[EM] Reply to Layton about the Kansas City Council:

[Martin Harper]

I Like Irving wrote:

> LAYTON wrote:
> 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.

Actually, you'll be interested to learn that it does compute.

> You agree with me that each
> electorate will be more proportional, but how can you say the legislature
> is going to be less proportional than the sum of its parts, that is, if all
> the electorates are proportional then the legislature must also be
> proportional.

Here's an example, for illustrative purposes. This is of course, just an
example: there are many many situations which could result in similar, or indeed
much worse, results. Also, while I will use Rep and Dem, the same could happen
with Male and Female, or Black and White, or indeed any other division you feel
is important.

Suppose that every district in Kansas has 60% Republican voters, and 40%
Democrat voters. There are three districts, and we're choosing 4 council members
from each district, for a total of 12. We also elect a mayor, who is the
thirteenth member.

Now, 60% of 13 is 7.8, so we should elect 8 Republicans, and 5 Democrats, if we
are to be proportional.

Now let's see what actually happens. The Republicans get their mayor elected,
with 60% of the votes. In each district, we give 2 seats to the Republicans, and
2 seats to the Democrats: this is a  proportional allocation, since 60% of 4 is
2.4. So, in total, the Republicans get 7 seats, and the Democrats get 5.

Now note that the individual districts are proportional, while the legislature
as a whole is not propotional. Now, you may claim that this nonproportionality
is unimportant, or won't happen in practice - but don't claim that it can't
exist...

Hope this helps,
Martin