# [EM] (P1) tweak up of AV meth. using random walk may fail

Craig Carey research at ijs.co.nz
Mon Jan 24 09:08:48 PST 2000

```In the this message, it shall be assumed that there is only one winner,
and also that the votes all sum to 1. (They have been rescaled).

Here's a definition of this maili list's 1 winner tweaked-AV method:

AV0 preferential voting method

For stage=1,2,3,... loop
If the number of candidates remaining is <= 1 then quit the loop.
Sort the remaining candidates by their 1st preference votes.
Identify those candidates which have less 1st preference votes than
the quota, and identify the candidate with the least 1st preference
Eliminate all the candidates that did not meet the quota and the
end loop.

The quotas are not defined in AVO

I shall call this method, where the quotas are undefined, the AV0 method.

The problem is, find the quotas that give the best method.
It seems to be just the sort of thing the UK electoral reform group
(I forget the name of it), should have done and finshed many years or

----------------------------------------------------------------------
AV1 preferential voting method

AV1 is AV0 except that the quotas are defined.

Quota =
1 / (number of remaining candidates)

Hence, if there are n candidates, then the quota at the 1st stage is
1/n (e.g. 1/10 if there are 10 candidates). The quota for the 2nd
stage =
1 / (n - 'number of candidates eliminated in 1st stage')

----------------------------------------------------------------------

Note that with the AV1 choice of quota, it is not necessary to explicitly
eliminate the candidate with the least 1st preference votes, since the
quota does that.

If there are 3 candidates and 1 winner, then the IFPP method is identical
to this AV1 method.

I have no doubt that this AV1 method is major improvement over the
Alternative Vote (i.e. the 1 winner STV method).

It seems to me that there are two unsatisfactory areas with AV1:

(i) Proving AV1 satisfies (P1) won't be simple most probably. I.e. it
it is not known to have good theoretical properties.

(ii) The quotas to drop into STV are missing. It can't be hard to make
some up for STV. Maybe the quota could be the smallest that always
eliminates at least one candidate (when at least one needs to be
eliminated). Anybody want to write in and say what those quotas
should be? (the derivations from first principles can come later).

(The names AV0 and AV1 can be renamed.)

-------------

At 16:16 24.01.00 , Blake Cretney wrote:
>(slow)
>Date: Mon, 24 Jan 2000 03:16:19 +0000
>Content-Type: text/plain; charset="iso-8859-1"
>
>Craig Carey wrote:
>
>>
>> Topic: Finding Quotas for Losers in the AV method using Random Walks may
>>  run into a slowness-of-computers problem.
>>
>> I tested a method to improve the Alternative Vote method (STV with one
>>  winner). I would have added quotas for losers that sometimes caused
>>  more than one candidate to be rejected at each stage.
>
...

>Since FPP passes P1, a quota of 100% would work, since this makes the
>method equivalent to FPP.  A quota of 0% obviously wouldn't, as this
>would be equivalent to AV.

A quota of 100% does not make the general AV1 method turn into FPP.

>
>The question is, what is the minimum quota necessary to meet P1?
>

I don't imagine that that problem is so easy to solve.
It might be easier to ask: what quotas would people best agree on?.

A 1 candidate AV1 election would have quota of 100%. Hence a winning
candidate can have a count of 1st preference votes that touched a
quota.

I deleted what Mr Cretney wrote because it is really hard to follow.

...
| That says the alteration is (AC {B+)--(CB {A+). That alters (AC)
| 'at and after' A, so (P1) is violated since A becomes a winner in
|  the 2nd.
| (The alteration satisfies (P1) wrt. candidate C, and it is not an
| allowed 'at and after' alteration wrt. candidate B).

Here's an example of where the AV method fails (P1) and AV/IFPP does
not

10  A
6  B A
5  C B
AV winner = B,    i = 10, j = 6, k = 5, q = 7
AV1 winner = A

8  A
6  B A
7  C B
AV & AV1 winner = A,  So the alteration = (AC {B+)--(CB {A+)

...
>not going to do that.  However, from experience with the method, I am
>confident that there are no significantly different ways to violate
>monotonicity, at least for 3 candidates.  If we set a quota higher
>than 1/3, we will eliminate all greater numbers of candidates anyway,
>so the possibility of more candidates does not alter the conclusion.
>
>So, a quota greater than 1/3 is safe.  But what about 1/3 or less.

A quota greater than 1/3 is not safe.

Reword Mr Cretney's arguments...

The following alteration example does not satisfy (P1) if is it just
the AV method. If the winner is picked using the AV0 method, then
the quota is made to be large enough to make the alteration satisfy
(P1).

1/3 + 3/v  A C
1/3 - 1/v  B A
1/3 - 2/v  C B
AV0: A wins, if (1/3-1/v < q), otherwise B wins

1/3 + 1/v  A C
1/3 - 1/v  B A
1/3 + 0/v  C B
AV0/AV:  A wins

May need to have all terms positive, so (1/3-2/v > 0) = (1 > 6/v),
so (v>6).
To make A win the 1st, and not B, must have (1/3-1/v < q), so
let v tend to infinity and the smallest quota has to be 1/3 or
more.

The quota can't be above 1/3 because that could reject all
candidates, for example, when candidates have the 1st preference
counts of 1/3 after being perturbed by (3/v, -1/v, -2/v), and v
is large enough to take all three under the quota.

So presumably that quota for first stage is 1/n.
The argument about (P1) only holds for 3 candidates.

...
>
>> This ought reduce
>>  the weight of Lord Alexander's criticisms of the method.

Might not have been true. Lord A. knew it was a bad method but
didn't seem to be that clear. The UK parliament has reformers

>
>Here's a quote from Lord Alexander (speaking about AV)
>
>> I find this approach wholly illogical. Why should the second
>> preferences of those voters who favoured the two stronger
>> candidates on the first vote be totally ignored and only those
>> who support the lower placed and less popular candidates get a
>> second bite of the cherry? Why, too, should the second
>> preferences of these voters be given equal weight with the
>> first preferences of supporters of the stronger candidates?
>
The 2nd sentence is hard to understand and it does not appear to
really be obviously true. It might be about a problem that appears
in the 3 candidate election that reapears in all others.

Regarding his last sentence, he is getting close to saying votes
ought be wasted. Ideally the 100th preference should if possible,
have the same power and influence as a first preference, contrary
to the nature of STV.

>He states further,
>
>> In addition, as all experts on electoral systems have
>> acknowledged, AV can operate haphazardly depending upon the
>> ranking of candidates on first preference votes
>

Haphazard means corners that protrude too much; making it compliant
with (P1) would tilt slopes and reduce the surface area of all the
boundaries to win-lose regions.

>If you agree with this, what do you hope to accomplish by finding a
>method that is part way between AV and FPP?  If the basic process
>employed by AV, successive elimination, is irrational, as it appears
>to be, why would you favour a method that employs this process?
>
>As well, I have never understood why you think that a method that
>says a change from
>C B A
>to
>C A B
>can make the winner change from C to A, is not behaving reasonably.

The alteration is: (C+B)-(CA+).

>If A gets more support, that can reasonably mean that A should win.

It doesn't matter what happens to A because the alteration is ruled
out by 'truncation resistance SPC' (and (P1)) because of what happens
to candidate C.

So the AV method doesn't always get the wrong winner through
emphasizing the wishes of supporters of a string of dead losers.

>---
>
>Blake Cretney

Mr G. A. Craig Carey,  research at ijs.co.nz,   Auckland, New Zealand
Snooz Metasearch: <http://www.ijs.co.nz/info/snooz.htm>