[EM] reply to Venzke's confusion

[Warren Smith]

> WDS: 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.

>Venzke:
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.

--WDS: Actually, it was YOU who, in at least two different instances, conducted
a computer sim of exactly this scenario, and it weas YOU who introduced
this scenario for that purpose.  Now if you want to accuse me of
noticing the obvious, look in the mirror...

>Venzke:
I don't understand the "proof" section at all. How do you determine that (1)
X will get 100% approval and (2) everyone else will get 0.5*f approval?

--WDS:
(1) Because everybody above f*(X's utility) is approved.  Therefore X is always approved.
(2) Because all utilities are random uniform in [0,1], so the probability
your random exceeds  f*(another given random)  is f/2  in expectation.

> WDS:
> Thm 2.
> Suppose each voter votes approval-style by approving of all candidates
> with more utility 
> than U_X*f+U_Y*(1-f), 

>Venzke:
I guess you wanted to divide by two here.

-WDS: nope.  f=0.5 does divide by 2.

> where U_X is X's utility (to that voter) and X is
> the best among {A,B} and Y the worst (in that voter's eyes) and A and B are two
> randomly selected candidates (but fixed once and for all after their
> choice, which happens
> immediately before the election; think of A and B as the two
> "major-party" candidates)
> and f is a constant (for example f=90%) with 0.5<f<1.
> 
> Then: the most popular among {A,B} gets elected with 100% probability in
> the V=large limit,
> i.e the same winner strategic plurality voting would elect.
> 
> Proof: the most popular among {A,B} gets over 50% approval.  Each other
> candidate gets (2-f)/3<0.5 approval in expectation.  Q.E.D.

>Venzke:
Sorry, you're wrong. Let's say f=50% and 50 voters have utilities 100 and 0
for A and B, and 50 voters have utilities 0 and 100 for A and B. Say
there is a third candidate C with utility 51 for every voter. Then C
wins with 100% approval.  

--WDS:
First of all, the probability of an exact tie is 0% in the V=large limit, so you are wrong.
Second of all I had assume f>0.5 in this so you were not allowed to take f=0.5
like you just did.
Third of all, the other (non-A and non-B) candidates are getting approved
iff they have utility >U_X*f+U_Y*(1-f)  where U_X and U_Y are random uniforms in [0,1]
with U_X>U_Y.  This happens in expectation with probability (2-f)/3.
For example if U_X = 0.666 and U_Y = 0.333 (to that voter)
and f=.7 then a candidate Z gets approved (by tht voter) if U_Z > .7*.666 + .3*.333 = .566.
This will happen 0.444 of the time.  And sure enough 0.444 < 0.5.

>WDS:
> These facts make some of Kevin Venzke's simulations look, in retrospect,
> pretty stupid.

>Venzke:
I don't agree, since I didn't assume Schulze voters would use (extreme)
favorite betrayal. Also, I came up with useful numbers regarding Approval.

--I nowhere said the reason your sims looked stupid was something about
favorite betrayal.  I said they looked stupid because you had simuklated, rather imperfectly
some scenarios you could have foretold the results for (perfectly) without
need of a computer.  Because of the theorems I just shwoed.  Furtermroe,
it is getting worse because every one of your attacks on those theorems has been wrong
as I just showed.

-wds