[EM] a mitigating feature of AER
Kevin Venzke
stepjak at yahoo.fr
Tue Jan 27 21:08:14 PST 2004
I recommend AER (Approval-Elimination Runoff, or "Approval AV") for cases
where Schulze(wv) is considered impractical (due to the need for a pairwise
matrix) or unintuitive (as a decision-making process).
AER is like IRV, except that the elimination order is set in stone at the
beginning based on "approval" (non-last rankings, or based on a cutoff).
This allows AER to be monotonic (although it is at the expense of Mono-add-top
and the LNH properties). It also permits two counts of the ballots to
suffice, if N^2 tallies (as in a pairwise matrix) become acceptable.
I prefer AER to Approval because of its superior ability to respect
pairwise victories. It satisfies Majority (like IRV). It doesn't meet any
Condorcet criteria (even the weaker "gross" forms), though. For instance:
5 C>A
4 C>B
6 A
6 B
21 total
C is the CW, but in AER's search for a majority, C is eliminated, since
C has the lowest approval. A is elected. Note, though, that Approval
might not have done any better in this scenario.
Here is the mitigating feature I've come up with:
When the AER winner wins, he is the majority favorite on the remaining
ballots. A majority favorite can't be beaten pairwise by any candidate
remaining on the ballots (intuitively). Who remains on the ballots? We
know that every candidate with greater approval must still be on the
ballots (not eliminated), because if they had been eliminated, the AER
winner must have already been eliminated, which is nonsense.
That means a few things, all related. I couldn't settle on just one:
1. The CW can only lose to a candidate with greater approval.
2. The AER winner has either a pairwise or approval victory over every
other candidate.
3. The AER winner pairwise beats every candidate with greater approval.
It's easy to show that if the CW loses in AER, some CW>AERW voters must
have approved both candidates. Let "C" represent the number of voters
preferring the CW and approving only him. Let "C,A" be the number
preferring the CW but approving both. Suppose there are no C,A voters,
yet A wins.
If C beats A pairwise, then C > A + A,C
We know A must have greater approval, so A + A,C > C + A,C; or, A > C
But if A > C, C can't exceed A plus some other number, so the scenario
isn't possible.
I'm not certain if this is too trivial to be called a "guarantee." But
in any case I think this kind of reasoning might help convince even
Condorcet fans that AER is not an unnecessarily random or senseless
method.
Kevin Venzke
stepjak at yahoo.fr
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