[EM] Two notes and a possibly interesting method from a friend
Kristofer Munsterhjelm
km_elmet at lavabit.com
Fri Jun 28 00:48:48 PDT 2013
On 06/27/2013 06:58 PM, Benjamin Grant wrote:
> Hi, first a quick note: I haven’t been commenting because real life
> stuff, work, etc has been keeping me busy, but I fully intend to go back
> and answer any posts sent to me via the list(s). If just that my time
> and focus comes in bursts and droughts. ;)
>
> Second note, I continue to thank all who are being helpful to me in the
> journey.
>
> Now, I asked my friend, who hasn’t read up on election stuff to come up
> with a good method – I was wondering what someone intelligent would come
> up with, with no prior exposure to election science.
>
> Note: the thought experiment I asked of him had many basic constraints,
> for example, the requirement that a voter be able to go and vote on a
> single day within ten minutes, and that there would be ten candidates,
> among others.
>
> This is the method he suggested:
>
> ·Present the people with the ballot of 10 candidates and ask them to
> pick their top three and their bottom three.
>
> ·Every time a candidate is picked in a person's top three, the candidate
> gets a +1. Every time a candidate is picked in a person's bottom three,
> the candidate gets a -2. The four candidates the person did not pick for
> either get +0. (Sidebar: For N number of candidates, you have MOD(N/3)
> positives, MOD(N/3) negatives, and the rest are left neutral.)
>
> ·At the end of the night, we add up the scores and the candidate with
> the highest score wins--even if the score is negative.
>
> It’s very interesting, and I in my newness to this all don’t immediately
> the warts, but since every method has them, I assume this one does too?
That's a weighted positional system. Every weighted positional system
except Plurality fails the Majority criterion.
Here's an example. Say we have ten candidates, so that the first three
in a ranking is supported, the next four are neutral, and the last three
are penalized. I'll mark the divisions with a |. Then:
74: A > B > C > | D > E > F > G > | H > I > J
26: H > I > G > | J > D > E > F > | C > A > B
There are 100 voters in total, so A must win by the Majority criterion.
In fact, A has greater than 2/3 majority support. But let's count the
score for A and G.
A gets +1 point from 74 voters and -2 points from 26 voters for a total
of 22 points.
G gets 0 points from 74 voters and +1 points from 26 voters for a total
of 26 points.
So A doesn't have the greatest score and thus can't win, contrary to the
Majority criterion.
I don't know how to generalize the method for n candidates because I
don't know what MOD(N/3) means. If it means the remainder after dividing
N by 3, then that doesn't match: 10/3 gives a remainder of 1, which
suggests you should have one positive and one negative, not three of
each. But I imagine it would in any case have a problem when n=2.
It would probably also have clone problems. Say H is cloned in the
example above. I'm going to clone H only once (into H and h) and keep
the limits where they are, since I don't know how to generalize the
number of positives. Still, you can probably adapt it to fit the general
system.
74: A > B > C > | D > E > F > G > H > | h > I > J
26: H > h > I > | G > J > D > E > F > | C > A > B
The problem here is that it pushes G off the positives list and so G no
longer wins. Even if you increase the number of positives, one just has
to add more clones to make the same example work.
If MOD(N/3) is just the integer division of N/3, then with 11 candidates
you'd still only have 3 positives and 3 negatives, so the clone problem
above works. And if MOD(N/3) is integer division, then for N=2, you'd
get 0 positive places and 0 negative places, so there would be no way of
assigning points to any candidate.
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