[EM] Proportional representation through Bucklin-STV/Asset
raphfrk at gmail.com
Mon Mar 29 03:13:19 PDT 2010
On Mon, Mar 29, 2010 at 4:02 AM, Abd ul-Rahman Lomax
<abd at lomaxdesign.com> wrote:
> I'd like to get specific comment on this method. I use the Hare quota,
> generally, with Asset methods, because it provides natural consequences for
> inability to compromise. The loss of a seat is not particularly harmful if
> this is a large-member election.
In practice, if there is a reasonably large number of voters, this
effectively means that at least 1 seat will be unfilled and the Hare
quota with one seat guaranteed to be unfilled is pretty much
equivalent to the Droop quota, but with 1 fewer seat.
In practice, the optimal quota, in terms of maximising representation
would be somewhere between the Hare and the Droop quota.
With 9999 voters and 5 seats, the Hare quota is 1999 votes (assuming
rounding down) and the Droop quota is . This means that all 5 seats
can be taken by 9995. Thus, if even 5 voters refuse to compromise,
then only 4 seats can be filled and so, with high probability, only
7996 will be represented. The Droop quota is 1667 and if that is
used, then 5 seats would mean that 8335 voters are represented.
This means that the Droop quota is likely to result in higher representation.
However, an alternative would be to use something like
0.75*Hare + 0.25*Droop = 1917
Filling all 5 seats would require 9585 votes. This could be
accomplished even if there are 414 holdouts.
Obviously, there is a trade-off. Maybe the rule would be
a*Hare + (1-a)*Droop
with a being decided by the previous election. If all 5 seats were
filled, then a can be made slightly larger. Otherwise, it is reduced
> If it is an election of a limited number of
> members from each district (which means loss of proportional representatin),
> then I presume a Droop quota would be used, because gaining a full slate
> would be important. On the other hand, one could use the Hare quota for
> district elections, then allow the Asset electors with votes remaining from
> a district election to amalgamate across the entire legislative
> jurisdiction, thus providing small-minority representation state-wide. I
> like that, isn't it interesting? District elections, but no loss of minority
Sounds reasonable. However, using Hare nationally would still mean
that there is 1 unfilled seat.
It also depends on how you calculate the Hare quota in each district.
It could be calculated nationally after all the votes are counted, or
calculated locally based just on the votes in that district.
Calculating it nationally would mean that each seat is equal to the
same number of voters. However, calculating is locally would make the
system easier to manage.
Also, if the quota is different between districts, then it creates an
incentive to game the system, by electors choosing which quota to use.
> 1. Q = V / N. (This can be done with the Droop quota to make it more
> deterministic. I oppose it for reasons I won't detail here.)
> 5. Because there have been no eliminations, all elections so far can be seen
> as rigorously correct and fair.(skipped original sections)
So this is just standard PR-STV, up to elimination?
> 10. The ballots are now treated as Bucklin ballots. The second rank is
> counted. ("Second rank" means "second active rank.") Seats are assigned
> whenever a candidate, in a round of counting, gains a quota of votes, and
> those ballots are devalued accordingly. In this case, an elected candidate
> might be in a lower position on the ballot. The candidate is marked as
> elected, but the higher position candidate remains active, and may attain
> election through votes from other ballots, to the extent that any voting
> strength is left. As a winner is found, any ballots counted for that
> candidate are devalued as before.
Since this method would be pretty difficult to hand count, it would
probably be worth using Meek's method.
- 2nd round
- A + C are elected
- ka = A's keep value
- kb = B's keep value
B and D would get a vote equal to
Effectively, assuming the nth (Bucklin) round, you scan for the first
n non-elected candidates on the ballot and they all get votes equal to
what the last of those candidate would get, if all the other remaining
candidates had a keep value of 0.
A gets ka
B would get 0
C gets (1-ka)*kb
D gets (1-ka)*(1-kb)
Thus B and D get (1-ka)*(1-kb).
This may not be entirely fair.
For example, if C is elected before A, then that ballot will not
actually help getting C elected. However, once the process moves to
round 2, then it will have to provide a share.
be treated differently than
once both A and C are elected.
> (There is another choice here, where a voter
> has added additional ranked candidates, and the first position has been
> elected, but my opinion is that it dilutes the power of the Asset method,
> which is to make the *most trusted candidate* the effective proxy for the
Another option would be to allow the voter to specify a proxy rather
than having to choose their first choice.
However, that reduces the incentive for them to vote first choice for
a trustworthy candidate, rather than an well connected candidate.
> I believe this is monotonic.
I am not sure. However, the lack of eliminations should help with that.
In another thread, Kristofer conjectured that if a method meets quota,
then it must automatically fail monotonicity.
This method might not meet the Droop proportionality criterion, as it
> The voter can safely vote for one candidate, or can vote for more than one.
Right, it allows the voter to choose between PR-STV (ish) and asset voting.
> I think this is precinct summable.
I don't think so, especially if there are many ranks. It would suffer
from the same difficulties as PR-STV.
> I'm not familiar with Proportional Approval Voting, but I'd guess it is like
> this, one rank, but no Asset, so it's deterministic and would use a Droop
PAV has 2 forms. The easiest is the sequential version (which is
basically reweighted score voting, but with only 2 ratings).
A ballot is deweighted by
[k + (number of approved candidates elected)]
k = 0.5 gives Sainte Lague
k = 1.0 gives d'Hondt
The first round selects the approval winner. All ballots which
approved that candidate are then deweighted by 1/(1+1) = 0.5 (assuming
This is precinct summable.
The standard PAV doesn't operate in rounds.
Each voter's "happiness" with the result is assumed to be
1 + 1/2 + 1/3 + ..... + 1/n
where n is the number of candidates elected that the voter approves of.
The result which gives the highest total happiness is selected.
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