With this mistake removed, this is my proposal again:
Ranked ballots, equal preferences ok. Count the first preference votes. .If any candidates have a Droop quota they are elected, and then reduce the values of the ballots which have elected members by an amount which sums to a Droop quota.
If more than one place remains unfilled, proceed to to the second round. Add the second preference votes to the first preferences (based on the value of the ballots after the any reductions that were made the previous round). If this gives any candidate a Droop quota, then elect the candidate with the highest tally. If there is a tie, then elect the tied candidate who had the bigger tally at the last round, if still tied then the round before that if there was one, otherwise the Condorcet winner of the tied candidates based on the ballots after the most recent devaluations. Reduce the value of the ballots that elected this winner by an amount that sums to a Droop quota. If there is still more than one place unfilled and if after the latest devaluing of ballots any candidates have a Droop quota, then elect the one with the highest tally (same tie-breaking proceedure) and so on.
If there is more than one place to be filled, then add the third preference votes to the tallies of first and second preferences and if that gives any candidate a Droop quota, then as before the candidate with the highest tally is elected and so on.
If proceeding in this way leads to the situation where there is one and only one more place to be filled, then based on the ballots after all the devaluations elect the Condorcet Winner.
A more simple, pure GB-style method would be to fill the last seat by Generalized Bucklin (GB) as well. In the following examples I will give both results.
This example is lifted from a June 2002 UK Electoral Reform Society article "Sequential STV - a new version":
2 seats, 5028 votes, 5 candidates.
1248: A>B>C>D>E
1236: B>C>D>A>E
1224: C>D>B>A>E
1212: D>B>C>A>E
36: E>A>B>D>C
36: E>C>D>B>A
36: E>D>C>B>A
Droop quota =1676.
First round: A: 1248 B:1236 C:1224 D:1212 E: 108
No candidate has a Droop quota, so the second preference-level votes are added, to give
Second round: A: 1284 B: 3696 C: 2496 D: 2472 E: 108
If any candidate has a quota, elect the one with the highest total. Here 3 candidates have a quota, B has the highest total so is elected. All those ballots which helped elect B, i.e. those that give first or second preference to B, are now all reduced in value by a Droop quota, so that they now sum 3696 - 1676 = 2020. So for example with the election of B and the subsequent devaluations, the 1248 ABCDE ballots become 1248/3696 x 2020 = 682 ACDE ballots. So after these devaluations the votes
now are :
682: A>C>D>E
1900 :C>D>A>E
662: D>C>A>E
36: E>A>D>C
36: E>C>D>A
36: E>D>C>A
On these ballots C still has a quota and so is elected, making the result BC (agreeing with both plain and Sequential STV).
The second example is taken from the same source, and is the same as the first except that all those voters who ranked E last now rank E second. The first round is the same.
First round: A: 1248 B: 1236 C:1224 D: 1212 E:108
No Droop quota, so add the second preference-level votes:
Second round: A:1248 B:1236 C:1260 D:1248 E: 5028
E is first or second on all the ballots! So E is elected, and there is now no point in devaluing ALL the ballots. On the original ballots, disregarding E, B is the CW and also the Bucklin winner so both the PR methods elect E and B.
Sequential STV also elects EB, but plain STV elects BC as before.
Chris Benham.
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