[Election-Methods] [Election Methods] Strongest pair with single transfer (method)

Chris Benham cbenhamau at yahoo.com.au
Mon Apr 28 20:55:36 PDT 2008

```Kevin Venzke stepjak at yahoo.fr
Thu Jan 18 15:25:59 PST 2007

Hi,

Here's an attempt at a method that behaves well in the three-candidate
scenario with preferences based on distance on a one-dimensional spectrum.
I would call it "strongest pair with single transfer" or "SPST". It
satisfies LNHarm and Plurality, and doesn't suffer from the worst kind
of burial incentive. It also satisfies Clone-Loser I believe, though not
monotonicity.

My idea was to come up with a method that, in the three-candidate case
with distance-based preferences on a one-dimensional spectrum, could
elect the inner candidate in the absence of a majority favorite. I also
wanted to avoid truncation strategy (Approval, Condorcet), gross Plurality
failures (as under MMPO), and the sort of burial strategy where you give
a lower preference to a candidate whose supporters are not ranking your
candidate.

Definition:
1. The voter may vote for one first preference and one second preference.
2. The "strength" of a candidate, or pair of candidates, is defined as
the number of voters giving such candidates the top position(s) on
their ballots in some order. (This is as under DSC.)
3. A pair of candidates has no strength, if it includes any candidate
who is not among the top three on first preferences. (I don't like this
rule, but it's needed for LNHarm.)
4. If the strongest candidate is in the strongest pair, or stronger than
the strongest pair, then this candidate wins.
5. Eliminate the strongest candidate. The second preferences of his
supporters may be transferred to the individual candidate strengths of
the two members of the strongest pair of candidates.
6. Now, the strongest candidate in the strongest pair is elected.

examples:
40 A>B
25 B>C
35 C>B

Strongest pair is BC; strongest candidate is A. BC is stronger than A,
so A is eliminated and 40 preferences are transferred to B's strength.
B wins.

35 A>B
25 B>C
40 C>B

Here BC is again the strongest pair, but C is the strongest candidate and
wins immediately unfortunately.

This method is a lot like DSC, but never requires more than N^2 numbers
to be counted, whereas DSC requires 2^N if you keep track of every set.
The elimination doesn't create IRV's counting issues, since with only
two preferences taken we can just count them all.

The burial strategy works like this: Say it's A, B, and C, with B as the
middle candidate. A is expected to be the strongest candidate. Then
voters with the preference order B>A have incentive to instead vote
B>C. This is because if BC is the strongest pair, A will be eliminated
and hopefully transfer preferences to B. But if AB is the strongest
pair, A wins outright. As a result of this strategy, it is possible that
(despite the LNHarm guarantee) A voters would decline to give a second
preference to B, so that B>A voters can't count on the A voters to
give a second preference to B.

It is only possible to eliminate the first preference winner, due to
LNHarm. It's only safe to eliminate a candidate who was going to win.
Otherwise it could happen that voters have incentive to weaken a pair
involving their favorite candidate, in order to prevent an elimination
that causes the favorite candidate to lose to the second preference.

Limiting pairs to the top three FPP candidates is necessary for LNHarm
when there are more than three candidates. Otherwise it could happen,
say, that BC is stronger than BD is stronger than A, A is eliminated,
and then C wins. Whereas if BC were weakened and BD were strongest, A's
elimination might result in B winning.

Monotonicity can be failed when the winner is not the FPP winner, he
gets more preferences, changing which pair is strongest, and causing
the other candidate in the pair to win.

I ran some simulations to try to measure this method against others. When
the only ballot types are A>B, B>A, B>C, and C>B, this method is identical
to DSC. When all 9 ballot types are allowed, this method seems to be
strictly more Condorcet-efficient than DSC, although not by much.

I found that IRV is more Condorcet-efficient than either, except in
the scenario where only the four ballot types are permitted, and the
proportions of the B>A and B>C ballots are divided by 5. There IRV is
worse because it wants to eliminate B.

(With the four ballot types, IRV can elect B as long as B doesn't have
the fewest first preferences. That particular scenario is important to
me, though. DSC can elect B unless, say, the A>B faction outnumbers the
B>A B>C factions, and also the B>C C>B factions.)

That's it for now.

Kevin Venzke

Kevin,
Soon after you raised this I asked if this met any criteria not met by IRV.
You said "no", and that it fails Later-no-Help.

I thought it might be interesting if applied to the IRV last 3.

49: A
48: B
03: C>B

But now  I see that like DSC it disastrously fails  Dominant Mutual Third, electing
A in the example.

So IRV remains my favourite  LNHarm method

Chris  Benham

Get the name you always wanted with the new y7mail email address.
www.yahoo7.com.au/y7mail

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