Re: [EM] Two notes and a possibly interesting method from a friend

[Kristofer Munsterhjelm]

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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[EM] Two notes and a possibly interesting method from a friend

[Benjamin Grant]

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?

 

What do you guys think of this?

 

-Benn Grant

eFix Computer Consulting

  b...@4efix.com

603.283.6601

 


Election-Methods mailing list - see http://electorama.com/em for list info