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

Dave Ketchum davek at clarityconnect.com
Tue Jun 9 17:58:47 PDT 2009


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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