# [EM] PR, Approval ballots, "power"

Kevin Venzke stepjak at yahoo.fr
Sat Jul 12 00:04:02 PDT 2003

```Adam,

--- Adam Tarr <atarr at purdue.edu> a écrit : >

Thanks for your response.  I hope these explanations will be clearer:

"Total Power" is an arbitrary number; I've been using 10,000.  Suppose there
are 5 seats to be filled and 100 voters.  In that case, every elected candidate
will end up with a fifth of the power (2,000).  Ideally every voter should
get (10000/100) or 100 power.

Let's say I am the only voter (of 100) who voted for a certain candidate.  If he is
somehow elected, I personally enjoy 2,000 power, or twenty times too much.  (The
idea is that this candidate is only responsible to me.)

Let's say I am one of 90 voters who vote for a certain candidate.  If he is elected,
I only enjoy (2000/90) or about 22 power, and am entitled to 78 more (from
some other candidate).  The idea is that this candidate doesn't represent me very
specifically, so he shouldn't cost me as much.

The value I presently see in this, is that we can equalize voter power by electing
five candidates who each represent a fifth of the voters, as well as by electing
five candidates who each represent all the voters, or some combination.  No doubt
there are other ways of doing this.

> >if I am a voter, my "power" for
> >a given set of winners is the sum of
> >
> >([TotalOfAllPower] / [NumberOfSeats]) / [NumberOfVotersForThisCandidate]
>
> >(Every ballot initially has strength: (TotalOfAllPower)/(NumberOfAllVoters).
>
> >2. The strength of every ballot who voted for this candidate is reduced by:
> >((TotalOfAllPower)/(NumberOfSeatsTotal)) / (VotersForThisCandidate).
>
> explain exactly what they are.

Maybe something resembling an example will help.  Let's say the candidate with
the most votes got 55 votes.  Every ballot starts with strength 100 (that is,
10000/NumberOfVoters).  For the first seat, the method sees that 5500 ballot strength
(100 strength times 55 ballots) are "available" for this candidate, so he's elected.

The ballots of everyone who voted for this candidate lose strength equal to
2000/55 (that is, power wielded by a single candidate, divided by the number
of supporters for the candidate), or 36.36.  Each of those ballots has 63.67 strength
remaining (to spend on the remaining four candidates to get seats).

I'm particularly interested in what may be lacking with this form of "decay."

>
> >It seems
> >a desirable quality that the Approval winner is always elected.
>
> Not necessarily.  Foe example, if the candidates are left of center, right
> of center, and center, and you're electing two seats, then electing center
> (the probably approval winner) will result in a lopsided slate of
> candidates.  Some voters are over-represented, and others are
> under-represented.  The same would be true with left, left-center, center,
> right-center, and right, with four seats.

I definitely notice this.  The reason I'm inclined to say it's good to elect
the Approval winner, is mainly so that it is not a liability to have the most

If you were only filling two seats, what do you think about electing the two
candidates who maximize the number of voters represented?  That's attractively
easy to do.

> If you wrote your simulation to actually step through every
> possible slate of candidates, and find the one that equalized voter "power"
> to the greatest extent possible, you would probably avoid these issues.

Well, that might take hours to calculate.  The "random trials" mode usually
out-performs the actual method, though.  I guess I'm not clear on how complicated
and thorough a PR method is allowed or expected to be.

By the way, my measurement of a "better" slate is if it has a lower sum of
the squares of the difference between each voter's ideal and received "power."
I wonder if that may be similar to the "PAV score."

>
> >Any thoughts?
>
> Only that you may as well look at the PAV links I provided earlier:  Here's

I've made efforts to read them.  It seems to be something similar but more
sophisticated.  In particular I don't understand the different decay measures,
but I'm content to be ignorant for now.

Kevin Venzke
stepjak at yahoo.fr

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```