[EM] Martin's Sixty-Forty Split Districts:

Martin Harper mcnh2 at cam.ac.uk
Tue Apr 3 16:20:58 PDT 2001

I Like Irving wrote:

> Martin,
> You are in error when you claim that the individual districts of your
> example are proportional.  This is not correct.

You are quite correct: I was misthinking "proportional" for "as proportional as
possible". My bad. :-(

As an aside, I happilly admit to knowing next to nothing about Kansas - I'm not
criticising your proposal at all - in fact it makes a good deal of sense to me. I
was just trying to show what I consider to be a flaw in your reasoning. Namely that
district sizes which give less proportional results may indeed give more
proportional results overall.

The original point was below:

> Donald wrote:
> >     I agree with your suggestion of an even number in the district.
> >     The election may be nonpartisan, but the voters and the candidates
> >will still line up according to the two major parties.  An even number of
> >members in a district will allow them both to have a more just
> >proportionality.
> LAYTON wrote:
> This isn't exactly right.  It's true that each electorate will be more
> proportional, but the body or legislature that is elected will be less
> proportional.
> Don: What you are saying does not compute.

Here's a better example of how Layton's comment computes...

support = % reps, % dems

five districts, with equal numbers of voters in each, and support as follows:
1: 60% Rep, 40% Dem
2: 53% Rep, 47% Dem
3: 51% Rep, 49% Dem
4: 49% Rep, 51% Dem
5: 47% Rep, 53% Dem
total: 52% Rep, 48% Dem

four seats per district
reps and dems both get two seats in each district - the final council is tied at
10-10. This is 50% Republican, with a proportionality error of 2%
the proportionality errors in each district are 10%, 3%, 1%, 1%, 3%, for a total of
18% and an average of 3.6%

five seats per district:
districts 1-3 elect 3Reps, 2Dems.
districts 4-5 elect 2Reps, 3 Dems.
final council is 13R-12D: This is 52% republican, with a proportionality error of
the proportionality errors in each district are: 0%, 7%, 9%, 9%, 7%, for a total of
32% and an average of 6.4%

Thus in both cases the sum is more proportional than the sum of its parts. In
particular, the five seat districts create a more proportional council, despite
being individually much less proportional than the four seat districts.

I've tried to make this example a little less extreme than the previous one - at the
expense, sadly, of simplicity. Whether it applies in any way to the situation in
Kansas is of course open to question. I note, though - that it seems reasonable in
principle - perhaps district 1 has a particularly bad set of Democrat candidates, or
is the wealthy part, so the republican tax cuts have a big following, or some
special promises have been made. Elsewhere, support is fairly evenly divided.

Whether, in general, it is better to have odd or even numbers of seats in each
district is beyond my ability to determine - but Layton appears to be able to do so,
and claims that odd numbers of seats will give a more proportional council in
general. Perhaps you should check out what he has to say?


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