# No subject

Thu May 8 11:58:05 PDT 2014

```4, etc... are elected, in the 0-info Approval case, it is best to vote for
the candidates that are above average.

Initially, I was not satisfied enough with this and went on to imagine
comparing all the combinations of candidates in a slate using Weber's
equation instead of candidates, similar to what Mike Ossipoff did.  I more
or less came to the same conclusion as I did above except with a more
vague argument.

Say there is an Approval vote where to top 3 are elected.  I'll take two
slates of candidates AEF and BEF, and try to put them head-to-head.  In
order to do this, the sum of the utilities of the candidates AEF are added

(While reading about PAV, I noticed that Joe Weinstein posted a message in
Jan/Feb 2001(?) saying that for a slate of candidates, the utility of a
slate is not the same of the utilities of candidates in the slate all
added up.  However, in order to keep things simple, I'll assume this is
not the case.)

Notice that both slates are the same apart from the fact that one slate
has candidate A while the other has candidate B.  Therefore, when applying
Weber's equation to the slates, the candidates ELF on the left cancels out
the EF on the right.  All the equation would do is then A - B, which is
the same as just comparing two individual candidates rather than two
slates.

At this point, I was quite satisfied that from this I could conclude that
it is best to vote for the candidates that are above average in 0-info
Approval.  Though, with hindsight, I think it is quite a flimsy argument.

However, I could go on and compare different slates of candidates that
involves the candidate I want to find out how I should vote for:

AEF - BCD
:
:
ACD - BEF

Notice above the EF and the CD.  As I am adding/subtracting when I turn
the above into a one line ballot vector, the EFs and CDs cancel each other
out.  And I can go on.

:
:
:
:
ACF - BDE
:
:

Here, the same thing has happened.  However, this time it is with CF and
DE.  Admittedly, I don't think this is as good as the very first argument

Unless someone can come up with something "cunning", I think full-info
Approval strategy is nasty to work out.  I initially thought of finding
the probability of candidates coming 1st and from this get the probability
of coming 2nd, then 3rd, then 4th etc...

But I then thought about Weber's equation.

Wij = Pi * Pj * (Ui - Uj)

Pi is the probability that candidate i wins.  Pj is the probability that
candidate j wins.  Therefore, Pi * Pj is the probability that candidates i
and j both win (i.e. there is a tie).

Why not push this a bit further?  As Mike Ossipoff mentioned, the
probability of a slate of candidates winning needs to be calculated.
Again, I'll take the candidates we had before and say that there is an
Approval vote where the top 3 are elected.

To work out the Weber score by using Weber's equation, I'll use Pi as the
probability that acf win and Pj as the probability that bde wins.  These
are easy to work out.

Pi = Pa * Pc * Pf
Pj = Pb * Pd * Pe

The utilities are also easy to work out by adding up the utilities of the
candidates in each slate.  Now these values can be plugged into Weber's

That was the easy bit.  The hard bit is that I THINK this needs to be done
for every combination of slates.  In this case it is 120 times, if I've
got that correct.

Say the top 200 of an Approval vote will be used to fill in 200 seats.
The number of combinations here is extremely massive.  Also, there would
at least be literally hundreds of candidate probabilities to use.

Referring to posts a while back when there was a discussion about having
very simple voting methods so that hand counting would be easy for a
multi-seat election.  Though just plain top-N Approval seems susceptible
to a very large faction of voters voting for their own block of
candidates, causing them to be virtually guaranteed to be elected, I don't
think it would do too badly.

What I picture in my head is the (stereotypical) left-right wing spectrum.
In one situation, there would be just a single normal distribution
histogram in the centre.  When a seat is allocated, a square of the
histogram randomly flashes to represent where in the spectrum the
candidate stands.  As it is a normal distribution histogram, the flashes
tend to be concentrated in the middle of the spectrum.

This could be an advantage.  As most of the candidates tend to be near the
centre, therefore there will be less disagreements.  Therefore legislation
could be passed with less argument and therefore more speed.  To
compensate for the "narrowness" of thought of the people elected,
elections could held more often.

On the other hand, there may be a country where two normal distribution
curves would be best in representing who is elected.  One curve is on the
left of the spectrum while the other is on the right.

I must admit, I don't know how good these mental pictures are.

Thanks,
Gervase.

```