[EM] Strategic voting in Condorcet & Range N-canddt elections

[Warren Smith]

>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
http://RangeVoting.org  <-- add your endorsement (by clicking
"endorse" as 1st step)
and
math.temple.edu/~wds/homepage/works.html