[EM] PR methods and Quotas

[Andy Jennings]

Kristofer Munsterhjelm wrote:

> Andy Jennings wrote:
>
>> Like Jameson and Toby, I have spent some time thinking about how to make a
>> median-based PR system.
>>
>> The system I came up with is similar to Jameson's, but simpler, and uses
>> the Hare quota!
>>
>
> How about clustering logic? Say you have an electorate of n voters, and you
> want k seats. The method would be combinatorial: you'd check a prospective
> slate. Say the slate is {ABC...}. Then that means you make a group of n/k
> voters and assign A to this gorup, another group of n/k other voters and
> assign B to that group, and so on.
> The score of each slate is equal to the sum of the median scores for each
> assigned candidate, when considering only the voters in the assigned
> candidate's group. That is, A's median score when considering the voters of
> the first group, plus B's median score when considering the voters of the
> second group, and so on. The voters are moved into groups so that this sum
> is maximized.
>

The median is not what you want for clustering like this, because it
basically ignores the scores of half the voters assigned to each candidate.
 That is, if I'm assigning 11 voters to each candidate, I can assign 6
voters who love that candidate and 5 voters who hate the candidate and still
have a very high median.


 Say there are 100 voters and you're going to elect ten representatives.
>>  Each representative should represent 10 people, so why not choose the first
>> one by choosing the candidate who makes 10 people the happiest?  (The one
>> whose tenth highest grade is the highest.)  Then, take the 10 voters who
>> helped elect this candidate and eliminate their ballots.  (There might be
>> more than ten and you'd have to choose ten or use fractional voters.  I have
>> ideas for that, but lets gloss over that issue for now.)  You can even tell
>> those 10 voters who "their" representative is.
>>
>
> I imagine you could eliminate the voters directly, though that would have
> some path dependence problems (which was why I suggested the above). Say you
> make use of highest tenth grade. Then you know which voters voted the
> candidate in question that high. Eliminate these. Find the highest tenth
> with those voters elminated, among uneliminated candidates. Again, you know
> the 10 voters who voted the next winner at that level or higher. Eliminate
> *them*. And so on down.
>
> Is that what you're suggesting?


Yes, this is what I'm suggesting.


> Then the last candidate is only the one with the best worst votes in the
> sense that there are only ten voters left.
>
> How about using the midpoint? That is, you find the 5th voter down, not the
> 10th. Then when you're down to the last 10 voters, the 5th voter down is the
> median. Doing so would seem to reduce it to median ratings in the
> single-winner case, since 100/1 = 100, so you'd pick the midpoint, i.e. at
> the 50th voter, which is the median.


True, but in filling the first seat, I don't think we should take a
candidate loved by 5 and hated by 95 as the first choice to represent
one-tenth of the population.



> But this system doesn't reduce to median voting.  Which got me thinking...
>>  Is there anything that special about the 50th percentile in the
>> single-winner case anyways?  I can imagine lots of single-winner situations
>> where it's more egalitarian to choose a lower percentile.  In a small and
>> friendly group, even choosing the winner with the highest minimum grade is a
>> good social choice method.  It's like giving each person veto power and
>> still hoping you can find something everyone can live with.  This is the
>> method we tend to use (informally) when I'm in a group choosing where to go
>> to lunch together.
>>
>
> I think the median is used because it's robust. If you assume unlimited
> ratings, the maximum and minimum could be altered by a single voter
> (whoever's at the min or max), as could the mean (by any outlier). However,
> the median is robust to distorted values - quite a number of voters would
> have to change their votes to alter the median.
>

With any finite number of voters, the median is still the score of one
voter, who can change the median by changing his vote.  But you are right
that if the scores follow a normal distribution, then he probably can't
change the median very much before he crosses another voter's score and is
not the median vote anymore.   But that's not true for a bimodal
distribution.

- Andy
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