Below Warren writes:
But what is the number of candidates N?
He responds:
really it is better to regard N as the cardinality of the subset
of candidates that the
public thinks have a chance of winning.
I would add any candidates that a significant percentage of the voters
wish belonged above, even if their current odds were less.
Note also that the topic is a particular election - various happenings
affect N:
Perhaps a major party nominates a loser; or stumbles into having
two candidates.
Perhaps a third party has a stronger then usual candidate, or is
seriously trying to grow.
Looking at some Ns:
1 - method matters not - Plurality would be good enough here -
or any time voters can completely express their desires as to
preferring one candidate.
2 - Ditto.
3:
Approval can satisfy some - who want to approve more than one
while expressing equal liking for each approved.
Range can vote for more than one, expressing liking via ratings.
Ranking such as Condorcet can expressing liking via rankings.
The N*N arrays that contain liking information can be of special
interest to those wanting to identify how well they are doing as third
parties, even if not yet close to winning.
3+ - such as Condorcet both give voters ability to more
completely express their desires and see how they are progressing/
failing.
I can read below that Approval is better than Condorcet for N=3 - and
be puzzled. Both can rank A=B while Condorcet can also rank A>B (or
B>A) when that is a voter's desire.
Dave Ketchum
On Jun 9, 2009, at 6:52 PM, Warren Smith wrote:
in major elections, we usually have a pretty good idea who the
frotrunners
A & B are.
If we genuinely had no idea and the V-1 other votes were totally
random, then probably
in the V=huge limit best Condorcet strategy would be honesty (though
I've never seen a proof) and best range strategy is
mean-utility-as-threshold approval voting.
If all voters do that, then compare system 16 vs system 2 here
http://www.math.temple.edu/~wds/homepage/voFdata
E.g regrets using random-normal utilities & 200 voters:
system|2 canddts 3 canddts 4 canddts 5 canddts
------+--------- --------- --------- ---------
Cond| 1.61631 2.18396 2.43847 2.57293
Appv| 1.61631 1.85211 2.40181 2.83800
so in this experiment approval voting does better than Condorcet
with N=3,4 candidates; Condorcet does better with N=5; and both same
with N=2.
--Let me elaborate a bit on this:
The above regret numbers were for basic Condorcet. If you change to
other
Condorcet methods like Black or Schulze, you will get somewhat
different numbers,
but you pretty much always find that
approval is better than Condorcet for N=3, they are about the same
for N=4,
and for N>=5 Condorcet starts being superior, increasing advantage
at larger N.
But what is the number of candidates N?
See, really, N is not the number of candidates -- really it is
better to regard
N as the cardinality of the subset of candidates that the public
thinks have a chance of winning. Because they are going to vote
strategically about that subset.
In (say) an 18-candidate election, usually we are not completely
clueless.
We usually have a lot of reason to believe it is going to be A or B.
Or maybe we only feel confident narrowing it to a 3- or 4- or
5-candidate subset.
I personally have never experienced any election (which I voted in)
in which I felt unable to narrow it down to 5-or-fewer frontrunners
where I had
high confidence the winner would be in that set. Indeed, I don't think
I've ever needed to go above 3.
Elections with effectiveN>5 are, I think, very rare.
So if we believe in the "Venzke model" that the
"effective N-value" is <=5, then we conclude approval is
better than Condorcet if N=3, about same if N=4, and worse if N=5,
but they're pretty close in all three cases.
--
Warren D. Smith
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