd have chosen a more varied range of methods, myself - I'm not sure how big
> the differences between the different condorcet completions is - and whether
> its big enough to make a diff.
My primary intent is to compare the social utilities of the popular Condorcet
completion methods (with a few non-Condorcet methods thrown in for fun). I
think it's mostly accepted that only Condorcet methods are best at resisting
strategic voting. No matter how big the differences between them are, I'm
curious to see which are best. I would certainly have included more methods if
they were simpler to implement, and I'll get around to it sometime; I'd
especially like to show how bad IRV is at satisfying sincere voters. Anyway,
don't be distracted by the non-Condorcet methods; I'm really just interested in
comparing the six I mentioned above.
> Incidentally (this is to Rob), what are the specifics of your Borda score? I
> would assume the conventional 24 points for 1st preference, 0 points for 25th
> preference. How are draws dealt with? This has always been a problem in
> Borda - draws are not traditionally allowed, because if you allow them, you
> create problems eg. A=B=C>D=E. D and E get 0 points, but what about A,B,C?
> To really test Borda, you should force the voter to choose between drawn
> candidates (ie randomise), because this is how the system works in practice.
My pairwise matrix P for the above ranked vote would be:
A B C D E
A 0 0 1 1
B 0 0 1 1
C 0 0 1 1
D 0 0 0 0
E 0 0 0 0
To calculate the Borda score for a candidate, I subtract its column total from
its row total, giving 2 for A, B and C, and -3 for D and E. This approach is
equivalent to giving average Borda counts to tied candidates, i.e. 3 to A, B
and C, and 1/2 to D and E. So I get the same average results as if I'd asked
each voter to choose randomly to break ties; I think it's the most logical
solution.
In addition to counting how often methods agree with one another's choices for
election winners, for each method I'm totaling the social utilities of the
winners it picks. For example, say there are four candidates in an election.
All voters' cardinal ratings are added up, giving A 24891, B 25027, C 25001 and
D 24706. If methods One and Two pick B as the winner and Three picks C, 25027
is added to the totals of One and Two and 25001 is added to Three. Method
Three didn't pick the cardinal ratings winner, but it didn't do too badly. I
think this measure will be a better indicator of a good method, but I still
keep track of method correlations (where agreeing with the CR winner more often
is best). The measures should largely agree anyway. I'm doing the following
three simulations:
elections candidates voters
99999 4 10000
99999 10 10000
99999 25 10000
Barring any more criticism or questions, I'll post the results soon.
=====
Rob LeGrand
honky98 at aggies.org
http://www.aggies.org/honky98/
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