On 14.01.2022 12:45, Daniel Carrera wrote:

Nice! I agree, that is interesting. Benham is known for being pretty
resistant to strategy in general, so I'm wondering if that strategy
resistance forces the remaining weakness to be concentrated around easy
strategies.

Could you try a method from each of the two other categories, and see if
the trivial strategy is so overwhelmingly succcessful on these too?

E.g. Smith//Plurality (Or Smith,Plurality) from the extremely easy to
manipulate category, and Minmax from the intermediate-to-high one.

-km

(re-sent as Reply to All)

Hi Kristofer,

There's a slightly easier way.

Each candidate is asked "How many other candidates defeat you?"

The one with the fewest defeats (the Copeland winner) moves to the right of
the line. If there is a tie, all tied candidates move right.

The candidate(s) on the right is (are) asked, "Which candidates on the left
side of the line defeated you?" And as they are named, they come to the
right of the line.

Each new right-side candidate is asked the same question, and any left-side
candidates named come to the right side.

When none of the candidates on the right can name any other candidates on
the left who have defeated them, the Smith set is complete.

On Thu, Jan 13, 2022 at 2:34 AM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:

Cool!

El vie., 14 de ene. de 2022 3:02 p. m., Ted Stern dodecatheon@gmail.com
escribió:

On Fri, Jan 14, 2022 at 6:33 AM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:

That was a good hunch. That seems to be the case for Minimax at least. For
Minimax the successful strategies are less skewed toward the easiest ones,
but for Plurality and Hare they seem to be more skewed. I ran a test with
N=4, V=99, C=5, and 20,000 elections.

Method  , 95% c.i.    , trivial, reverse, moderate, majority
Plurality, 0.5807-0.5945, 1.000  , 0.0000 , 0.00000 , 0.318
MiniMax, 0.4032-0.4157, 0.900  , 0.0768 , 0.02284 , 0.210
Benham, 0.0449-0.0505, 0.979  , 0.0188 , 0.00209 , 0.133
Hare    , 0.0660-0.0727, 1.000  , 0.0000 , 0.00000 , 0.137

Incidentally, all the 95% intervals agree with Table 1 of JGA so this makes
me feel more confident that my program is working correctly. In any case,
for Hare and Plurality 100% of the successful strategies were the "trivial"
case. For Benham there are 20 non-trivial strategy successes (18
"reverse" + 2 "JGA"). If Hare had the same trivial/non-trivial ratio as
Benham, you'd expect to see 30 non-trivial successes for Hare. That's large
enough that I should have seen it if it was there. So I feel fairly
confident in saying that Hare and Plurality are both even more heavily
skewed toward the trivial strategy than Benham is.

Lastly, there is Minimax. That one was less heavily skewed toward easy
strategies. Of the successful strategies, 90% were trivial (vs 98% for
Benham) and out of the non-trivial ones, 77% were "reverse" (vs 90% for
Benham). So there seems to be a signal here, but even Minimax seems to be
very heavily skewed toward very easy strategies.

So... in other words... how often is the strategy easy? (i.e. trivial or
"reverse")

Plurality --> 100%
Minimax --> 97.7%
Benham --> 99.8%
Hare --> 100%

Wow.

Cheers,

Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University

On Fri, Jan 14, 2022 at 6:33 AM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:

I forgot you specifically asked about Smith//Plurality. I'm not familiar
with the '//' notation but I'm going to guess that the rule is "find the
Smith set and then find the plurality winner within that". If so, then that
method is like Minimax in that it is less skewed toward the simplest
strategies:

Smith//Plurality

• 59% susceptible
• 89% of successful strategies are the trivial one
• 79% of the non-trivial strategies are the "reverse" strategy and 21% are
  the JGA-style search.

Minimax

• 41% susceptible
• 90% of successful strategies are the trivial one
• 77% of the non-trivial strategies are the "reverse" strategy and 23% are
  the JGA-style search.

Hare

• 6.9% susceptible
• 100% of successful strategies are the trivial one

Benham

• 4.8% susceptible
• 98% of successful strategies are the trivial one
• 90% of the non-trivial strategies are the "reverse" strategy and 23% are
  the JGA-style search.

And I should have realized that for "Plurality" it is trivially true that
100% of successful strategies are trivial since only the first choice on
the ballot matters.

Cheers,

Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University

This seems very interesting. If I understand it correctly, Daniel is
saying that when a voting method can be subverted by tactical voting in
such a way that candidate c is elected in place of the rightful winner
w, the subversion can nearly always be accomplished if all voters who
prefer c to w simultaneously compromise on c and bury w.
   I'm surprised that "false cycles" don't come into it. Should I
conclude that artificially placing a candidate second hardly ever
achieves anything not achieved by compromising and burial?
      CJC

On 15/01/2022 05:32, Daniel Carrera wrote:

On 15.01.2022 15:04, Colin Champion wrote:

Yes, that's correct.

The setting may be a bit misleading: in the game that the strategy's
being calculated on, there's first an "honest" round, and then factions
who prefer someone else to the winner get to try to make that someone win.

In particular, there's no hidden information. In a real election, voters
who prefer some Y to X might not know that Y is the candidate they
should be compromising for, and that e.g. trying to compromise for Z
instead would backfire. So adjusting lower ranks may still be useful in
such a situation, or when the voters are aiming for destructive strategy
("anyone but T") rather than constructive.

Since the simulator tests the compromise+burial strategy first, it also
doesn't determine whether a less stark strategy would also have worked.
In a real setting, the candidates may not enjoy total loyalty from Y>X
voters.

As an argument in favor of Range, Warren sketches a scenario where the
frontrunners are X and Y and everybody either votes X>...>Y or Y>...>X.
As a consequence of the majority criterion, either X or Y wins; and then
he says that this probably will lead to two-party domination because the
strategy is self-stabilizing . The results from the strategy
calculations could be used to back up that argument.

However, it may well be that if everybody but one voter is doing so, and
a third party voter does A>X>...>Y instead of X>...>Y, then X still
wins; and so on up until enough tird party voters put A first, after
which A wins. And if the X>Y voters can't all be relied on to put X
first no matter what, then such weaker strategies may grow until there
are no longer only two frontrunners.

Dynamic strategy (my response to your response to...) is much harder to
model than static. But I would expect that methods that resist static
(one-shot) strategy better would also resist dynamic strategy better.

-km

Thank you Ted!  And Kristofer and Forest!

I'm seeing that the secondary hand-written signs -- that show the win
counts -- are not needed.

Instead, each person who represents a candidate holds all the
hand-written pairwise win signs for which that candidate wins. The other
name on the sign indicates who lost that pairwise comparison.

As Forest suggests, it would be easy for the "candidates" to sort
themselves according to how many signs they are holding.

Then, as Ted suggests:

That makes sense.

Yet I believe that lots of audience members would give more trust to a
method that eliminates candidates one at a time. This is the simplicity
advantage that instant-runoff voting (IRV) has. In contrast, starting
with the Copeland winner might seem like magic to people who fear math.

Would eliminating from the bottom always work?

In the Wikipedia example below, it does work. It identifies B and E as
outside the Smith set, so the remaining candidates are in the Smith set.

5 wins for A

5 wins for D

4 wins for G

3.5 wins for C

2.5 wins for F

1 win for B

0 wins for E

(Half indicates a tie.)

So, my question is: Does elimination from the bottom always work?

My guess is the answer is yes.

Again, thank you for your help!

Richard Fobes

On 1/14/2022 3:02 PM, Ted Stern wrote:

On Sat, Jan 15, 2022 at 8:41 AM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:

It is also worth noting that while real elections don't have perfect
information, they generally do have some information. Most elections are
repeats of previous elections. Candidates change, but party platforms and
political affiliations don't as much. It would actually be interesting to
rethink the "elections" and "candidates" from the simulation as instead
representing relatively fixed voters and parties. We can grab
(electorate,parties) where there was a successful strategy, and have it
repeat the election over and over again, each time allowing coalitions to
shift their votes, until we reach a Nash equilibrium. I'm not exactly sure
what I would do with that information though.

On a mostly unrelated note, I just ran Smith//IRV. Unsurprisingly, it's
similar to Benham, but there is more room for the JGA strategy:

• N = 4, V = 99, C = 5
• Smith//IRV
• 4.7% of elections have successful strategies
• 94% of successful strategies are the trivial one
• The remaining strategies are split 38% "reverse" and 62% "JGA".

As an argument in favor of Range, Warren sketches a scenario where the

I'm not sure I understand how this is an argument in favor of Range. The
JGA paper doesn't include Range, but it includes Approval and Approval is
one of the easiest to manipulate systems. I would guess that Range would be
similar. Warren's scenario seems to me like a likely end result of the sort
of repeat elections leading to a Nash equilibrium that I suggested earlier.
If we make the simplifying assumption that every election that can be
manipulated is an electorate that will one day end up in a two-party
system, it would follow that the systems that are hardest to manipulate are
the ones most likely to avoid going down that path.

However, it may well be that if everybody but one voter is doing so, and

That would be my guess too.

--
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University

On 15.01.2022 21:50, Daniel Carrera wrote:

Yeah, I'm not sure either. What we'd want to have is a system where the
honest winner is the equilibrium winner as often as possible, although
Gibbard shows that no method is strategy-proof except combinations of
duple (random pair) and random ballot.

I've at times thought that a more realistic approach would be to use
evolutionary stable strategies, e.g. suppose everybody is honest, then
some small faction defecting can't grow in power to the point where
everybody feels the need to vote tactically. But that's even harder to
model, and some societies are used to tactical voting at the start (e.g.
US Plurality elections), and might thus not fit that very well.

94% vs 98% is not that much of a difference, but any little bit helps
(as long as it's not too much at the expense of some other desirable
properties of the method). Increasing the need for complex strategy is
also good, of course, because it's easier to imagine trivial or reverse
than some tailor-made strategy.

Now I'm wondering if the strategically resistant methods mainly increase
their resistance by making it impossible for trivial strategy to work in
cases where it would otherwise work. From my experimentation, I've found
out that reversal symmetry usually makes for a susceptible method, and
that rev. sym. and Condorcet together are incompatible with dominant
mutual third burial resistance.

Perhaps it'd be easier to reason about these implications based on the
trivial strategy. Something like: suppose you vote X>...>Y. If there is
a benefit to voting X first, this must also weaken Y's position, because
with the reversed ballots, Y would be helped by this upranking. (That's
only for my stronger sense of reversal symmetry, though.)

I'd have to think more about it, but I think trying to reason about what
kind of properties reduce the effectiveness of trivial strategy could be
a good approach.

He has a few arguments that generally boil down to "if everybody goes on
a burial/compromising spree, then ranked voting does badly but Range
does not". I think his argument in this particular case is that if
voters are all X>...>Y or Y>...X, a common consensus candidate might
still win in Range (e.g. 52 voters X: 10, A: 9, Y: 0, 49 voters: Y: 10,
A: 9, X: 0, then A wins). So the idea is that the voters, used to two
party rules, don't want to risk honesty and instead vote the lesser evil
top and the greater evil bottom all the time, after which the majority
criterion does the rest.

Similar reasoning can be found in https://rangevoting.org/WVmore.html
which seems to argue that DH3 is a problem even for Condorcet methods
that pass DMTBR (like Smith//IRV, Benham, etc):

In any case, as you say, strategy-resistant methods like the Smith-IRV
hybrids will reduce the chance that strategy works, and so would give
third parties some more room in which to grow.

It would be interesting to do a test with Range; I imagine that it would
be very susceptible to strategy, similar to Approval. And I would also
imagine that STAR would do considerably better.

-km