Hi Forest, two replies in one:

I call this BTPA. (But I used beat-or-tie, not strictly beat.) It is quite different from BTP or BTA, and appears extremely similar to Condorcet//DSC, especially with 3 candidates. For my taste this is not very good. It's also pretty close to fpA-fpC though, with 3 candidates.

Basically with BTPA your score as a candidate is capped by your implicit approval, and you can be deprived of it by weaker candidates who beat you pairwise. In contrast it seems to be important to BTP's properties that candidates can score off of ballots that didn't vote for them, which only voted for candidates that were beaten by them.

Also, it seems to me to be an advantage of BTP over BTA or BTPA that you don't need any implicit approval concept.

Regarding HBH and Weinstein, I'm having a lot of trouble getting this to work reasonably. For one thing, I can't find a way to interpret single-round Weinstein DSV Approval so that it is monotone. Also, if approval can extend into the bottom rank, it violates Plurality a lot. I can disallow that, but this creates a lot more monotonicity issues.

(One could suggest that the approval cutoff be placed prior to the rank that exceeds 50% of the random ballot odds, but this doesn't seem workable because it means a majority favorite can't receive his votes.)

Here's a Weinstein mono-raise example for your consideration.

0.311: D>C>A>B approves DCA
0.270: A>B>C>D approves AB
0.186: C>B>D>A approves CBD
0.143: B>D>A=C approves BD
0.087: B>D>A=C approves BD

D wins. Now change the 0.143 bloc to vote D>B at the top instead of B>D. I find that this causes the .311 bloc to stop approving A (since D carries more weight now), and the .270 bloc to start approving C (since B carries less weight). So now C wins.

Kevin

On 24/12/2020 22.19, Kevin Venzke wrote:

DSV tends to be nonmonotone in general: in a way, IRV is Plurality DSV.
Brams made a simlar point about a ministerial apportionment method: the
greedy approach could make someone wish they'd been asked to pick a
position later. He solved the problem with a negotiation mechanic: if
party X wanted to switch a minister seat with party Y and both agreed,
then it would be done. So some kind of either fix to get out of a local
optimum, or lookahead to avoid getting into that optimum to begin with,
seems to be needed.

About fpA-fpC, one way to get an idea of what a four-candidate fpA-fpC
method may be like could be to investigate elections where clone
independence demands that some candidate wins (e.g. you can get to the
election by cloning A, and B wins in the original three-candidate
fpA-fpC election; then A must win in the new election).

One of the problems of extending fpA-fpC is that there's no obvious
cloneproof way to do first preferences (apart from DAC/DSC). Perhaps
such "clone continuations" could give some ideas.

(But fpA-fpC could also be an uncontinuable dead end. I'll have to check
when I get more RAM to run my manipulability simulations on.)

Kevin,

Thanks for holding my feet to the fire!

...Weinstein approval comment in line below. We seem to understand it
differently in two ways...

On Thursday, December 24, 2020, Kevin Venzke stepjak@yahoo.fr wrote:

But .311+ .186 +.270 > .50 so only D & C can be approved. "the total
probability of approved alternatives on a ballot cannot exceed 50 percent"

But .186 +(.143+ .087) + .311 > .50 so only C and B can be approved.

That said, it seems like the approval cutoff rising on a faction where D
was approved and B not, could lead to D falling out of approval, etc.

I'm starting to think of the points awarded as a kind of updating of
approvals in light of the pairwise information; an alternative that beats
every alternative ranked above it is a threat to be reckoned with, so
perhaps the new info might justify lowering the approval cutoff adjacent to
the threat ... below it if the threat is ranked ... above it otherwise.

Running with this idea ... run one pass of a bubble sort from the bottom
ranked (but not truncated) alternative upward (Llull applied to the ballot
order of the ranked alternatives). This Llull winner becomes the
(inclusive) approval cutoff.

A refinement of this idea starts with a (tentative) explicit approval
cutoff.  If the Llull winner falls below the tentatve cutoff, then the new
cutoff is immediately above the Llull winner.  In every case the new cutoff
is obtained by moving from the tentative cutoff to a position immediately
adjacent to the Lull winner (stopping as soon as adjacency has been
achieved).

This seems to be taking us back to Rob LeGrand ...

Joyeux Noel!

Forest

Hi Forest,

Real quick:

Le vendredi 25 décembre 2020 à 00:15:20 UTC−6, Forest Simmons fsimmons@pcc.edu a écrit : 

I considered that initially. But what about majority favorites? They wouldn't receive any of their top rankings as approval.

I could see permitting a method that allows ER to perform the transition in two steps. However, for Mono-raise, two steps is strictly a harder test. If you change the winner taking a single step, you will still change the winner when taking two steps to get to the same place.

Kevin

Hi Kristofer,

Le jeudi 24 décembre 2020 à 20:56:20 UTC−6, Kristofer Munsterhjelm km_elmet@t-online.de a écrit :

I took a look at this problem. You can use a different route to get to the same 3-candidate scores, while trying to make sense of what the fpA-fpC metric represents. I think it's a measure of doubt as to whether the A>B win is genuine. An A-top ballot is definitely a genuine expression of A>B. A C-top ballot may not be. A B-top ballot is irrelevant. (A C>A=B or C>B>A vote would also be irrelevant. There could be a possible refinement to be found in this point, though so far I haven't had much luck.)

My best attempt to extend the method has better monotonicity than C//IRV, has a few Plurality failures, and appears to satisfy Schwartz without imposing it explicitly... I'm not sure if that's correct though. It could be that with a large number of candidates it becomes possible to make a scenario that fails Schwartz.

Here is the definition:

Define singlescore(a,b,c) like this:
If candidate C doesn't beat A, return positive infinity.
Otherwise, return:
( number of votes preferring A to both of B and C ) minus ( number of votes preferring C to both of A and B).

Define combinedscore(a,b) like this:
The lowest possible singlescore(a,b,x) where a/b/x are distinct candidates.
If there are no values to consider, return positive infinity.

Define winscore(a,b) like this:
Undefined if A doesn't beat B pairwise. Otherwise combinedscore(a,b).

Define candidatescore(a) like this:
The greatest value of winscore(a,x) given some other candidate x.
If there are no values to consider, return negative infinity.

Elect the candidate with the greatest candidatescore.

There are perhaps some options available to change whether min or max is used in various places, but the above seemed to work the best. For example combinedscore could return the greatest value (though this seems illogical to me), and the candidatescore could use the least value of winscore as the measure of a candidate.

I originally intended to run the scored contests through the River mechanism, but that doesn't match how fpA-fpC simply elects the candidate with the best score. My extension interprets the score to belong originally to a contest, not the candidate. There could be implications for clone independence and Plurality.

The Plurality failures aren't common, but here is an example of one:
0.331: D
0.307: B>C
0.270: C>A
0.089: A

All candidates are in Schwartz. There are only a couple of positive singlescores: singlescore(b,c,a) is the highest (signifying that the B>C win was not fabricated by A voters, who don't even express it), followed by (d,b,c) indicating that D>B was not fabricated by C voters. But none of the combinedscores are positive. The final candidate scores are:
B: -.024 for singlescore(b,c,d)
D: -.028 for ss(d,b,a).
C: -.037 for either (c,a,b) or (c,d,b)
A: -.181 for (a,b,c)

So B wins. The algorithm seems to get lost in specific details, I'd say, and doesn't really assess candidates' overall merit. But maybe there's a way to redesign it.

Kevin

A very late reply, but:

On 26/12/2020 22.25, Kevin Venzke wrote:

I think this method also fails DMTBR, if I've implemented it right.

Before:

1: A>C>D>B
1: A>D>B>C
1: B>C>D>A
1: C>A>B>D
1: D>B>A>C

A wins. Now let a C>A voter bury A:

1: A>C>D>B
1: A>D>B>C
1: B>C>D>A
1: C>B>D>A  (A is buried here)
1: D>B>A>C

and A and C tie for first.

Replacing min and max with leximin and leximax where possible helps, but
doesn't seem to fix that particular problem instance.