Unifying DMTC and DMTCBR, and method X criterion
Here's a way to unify DMTC (elects a DMT candidate) and burial resistance:
Suppose that candidate A has more than 1/3 of the first preferences, and
beats B pairwise. Then B must not win.
This works because B>A voters can't influence A's first preferences, nor
can they influence whether A beats B pairwise. A neat trick is also that
this can never lead to a cycle, because there can't be more than two
candidates with more than 1/3 first preferences, and A can't both beat B
and B beat A at the same time.
In that spirit, here's a weak criterion that's passed by method X:
Let an election restricted to set S be the election after everybody but
the candidates in S are eliminated. Then if, in every restricted
election involving A and B, A has more than 1/|S| of the first
preferences, and A also beats B pairwise, then B must not be elected.
Here |S| is the cardinality of set S, i.e. the number of non-eliminated
candidates in the restricted election in question.
Method X passes this criterion because when B's finding a maximizing
elimination path, no matter who's eliminated, A always has more than
1/|S| first preferences. Therefore A can never be eliminated, so B is
either forced to be eliminated (in which case he obviously can't be
elected) or his final matchup is against A, who beats him pairwise. On
the other hand, A can do no worse than following the same path, leading
to a matchup against B, which he wins.
It's weak because we need the criterion to hold for all restricted
elections, not just one of them. So the DMT set has to be
well-distributed: all its candidates must have about the same support.
But in return, if the criterion bars B from being elected, voters may
flip pairwise preferences between DMT members and non-DMT members all
they want; it'll never help. Furthermore, a voter lowering his last
ranked DMT set member under a bunch of non-DMT candidates also never helps.
Weak as it is, it could be useful in method design, just like clone
independence. I'll take every handhold scaling this cliff :-)
-km
Great criterion as far as it goes ... great at not rewarding buriers, but
bad at electing buried candidates ... if I understand it.
Hence the need for a sincere runoff:
40 A>B(Sincere A>C)
35 B>C
25 C>A
The sincere CW is C, which cannot be elected because 25 is less than 100/3,
if I understand the proposed critersion.
But a top three sincere runoff of the form
A vs (B vsC)
will elect C assuming rational voters informed of the true preferences.
On Sun, Aug 13, 2023, 7:51 AM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:
Here's a way to unify DMTC (elects a DMT candidate) and burial resistance:
Suppose that candidate A has more than 1/3 of the first preferences, and
beats B pairwise. Then B must not win.
This works because B>A voters can't influence A's first preferences, nor
can they influence whether A beats B pairwise. A neat trick is also that
this can never lead to a cycle, because there can't be more than two
candidates with more than 1/3 first preferences, and A can't both beat B
and B beat A at the same time.
In that spirit, here's a weak criterion that's passed by method X:
Let an election restricted to set S be the election after everybody but
the candidates in S are eliminated. Then if, in every restricted
election involving A and B, A has more than 1/|S| of the first
preferences, and A also beats B pairwise, then B must not be elected.
Here |S| is the cardinality of set S, i.e. the number of non-eliminated
candidates in the restricted election in question.
Method X passes this criterion because when B's finding a maximizing
elimination path, no matter who's eliminated, A always has more than
1/|S| first preferences. Therefore A can never be eliminated, so B is
either forced to be eliminated (in which case he obviously can't be
elected) or his final matchup is against A, who beats him pairwise. On
the other hand, A can do no worse than following the same path, leading
to a matchup against B, which he wins.
It's weak because we need the criterion to hold for all restricted
elections, not just one of them. So the DMT set has to be
well-distributed: all its candidates must have about the same support.
But in return, if the criterion bars B from being elected, voters may
flip pairwise preferences between DMT members and non-DMT members all
they want; it'll never help. Furthermore, a voter lowering his last
ranked DMT set member under a bunch of non-DMT candidates also never helps.
Weak as it is, it could be useful in method design, just like clone
independence. I'll take every handhold scaling this cliff :-)
-km
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On 8/15/23 02:24, Forest Simmons wrote:
Great criterion as far as it goes ... great at not rewarding buriers,
but bad at electing buried candidates ... if I understand it.
Hence the need for a sincere runoff:
40 A>B(Sincere A>C)
35 B>C
25 C>A
The sincere CW is C, which cannot be elected because 25 is less than
100/3, if I understand the proposed critersion.
Let's see: A has more than 1/3, and A beats B pairwise. So B is
disqualified. B has more than 1/3 and beats C pairwise, so C is
disqualified. Hence A must be elected.
That's right. Here the burial actually succeeds because the A voters
prefer A to C, the sincere CW. This proves that the burial resistance
isn't absolute. But note that this also happens to Smith,IRV and Smith,IFPP.
IRV and IFPP themselves elect A even in the sincere scenario.
But a top three sincere runoff of the form
A vs (B vsC)
will elect C assuming rational voters informed of the true preferences.
I'm not entirely sure about the notation. Could you reacquaint me with
the concept of a sincere top-k, k>2 runoff?
But on a more general note, I would say that it's not always necessary
to require that a criterion that removes the incentive for some kind of
strategy, to be able to gracefully recover if the strategy is used anyway.
Monotone methods all make pushover strategy irrelevant. However, if a
particularly monotone method were to elect B, then A>B>C voters decide
to downrank A to last place (pushover) and then it switches to electing
C instead, I wouldn't consider that a particularly severe weakness with
the method. Being able to recover the sincere winner anyway would be a
nice deluxe option, but it's not make-or-break, I wouldn't say.
Or consider Warren's attempt to generalize DH3 to every Condorcet method
by saying that if all the factions go on a burial spree, then every
Condorcet method will at the end elect the dark horse. He says something
like:
Consider
37:C>A>B>D
37:C>B>A>D
32:A>B>C>D
32:A>C>B>D
31:B>A>C>D
31:B>C>A>D
Every method elects C. But the A and B voters don't like that, so they
uprank D to second place:
37:C>A>B>D
37:C>B>A>D
32:A>D>B>C
32:A>D>C>B
31:B>D>A>C
31:B>D>C>A
Now A wins in Minmax, Schulze, etc. Simultaneously, the C voters are
saying they need protection against either the A or B voters doing so,
so they too bury A and B under D. However, they didn't expect that both
would be doing it at the same time, so what happens is:
37:C>D>A>B
37:C>D>B>A
32:A>D>B>C
32:A>D>C>B
31:B>D>A>C
31:B>D>C>A
And now D is the CW: cue the explosion stock effect.
Under a method that passes my aforementioned criterion, this can't work,
because post-burial, the criterion bars A and B from being elected, and
C has the highest Minmax score of the remaining two, hence the burial
does nothing.
Warren could then argue that "but if everybody just does what's
intuitive, then they all rank D second, and then there's still a big
boom!". But I would think that knowing that burial doesn't work and
might easily backfire would tend to temper such ideas.
-km
On Wed, Aug 16, 2023, 4:39 AM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:
On 8/15/23 02:24, Forest Simmons wrote:
Great criterion as far as it goes ... great at not rewarding buriers,
but bad at electing buried candidates ... if I understand it.
Hence the need for a sincere runoff:
40 A>B(Sincere A>C)
35 B>C
25 C>A
The sincere CW is C, which cannot be elected because 25 is less than
100/3, if I understand the proposed critersion.
Let's see: A has more than 1/3, and A beats B pairwise. So B is
disqualified. B has more than 1/3 and beats C pairwise, so C is
disqualified. Hence A must be elected.
That's right. Here the burial actually succeeds because the A voters
prefer A to C, the sincere CW. This proves that the burial resistance
isn't absolute. But note that this also happens to Smith,IRV and
Smith,IFPP.
Can you think of a method that is so burial resistant that a sincere runoff
restricted to Smith would not appreciably improve its Sincere CW efficiency
(given |Smith|>2)?
IRV and IFPP themselves elect A even in the sincere scenario.
But a top three sincere runoff of the form
A vs (B vsC)
will elect C assuming rational voters informed of the true preferences.
I'm not entirely sure about the notation. Could you reacquaint me with
the concept of a sincere top-k, k>2 runoff?
Basically you specify a tournament schedule in the form of a binary tree.
The voters start at the root node and by majority decision decide which
daughter branch to pursue. Recursively elect the method winner of the sub
tree the branch leads to. The boundary condition for the recursion is that
the winner of a leaf is the leaf itself.
For anything other than an election in the Society of Game Theoretic Nit
Pickers, the method should be restricted to 3 candidates, for example the
top three finishers of IRV restricted to Smith.
Suppose that when IRV is restricted to Smith, the finish order is
S1>S2>S3>...
The sincere runoff tree should be
S1 vs (S2 vs S3)
If one of these three (say C) pairwise defeats each of the other two, then
rational voters who are aware of the other voters' preferences, will elect
C.
Otherwise (still assuming rationality and preference awareness) candidate
C1 will be elected ... as can be easily (if tediously) shown in an
exhaustive (and exhausting) case-by-case analysis.
fws
But on a more general note, I would say that it's not always necessary
to require that a criterion that removes the incentive for some kind of
strategy, to be able to gracefully recover if the strategy is used anyway.
Monotone methods all make pushover strategy irrelevant. However, if a
particularly monotone method were to elect B, then A>B>C voters decide
to downrank A to last place (pushover) and then it switches to electing
C instead, I wouldn't consider that a particularly severe weakness with
the method. Being able to recover the sincere winner anyway would be a
nice deluxe option, but it's not make-or-break, I wouldn't say.
Or consider Warren's attempt to generalize DH3 to every Condorcet method
by saying that if all the factions go on a burial spree, then every
Condorcet method will at the end elect the dark horse. He says something
like:
Consider
37:C>A>B>D
37:C>B>A>D
32:A>B>C>D
32:A>C>B>D
31:B>A>C>D
31:B>C>A>D
Every method elects C. But the A and B voters don't like that, so they
uprank D to second place:
37:C>A>B>D
37:C>B>A>D
32:A>D>B>C
32:A>D>C>B
31:B>D>A>C
31:B>D>C>A
Now A wins in Minmax, Schulze, etc. Simultaneously, the C voters are
saying they need protection against either the A or B voters doing so,
so they too bury A and B under D. However, they didn't expect that both
would be doing it at the same time, so what happens is:
37:C>D>A>B
37:C>D>B>A
32:A>D>B>C
32:A>D>C>B
31:B>D>A>C
31:B>D>C>A
And now D is the CW: cue the explosion stock effect.
Under a method that passes my aforementioned criterion, this can't work,
because post-burial, the criterion bars A and B from being elected, and
C has the highest Minmax score of the remaining two, hence the burial
does nothing.
Warren could then argue that "but if everybody just does what's
intuitive, then they all rank D second, and then there's still a big
boom!". But I would think that knowing that burial doesn't work and
might easily backfire would tend to temper such ideas.
-km