[EM] Single-candidate DMTBR idea

Kristofer Munsterhjelm km_elmet at t-online.de
Sat Mar 12 16:42:02 PST 2022

```On 3/12/22 12:15 PM, Kevin Venzke wrote:
> Hi Kristofer,
>
>>> Tricky, to reduce a scenario to this state without breaking mono-raise.
>>
>> We could of course just stitch something together, e.g. if there're
>> exactly two candidates above 1/3 fpp, elect the candidate who pairwise
>> beats the other, otherwise just elect the Plurality winner. This should
>> be monotone because raising A doesn't harm A when A has >1/3 fpp, and
>> raising B to >1/3 fpp gives him a second chance against A (if B beats A
>> pairwise).
>
> Yes, I realized after posting that that would probably work, and was racing to
> prove it. So you can have any number of candidates, but you can only have a
> runoff if two candidates each have 1/3+ FPs.
>
> This means we have a 4+ candidate method that satisfies Mono-raise and both LNH,
> and is not just FPP. So this should be an answer for Craig Carey here. I wonder
> if he was aware of it, or what he would have thought about it.
>
>> Actually, now that I think about it, I think the rule I provided
>> combined with Condorcet *is* fpA-fpC (disregarding ties for now).
>
> I think so, yes.
>
> I tried to compare C//(this IFPP expansion) with fpA-max(fpC). Note that the way
> I define the latter doesn't block a Condorcet loser from winning. But it seems
> like this method is better than using the IFPP expansion, as the overall burial
> incentive is less.
>
> Both methods appear to satisfy Mono-raise, Plurality, and DMTCBR.

Could you try to do Smith,(method) for both of these? That may be closer
to comparing apples to apples - it should keep the Condorcet loser from
winning, at least.

I'm not quite sure which method you mean by "this method", but if
fpA-max(fpC) is better than the plurality-runoff one, that would also
make sense as it's less obviously stitched together.

In passing, I could also say that if you have a pairwise tied two-set
(weak DMTC?), e.g. something like

33: A>B>C
1: A>C>B
8: B>A>C
32: B=A>C
26: C>B>A

then breaking the tie by first preferences should preserve both
mono-raise and DMTCBR. This is unlike fpA-fpC, which would just call
this a tie.

That would mean that if the smallest DMT set has >2/3 support and
consists of two candidates, then breaking the tie by first preferences
within that coalition should be okay. While that's not very useful (it
can only happen with an exact tie), maybe the pattern could provide some
ideas about how to extend DMTCBR to DMTBR.

>> It's not very elegant: the seams are very obvious. But perhaps elegance
>> can come later... or perhaps it will be induced by turning DMT candidate
>> BR into full DMTBR.
>
> Incidentally, do you know if there is a DMTBR (not just DMTCBR) satisfaction
> proof known for any method? Some form of C//IRV maybe, but it doesn't seem like
> an easy thing to show.

IRV passes both LNHs and so burial has no effect. But to make the DMTBR
proof more explicit: there are two possibilities. Either two candidates
from the innermost DMT set last until the final round, or one of them
does and the other is eliminated. They can't both be eliminated because
when all but one candidate is, the remaining candidate has >1/3 fpp and
so will be protected from elimination until the final round. And in the
final round, whoever beats the other pairwise wins.

The DMT criterion says that a candidate from the innermost DMT set
should be elected. Suppose first that the innermost DMT set is two
candidates or more. Then two of these will be preserved until the final
round, and no matter who wins, the DMT criterion is satisfied. If there
is only one candidate (DMTC), then likewise that candidate is preserved
until the final round, and since by definition he beats the other one
pairwise, he wins.

DMTBR is satisfied because if you prefer A to W, then you're going to
rank A higher than W. If you now try to bury W under C, your ballot will
be of the form ...>A>...>C>...W. But the part of your ballot that gives
weight to C relative to W isn't exposed until A has been eliminated, so

(For IRV this holds even when C is in the innermost DMT set, but this
isn't true for Condorcet methods.)

I'm of the impression that if method M passes DMT and DMBTR, then
Smith,M and Condorcet//M also do. Obviously this is true for DMTCBR
because the DMT candidate is both the Condorcet winner and the winner of
method M, so engineering a cycle won't switch the winner away from the
original CW to someone else.

For Smith,IRV (and Condorcet//IRV even moreso), I need to show that when
the IRV winner is some candidate inside the innermost DMT set but this
candidate is not the CW or not in the Smith set, then the voters who
prefer the IRV winner to the combined method's winner can't change the
latter to the former. And it's not completely clear to me just how to do
that. Maybe I'll find something out after exploring a bit with my linear
programming solver, but I feel there's a simple argument that I'm missing.

(I think the LCR scenario shows that full burial immunity is impossible.
But the innermost DMT set in that election is {L,C,R} so there's nobody
to bury *under*.)

>>>> Does it also apply to the generalization where you just take the two
>>>> candidates with the most first preferences? I'm not sure.
>>>
>>> In the three-candidate case, electing the pairwise winner between the top two
>>> candidates is basically IRV.
>>>
>>> Without the 1/3 limit it could happen that the FPW gets more votes and changes
>>> who the second place candidate is. He might not beat the new one.
>>>
>>> This is interesting though. The "obvious" way to expand IFPP to many candidates
>>> is to eliminate candidates with a below-average vote count. But it seems like
>>> the 1/3 rule was the important thing, as it's what enforces that always either
>>> one or two candidates are eligible to win, and these candidates can't be harmed
>>
>> Yes, that also explains where the "third" in dominant mutual third comes
>> from. Like with Droop proportionatliy, it's the smallest quota so that
>> only two candidates can exceed it. And that would also suggest that (at
>> least by this approach), third is the best we can do; there's no, say,
>> dominant mutual quarter for Condorcet.
>
> Yes. I've been sitting here trying to get a 25% rule to "work" (defining that
> very generously), but there is a lot of trouble regulating who is allowed to
> benefit from various vote changes when three candidates are eligible.

Since the trick about 33% is that you can only have two, perhaps there's
an impossibility proof along the lines of IIA, that you have three >1/4
fpp candidates and no matter who wins, it's possible to make someone
else the CW through burial?

I'd have to think more about it. A method like plain IRV has no problem
ensuring burial immunity for such a group because it doesn't care about
burial at all. So the problem must be induced by Condorcet compliance
itself.

-km
```