[EM] (P1) tweak up of AV meth. using random walk may fail
Blake Cretney
bcretney at postmark.net
Sun Jan 23 19:16:19 PST 2000
(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.
Tell me if I understand the proposal. A particular fraction of the
vote is set as the quota. Any candidates who receive less than the
quota in first place votes are eliminated on the first round. If all
candidates receive less than the quota, the plurality winner is
chosen.
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.
The question is, what is the minimum quota necessary to meet P1?
To create a monotonicity violation, we need a situation like the
following
i A C B
j B A C
k C B A
For simplicity, I am going to use i, j and k as fractions of the
vote, and q will be the quota. i,j,k<.5
There will be a P1 violation in the following circumstance. B has to
win. Then, if some A-1st voters vote C B A, A will win. That would
be a clear violation of monotonicity, and therefore P1. This will
happen if in the original election, C was eliminated first, causing B
to win, but in the modified election, B was eliminated first, causing
A to win.
Since in the original election, B was not eliminated first off, it
must be that
j>=q
In the modified election, A and C must both survive the first round,
so there must be enough votes between i and k to allow both candidates
to survive quota.
i+k>=2q
Therefore,
i+j+k>=3q
1>=3q, remember that i+j+k=1
q<=1/3
So, either there is a more efficient way of creating a monotonicity
violation than the one I show, or a violation of monotonicity implies
a quota of <= 1/3. Anyone who wants a real proof will have to go
through all the possible ways monotonicity could be violated. I am
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.
Here I will go back to numbers of votes rather than fractions.
1/3v + 1 A C B
1/3v B A C
1/3v - 1 C B A
q=1/3
In this example, C will be eliminated and B will win. However, a
single vote transferred from A B C to C B A, will cause a 3-way tie.
This allows A a chance of winning. Its situation has improved, and
monotonicity has been violated. But if we want to avoid ties.
1/3v + 3 A C B
1/3v - 1 B A C
1/3v - 2 C B A
The winner is B.
But, transferring two votes from A B C to C B A.
1/3v + 1 A C B
1/3v - 1 B A C
1/3v C B A
A wins.
It's clear that as v increases, 1/3v-1 will increase towards 1/3, so
any quota less than 1/3 will cause a violation, without the need for
ties.
So, if we use a quota that says that candidates must have more than a
third of the first place votes, we can ensure there is no violation of
monotonicity, or of P1. Of course, I don't advocate such a method.
> This ought reduce
> the weight of Lord Alexander's criticisms of the method.
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?
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
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.
If A gets more support, that can reasonably mean that A should win.
---
Blake Cretney
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