[EM] Strongest pair with single transfer (method)

Kevin Venzke stepjak at yahoo.fr
Sat Jan 20 08:48:00 PST 2007

```Hi,

--- Chris Benham <cbenhamau at yahoo.com.au> a écrit :
> Kevin,
> My last message was a copy of one I sent to EM. You may want to

Oops. Thanks. Here it is below, with one change and one addition:

--- Kevin Venzke <stepjak at yahoo.fr> wrote:
> Chris,
>
> --- Chris Benham <cbenhamau at yahoo.com.au> a écrit :
> > Kevin,
> > Interesting. What (if any) harm would be done by
> > applying this to the three candidates remaining
> > after the rest have been IRV-style eliminated?
>
> You lose LNHarm, because your lower preferences can
> affect what competition your favorite candidate faces
> afterwards.

I thought this over and don't feel this explanation can be right.
If a given preference of yours will make the final 3 in IRV, then your
lower preferences have absolutely no effect on which other candidates
make the final 3.

I am inclined to think it makes no difference to LNHarm whether there
are only 3 candidates, or those 3 were acquired using IRV. That just
strikes me as hard to believe...

Actually, only the day before I posted the method description, did I
realize that LNHarm required the method to consider only the three
pairs involving the top three candidates. So I haven't put a lot of
thought into narrowing the field down to 3, because I wasn't thinking
it was necessary...

> > Is there any actual criterion that this method
> > meets but IRV doesn't?
>
> No. It fails LNHelp, which might count for something. And the burial
> scenario seems to have something working against its
> plausibility, in that if say A is expected to be the strongest
> candidate, it's likely that B is closer to C than to A. We might guess
> that the potential buriers, the sincere B>A faction, might not be very
> large. In that case it should be difficult for the A faction to know
> whether B voters are sincerely or insincerely planning to rank C in
> second.

Additionally, it's worth pointing out that if the burial strategy is
successful, it merely elects the CW as we would want the method to do
anyway. The downsides are just
1. that A voters might desire to defensively truncate despite LNHarm
2. that A voters do this, and CW supporters bury, causing the third
candidate to win.

> Woodall showed that LNHarm, Plurality, and Condorcet(gross) are
> incompatible. My approach for awhile was to weaken
> the latter two criteria and try to use a proof style to deduce what
> the results of my desired method would have to be in various
> situations. Then I would have to come up with a prescription of how
> to get those results (i.e. an actual method definition).
>
> I still think this approach is promising...
>
> Woodall's proof goes something like this:
>
> 3 A>B
> 3 B>A
> 4 C
>
> By symmetry B wins no more than 50% of the time.
>
> 3 A>B
> 3 B
> 4 C
>
> By LNHarm, B can't win with more than 50% probability. By this and
> Plurality, C's win odds must be above 0%.
>
> 3 A>B
> 3 B
> 4 C>B
>
> By LNHarm, C must still have positive win odds. But by Condorcet(gross)
> B must win.
>
> Here it seems like the most likely way to get past this is to weaken
> Condorcet(gross) such that in the last scenario, B doesn't get to win
> because B voters didn't create a majority defeat of one of the other
> candidates over the other.
>
> I don't know how to word this, though, or if it's possible anyway. In
> this scenario:
>
> 40 C>B
> 25 B>C
> 35 A>B
>
> I'm suspecting a LNHarm method can't elect B without at least having
> MMPO-style burial incentive. I'm almost certain it would have to fail
> Plurality.
>
> Kevin Venzke

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