Andy Jennings wrote:
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.
Well, yes, but the same thing holds for median ratings in general. If
you want to find someone who represents the whole population, median
ratings can pick someone who is loved by 51% and hated by 49%, rather
than someone that 80% think are okay (and I think Warren have made
arguments to the effect that this makes Range better than median).
The question then is: what makes that logic okay when you're electing a
single representative for the whole population, but not okay when you're
electing one of ten representatives for 10% of the population? Is it the
fluid nature of the clustering - that the optimizer could try to
artificially inflate the scores by packing "hate A" voters into the A-group?
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.
I guess you could be more gentle by placing the point at 50% (1/2) for
one winner, 1/3 for two, 1/4 for three ... 1/11 for ten. That would be
more Droop-like and less Hare-like. But then you can't simply eliminate
those who contributed to the voting, I think.
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.
He can't alter the median to an arbitrary extent, however. An outlier at
the mean can do so by setting his score arbitrarily high (or low), and
the max or min voter can do so, but to a limited extent, by raising his
score (if he's max) or lowering his score (if he's min). If the median
voter alters his score by too much, he's no longer the median voter.
That may change the median result by some amount (unless the new median
voter expresses the same score as the old one used to), but it's limited.
Ah, there is a term for this reasoning.
https://secure.wikimedia.org/wikipedia/en/wiki/Breakdown_point#Breakdown_point
I haven't investigated it in detail though :-)
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