[EM] More results about... correction

Kevin Venzke stepjak at yahoo.fr
Tue Dec 6 15:01:22 PST 2005


Hello,

--- Kevin Venzke <stepjak at yahoo.fr> a écrit :
> Warren,
> 
> --- Warren Smith <wds at math.temple.edu> a écrit :
> > Consider the following cheapo model of simulating an election. Each
> > candidate to
> > each voter has a utility which is an independent uniform random in the
> > interval [0,1].
> > There are some fixed number C of candidates and some number V of voters
> > where we shall assume V is LARGE:
> > 
> > Thm 1.  
> > Suppose all the voters magically know the identity X of the
> > max-summed-utility candidate.
> > Suppose each voter votes approval-style by approving of all candidates
> > with more utility 
> > than f*U_X, where U_X is X's utility (to that voter) and f is a
> constant 
> > (for example f=10% or f=90%) with 0<f<1.
> > Then: with 100% probability in the V=large limnit, X will win the
> > election.
> > 
> > Proof: X gets 100% approval. Any other candidate gets approval 0.5*f
> > approximately. Q.E.D.
> > 
> > Note 1a.  Actually the above proof works for ANY pre-fixed candidate X,
> > who instead could be the Schulze winner, the worst-summed-utility
> > candidate, or whoever.
> > X still will be elected.
> 
> Congratulations. You just proved that if all the voters can guess which
> candidate I'm thinking of, this candidate will get 100% approval, no
> matter how likely they think it is that this candidate will win.

Actually, I wasn't fair here. He proved that the above is true *if the 
voter "votes approval-style by approving of all candidates with more
utility 
than f*U_X"* when X is the candidate I'm thinking of.

But still. Why suppose that voters use such a silly strategy? They're
voting
as though X losing is worth 0, no matter who he loses to. You shouldn't
assume that X is everyone's favorite candidate.

> > Thm 2.
> > Suppose each voter votes approval-style by approving of all candidates
> > with more utility 
> > than U_X*f+U_Y*(1-f), 
> 
> I guess you wanted to divide by two here.

I read this wrong. I was assuming f was 50% and not present in the
expression.

Sorry for my mistakes, Warren.

Kevin Venzke



	

	
		
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