# [EM] reply to Venzke's confusion

Kevin Venzke stepjak at yahoo.fr
Wed Dec 7 07:32:19 PST 2005

Warren,

--- Warren Smith <wds at math.temple.edu> a écrit :
> > 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...

If you were trying to mimic my simulation, you messed up. You have
everybody voting as though their expectation is f*U_X. That is the same
as everybody believing they get nothing if X loses.

Your other theorems may be right given some assumptions, but this one
surely isn't.

> > 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),
> > 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.

But you selected U_X, U_Y, and f so that expectation is above .5.

Let me give you a different analysis, and you tell me what you don't
agree with: On average, U_A and U_B are approx. 0.500. Say that f is
as close to 50% as makes no difference. Then A gets approx 50% approval
and B gets approx. 50% approval. The average utility of a third candidate
is also approx. 0.500. That means that on average, *every* candidate
gets approx. 50% approval.

The more that (U_A + U_B) < 0.500, the more approval other candidates
can expect to get, of course.

I guess the result of your theorem should be that with 100% probability,
it is a tie among all candidates.

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

I don't follow you. In the above scenario (the "two evils" scenario),
you say A or B wins all the time. No simulation with realistic conditions
will have A and B winning all the time.

Whenever I have given a counter-example to something, you complain
that this wouldn't happen if we had enough people voting randomly.
So what you're trying to demonstrate is quite different from what I
was trying to find.

>>Venzke:
>Two problems:
>1. You didn't show burial here. Burying involves order reversal.
>2 . How did you show that the second scenario is superior?
>
>--WDS:
>(1) Aha.  Well in that case, ranking B (say) coequal bottom is >not a
>"burial"
>by your definition.  It seems like burila to me, but if that is how you
>define it,
>then fine, it isn't.

I call it "not voting for the other frontrunner." I agree that this

>>Venzke:
>>Isn't this identical to Thm 7?
>
>--WDS: no.  5/12 is not 1/2.

Please explain the differences in the Condorcet scenarios you set up
for Thm 7 and Thm 9. The only difference I can see is that the 1/2 figure
(probability of A being ranked above some Y) changed to 7/12. (The
symmetric completion shouldn't make a difference.)

>>51 X=Z
>>49 Y=Z
>>
>>Z wins even if we say that A and B are X and Y.
>
>--WDS:
>This counter example is failing to use full rankings (forbidden).

I could have added in ">..." but I thought it was clear that the lower
rankings didn't make any difference.

Kevin Venzke

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