[EM] Why I Prefer IRV to Condorcet

Kristofer Munsterhjelm km-elmet at broadpark.no
Thu Dec 11 02:11:26 PST 2008

Chris Benham wrote:
> Kristofer,
> You wrote (Sun.Nov.23):
> "Regarding number two, simple Condorcet methods exist. Borda-elimination
> (Nanson or Raynaud) is Condorcet. Minmax is quite simple, and everybody
> who's dealt with sports knows Copeland (with Minmax tiebreaks). I'll
> partially grant this, though, since the good methods are complex, but
> I'll ask whether you think MAM (Ranked Pairs(wv)) is too complex. In
> MAM, you take all the pairwise contests, sort by strength, and affirm
> down the list unless you would contradict an earlier affirmed contest.
> This method is cloneproof, monotonic, etc..."
> Raynaud isn't Borda-elimination. It is Pairwise Elimination, i.e. 
> eliminate the loser of the most decisive or strongest pairwise result
> (by one measure or another) until one candidate remains. You may have
> instead meant to write "Baldwin",though some sources just talk about
> 2 different versions of Nanson.

> Simpler and much better than any of those methods are  Condorcet//Approval
> and  Smith//Approval and  Schwartz//Approval ,in each case interpreting
> ranking as approval and so not allowing ranking among unapproved candidates.

I haven't had the time to reply to the longer posts here yet, but of 
course, you're right. I meant to say Baldwin, not Raynaud. The 
difference between the two is that one eliminates the loser, while the 
other eliminates all below average - somewhat like the difference 
between Hare and Carey. Nanson's the average, and Baldwin's the 
loser-elimination, unless I'm mistaken.

If you don't like approval cutoffs (implicit or explicit; you probably 
have no problem with the implicit ones, but I'm using the general "you" 
here), perhaps Smith//Range would work (or for that matter, UncAAO with 
range as partial approval, though it's not as simple). One of the 
problems with Range is that there's a great incentive to equal-rank 
(bottom or top). Interpreting the ballot as a ranked ballot to determine 
the Smith (or whatever) set, then breaking ties by whoever has the 
greatest score, might ameliorate both Range and Condorcet's problems: 
you can't bury without decreasing your score for the tiebreak, and you 
can't maximize without throwing some discriminatory power (that's useful 
for the first stage) away. Though, on the other hand, it might just lead 
to ballots like:

X: 100, Y: 99.999, Z: 99.998, W: 0.003, A: 0.002, B: 0.001, C: 0

But if Smith//Approval is good, then that won't be any worse - well, it 
might, somewhat, since you can't have "equal scores but different rank" 
in this version.

I guess one of the advantages of implicit Smith//A is that you can't 
bury disapproved candidates. Smith//Range would lose this advantage.

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