How would you design a voting system that produces the following result: the ratio of meat-eaters to...

The Hare proportional voting system is supposed to solve your problem.
Many years ago, the city of Cambridge, Massachusetts, elected its city
council members by this system (and as far as I know it still does).

In the Hare proportional system, there are some number of seats, K,
with N candidates running to fill those seats.  Assume N > K, otherwise
your problem is not how to select from among the candidates.  Each
voter's ballot ranks the candidates from 1 (favorite) to N (least
favorite).

Initially, each ballot goes into the pile for the candidate who was
ranked 1 by that voter, so each candidate now has a pile of ballots
with his or her name in slot 1.  The K winning candidates are
determined in a series of elimination rounds.  In each round the
candidate with the fewest ballots in his or her stack is eliminated
from the race.  The ballots in that stack are then redistributed to the
remaining candidates.  I think the rule is that each ballot goes to the
most "favored" candidate on that ballot who is still in the race.  On
the other hand, any candidate who has garnered 1/N of the vote is
considered to have been elected, and any votes that candidate got in
excess of 1/N of the voters are redistributed to the other candidates.
(In practice I think there are some technicalities, for example some
voters don't rank all N candidates, and there are some issues about
whether the "less favorite" choices on the "excess" ballots are
representative of the choices of all voters who voted for the winning
candidates; but this is the general idea.)

[ Editor: "1/N" here presumably means 1/K (one seat's share): in the
example below, with 10 seats, each elected candidate takes 1000 of the
10000 ballots. ]

What this means for your meat-eaters and fish-eaters voting for 10
seats is suppose there are 10000 voters of whom 2500 are fish-eaters and
rank all fish-eating candidates above all meat-eating candidates.  Then
at each point in the voting process, those 2500 ballots are in the
stacks of fish-eating candidates until at least two fish-eating
candidates get elected; each elected candidate takes exactly 1000
ballots out of circulation, leaving 500 ballots among fish-eating
candidates.  I think after this point you could end up with either 2
fish-eating winners or 3 fish-eating winners, depending on the exact
details of who voted for whom, but either one is as close as you can
get to the ideal 25% of seats filled with fish-eaters.

-- David A. Karr <karr@cs.cornell.edu>

Ireland uses a proportional representation system for all elections.
I'll describe it in terms of the differences to this system.

Initially, the number of valid votes is counted, V. This is the number
of papers which are correctly filled in (1..n for n <= N or just 'X'
for one candidate). The quota, Q is set to (V/(K+1))+1. You can see
that there can be no more than K quotas in the total poll.

After each round, the first thing done is to see if any candidate has
reached the quota. If they have, they are deemed to be elected. The
number of votes that they have above the quota is distributed to the
next preference on the paper. This is done in proportion, such that
when a candidate gets (x+1)Q quotas, a transferred votes counts as 1/x
votes. This preserves that total number of valid votes.

If no candidate has reached the quota, the candidate with the lowest
vote is eliminated and their votes distributed.

These distributions are called 'counts'.

When a ballot paper does not specify enough preferences, it becomes a
'non-transferrable vote' at the point where the missing preference
would be needed. It plays no further role in the voting.

Note that a `count' may combine several redistributions and
eliminations. Where the number of votes to be distributed is less than
the difference between the lowest two candidates, the lowest is
eliminated and the distribution of their votes is done in the same
count.

Also, if the result of further counts is predetermined, candidates can
be `deemed elected without reaching the quota'. For example, where the
next operation would be to eliminate a candidate and there is only one
more candidate than seats left.

[ Editor: Bryant continues, on Karr's fish-eater example: ]

That's the idea, but in practice it may not work so well.  For example,
there are 3 seats (quota is 25%) and 6 candidates. The candidates get
votes thus:
        A: 16%  2nd=E  3rd=F    (independent)
        B: 11%  2nd=E  3rd=F    (independent)
        C: 21%  2nd=B           (party 1)
        D: 23%  2nd=C           (party 1)
        E: 23%  2nd=F           (party 2)
        F: 6%  2nd=E            (party 2)

Note that, given A and B are unelectable, party 2 has 16+11+23+6=56% of
the vote (i.e. two candidates' worth).

Counts are:
        1: A=16 B=11 C=21 D=23 E=23 F=6         eliminate F
        2: A=16 B=11 C=21 D=23 E=29             E elected, distribute surplus
        3: A=16 B=11 C=21 D=23                  eliminate B
        4: A=16      C=21 D=23                  C and D elected

Party 1 has got two candidates elected with only 44% of the vote. If
party 2 had a balanced vote, the voting would have been:

        A: 16%  2nd=E  3rd=F    (independent)
        B: 11%  2nd=E  3rd=F    (independent)
        C: 21%  2nd=B           (party 1)
        D: 23%  2nd=C           (party 1)
        E: 16%  2nd=F           (party 2)
        F: 13%  2nd=E           (party 2)

Counts are:
        1. A=16 B=11 C=21 D=23 E=16 F=13        eliminate B
        2. A=16      C=21 D=23 E=32 F=13        E elected, distribute surplus
        3. A=16      C=21 D=23      F=20        eliminate A
        4.           C=21 D=23      F=32        F elected, distribute surplus
        5.           C=21 D=23                  D elected

Now party 2 has two seats instead of one.

-- Charles Bryant <ch@chch.demon.co.uk>

[ Editor: in the second table, B's 11% going to E gives E = 16+11 = 27
at count 2, not 32.  E's surplus over the 25% quota is then 2, so F
has 13+2 = 15 at count 3, fewer than A's 16, and F rather than A would
be eliminated next. ]

Cambridge still uses this system.  However, it has a quirk; when a
candidate is elected, a number of actual ballots which suffice to elect
that candidate are removed, instead of all such ballots being
proportionately reduced.  As a result, a recount of the same ballots
could lead to a different set of winners.

-- David Grabiner <grabiner@math.harvard.edu>