[EM] British Colombia considering change to STV

[Raph Frank]

On Wed, May 6, 2009 at 1:24 AM, Dan Bishop <danbishop04 at gmail.com> wrote:
> The one thing you haven't mentioned is surpluses.  The
> symmetric-completion-compatible way of dealing with them is weight the
> ballot by the average of the retention fraction for the top-ranked
> candidates.  For example, given the ballots:
>
> 4: A>B
> 3: A=B
> 5: C=D
>
> and a quota of 4 votes, we'd have
>
> A elected with 5.5 votes (excess of 1.5) -> retention fraction = 1.5/5.5 =
> 3/11
> B, C, D not met quota -> retention fraction = 1

Right, assuming that voting strength is shared for the election step

4: A>B -> A(4)
3: A=B -> A(1.5) and B(1.5)
5: C=D -> C(2.5) D(2.5)

A: 5.5
B: 1.5
C: 2.5
D: 2.5

> So the A=B ballots would be weighted by (3/11+1)/2 = 7/11, so the
> re-weighted ballots would be:
>
> 4*3/11: (A>)B
> 3*7/11: B
> 5: C=D
>
> 3: B
> 5: C=D

That is kinda what I said:

3) A=B

is replaced by

1.5) A>B
1.5) B<A

1.5*3/11: (A>)B
1.5: B(>A)

gives

7/11*3: B

Ofc, your way works for more than 2 without an explosion of options.

Also, averaging the keep values doesn't keep the A total equal to the
quota.  Under Meek, the ballot isn't deweighted, some of the vote
strength is kept and the rest is passed to the next ranking.

However, in this case, it is some kept and the rest passed to the equal ranking.

Effectively with equal ranks,

1/N of the vote strength goes to each of the N equal choices

they keep part of it based on their keep factor

the rest goes to the remaining candidates at that level.

I think this gets applied recursively.

For example, assuming that the keep factors were

A: 0.1
B: 0.3
C: 0.7
D: 1

and the ballot was

(A=B=C)>D

How much vote weight goes to each candidate?

Clearly in 'round 1', they get

A:
Keeps: 0.1*0.33 = 0.033
Passes to 'next' rank: 0.297

Pass gets split in 2 and passed to
0.1485: B>C>D
0.1485: C>B>D

and so on