Donald wrote:
> There is the possibility that the Hare set of elected members
> will be different from the Droop set of elected members.
>
> When this happens, the question must be asked: `Which set of
> members is the correct set of members to hold office?'
>
> The answer is the Hare set because the Hare set is more
> proportional than the Droop set.
To illustrate this point, I offer the following example:
In an STV-PR election with 120 voters for five places, the HARE quota would be 24
votes (= 120 / 5).
Suppose the election is contested by two parties (R and S) with three candidates each
(A, B, C).
With first preference votes as shown, Ra and RB would be elected at the first stage.
Stage 1
RA 24 elected SA 19
RB 24 elected SB 19
RC 16 SC 18
64 56
Of the four continuing candidates, candidate RC has the fewest votes and so is
excluded. No matter
what preferences are marked on those papers, SA, SB and SC will then be elected.
Stage 2
RA 24 elected SA 19 elected
RB 24 elected SB 19 elected
RC 16 excluded SC 18 elected
64 56
Applying the Hare quota, supporters of the larger opinion group (party), who
constitute an absolute
majority, elect only two candidates, while the smaller group elects three.
For this election, the DROOP quota would be 20 = (120 / (5 + 1)). With first
preference votes as
before, RA and RB would be elected at the first stage, each with a surplus of 4 votes
above the
quota.
Stage 1
RA 24 elected SA 19
RB 24 elected SB 19
RC 16 SC 18
64 56
If we (reasonably) assume that these surplus votes transfer to the remaining candidate
of party R,
candidate RC will be elected, and SA and SB will take the two remaining places.
Stage 2
RA 20 elected SA 19 elected
RB 20 elected SB 19 elected
RC 24 elected SC 18
64 56
The outcome with Droop quota is that the supporters of the larger group elect three
candidates and
the smaller group elects two.
Perhaps Donald should reconsider the assertion he made above.
Perhaps Donald should answer his own question: "You do support proportionality, don't
you???"
James
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