[EM] Proportional Representation via Approval Voting (fwd)
Forest Simmons
fsimmons at pcc.edu
Fri Jan 19 12:56:44 PST 2001
Craig, without trying all 56 possible subsets of size three I did verify
that under PAV the combination ABH wins against AEH, and even more so when
we assume that the voters approved half of the candidates.
I think this is reasonable for the following reasons.
The two combinations disagree only on whether B or E should be in the
winning circle.
It is true that B and E tie in the rankings; the voters are fifty-fifty on
that question. But the rankings where B beats E have greater separation
between the two than where E beats B. So although B and E have a Condorcet
tie, so to speak, B soundly beats E in a Borda Count comparison.
Of course, this by itself doesn't tell us which would be better, we need
to know which one better compensates for the two 10% factions of the
population not covered by the {A,H} combination.
In this regard, B is first choice of one of these factions and third
choice of the other, while E is second choice of one of the factions and
last choice of the other.
In other words, if B is included in the winning circle, then 100% of the
voters have representatives from their top three choices. If B is
replaced by E, then 10% of the voters do not have a representative from
their top seven choices.
Which do you think is better?
Forest
On Thu, 18 Jan 2001, LAYTON Craig wrote:
>
> I made up a fairly random (ordinal ranking) voting pattern with 8
> candidates. I assure you, it was the first (and so far only) example I
> tried, so it isn't contrived in order to prove a point. The eight
> candidates are ranked by an electorate of 100 voters in the following way;
>
> 30 A>B>C>D>E>F>G>H
> 10 B>F>G>D>A>H>C>E
> 5 C>H>D>F>G>A>B>E
> 5 D>B>A>H>C>E>G>F
> 15 E>D>A>F>H>B>G>C
> 10 F>E>B>G>A>D>C>H
> 5 G>A>E>B>H>C>D>F
> 20 H>G>F>E>D>C>B>A
>
> There are to be three winners.
>
> In STV with a droop quota, candidates A,E,H are elected.
>
> In PAV I assumed that every voter's first three choices were approved. Using
> the divisors in Michael Welford's explaination, candidates A,B,H are
> elected.
>
> The results varied quite a bit between the two systems. Although in STV,
> A,B,H was very close to the elected combination, in PAV, A,E,H was not
> (there are at least two combinations with a significantly better score)
> I then invented an ad-hoc formula for assigning utility values to the
More information about the Election-Methods
mailing list