# [EM] Your PAV vs STV example revisited

Forest Simmons fsimmons at pcc.edu
Fri Jan 19 17:43:50 PST 2001

```Craig, I still haven't checked all 56 combinations for the case where the
approval voters approve their top four of the eight candidates, but the
ones I didn't check probably wouldn't beat the ones I did.

The winners are ABF and BDE.  Each of these combinations covers 100% of
the voters with at least one representative (from their top four
preferences), as well as 75% of the voters with two representatives (from
among their top four).

Compare this with AEH which covers only 90% of the voters from among their
top seven choices, and only 45% of the voters with two representatives
from among their top four.

Of the tied PAV winners I would say ABF is better because it covers 95% of
the voters with at least one vote from their top three preferences, and
55% of the voters with two repsentatives from their top three choices.

Compare this with STV winner AEH which covers 90% of the voters from among
their top seven choices, and 20% of the voters with two representatives
from their top three choices.

The ordinary AV winner (based on top 4 preferences) would be BDF, which is
contained in the union of the tied PAV winners ABF and BDE, but has only B
in common with both of them, so PAV does give something different from AV,
yet keeps some candidates with widespread popularity.

I haven't carried out the full Condorcet calculation, but it looks to me
like it could give a three way tie to ABD, which is also contained in
the union of the tied PAV winners, again showing the close association of
the PAV winners with candidates having broad popularity.

Note that B is in both tied PAV winners, the ordinary AV winning circle
(whether based on top three or top four), and is in a tie for the
Condorcet winner (check me on this one), and yet STV threw it out.

If I'm not mistaken, a complaint about existing PR methods is that they

Do we have something worth pursuing here?

Forest

On Fri, 19 Jan 2001, Forest Simmons wrote:

> 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

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