[EM] Meek-based maths for equal ranked PR-STV

Raph Frank raphfrk at gmail.com
Thu May 7 02:33:48 PDT 2009


On Thu, May 7, 2009 at 4:57 AM, Dan Bishop <danbishop04 at gmail.com> wrote:
> 1: A>B
> 1: B>A
> 9: C>A>B
> 1: C>E>F
> 9: D>B>A
> 1: D>E>F
> 3: E>F
> 4: F
> 1: A=B

Under approval based elimination, the process would be

Round 1:
Quota: 7.5
C+D elected

Round 2
A:
Elect: 3.75 (A=B is worth 0.5)
Elimination: 4.25 (A=B is worth 1.0)
B:
Elect: 3.75
Elimination: 4.25
E: 3.5
F: 4

Anyway, if E is eliminated, the F's score goes to 7.5.  Technically,
that doesn't count as being elected.

The next step would then be randomly eliminating A or B, and then the
other will also be at 7.5.

One of them would then be randomly eliminated.

That gives a result of

C+D and
0.25: A
0.25: B
0.50: F

The key point is to allow A+B to work together for the elimination
stage so that E is eliminated.

If the voter had voted A>B, the results would be
0.5: A
0.5: F

and for B>A
0.5: B
0.5: F

> gives a worse outcome (C+D+F) to the A=B voter than if they had voted either
> A>B (C+D+A would win) or B>A (C+D+B would win).]

I am not sure that is actually correct.

Using A>B

Round 2 (after C+D election)

       A  4.25
       B  3.25
       C  7.5
       D  7.5
       E  3.5
       F  4

B is eliminated and A gets all of B's votes

A: 7.5
B: 0
C: 7.5*
D: 7.5*
E: 3.5
F: 4

The issue her is that the quota isn't being used correctly.  The Droop
quota is 8 votes and if the Hagenbach-Bischoff quota is used, then the
rule is that a candidate must exceed the quota to win (to mirror the
fact that the Droop quota adds 1).

The example assumes that if E is eliminated first, then F wins, and if
A (or B) is eliminated first then the other wins.  Thus, elimination
order matters.  Elimination order can matter, but I don't think it
matters here.

Also, I think my suggestion to have an election and an elimination
score resolves that problem anyway.  If you vote (A=B), your vote is
used at full strength for keeping both A and B in the election.

There is an additional issue with how exactly to do tie breaking in
PR-STV.  My personal favourite is to pick one ballot and reduce its
weight to (1-delta).  In effect, the winning candidate is the one who
receives the least voting weight from the drawn ballot.

Given the low probability of a tie, this is probably not a big issue,
and a coin toss would likely do.



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