[EM] Manipulability stats for more poll methods (fixed footnotes)

[Michael Ossipoff]

Of course, in Approval, if there aren’t other perceived reasons for
choosing whom to approve, one could approve above the mean…if you have a
feel for what’s average among the candidates. I guess that’s the usual
assumption for simulations.

If the candidate-lineup is so good that you do above-mean voting, then
you’re indeed fortunate.

If you’d do that, but you don’t have a feel about the average, & don’t
perceive cardinal-merits, then of course you could just approve the best
half of the candidates.

Maybe that was the assumed Approval strategy to which you were referring.

Approval is particularly perfectly matched for an election with
unacceptable candidates:

Just approve (only) all of the Acceptables.

On Sat, May 4, 2024 at 19:27 Michael Ossipoff <email9648742 at gmail.com>
wrote:

> There’s no reason for the renormalization. Among A, B, C & D (in that
> order of magnitude) if B is at the mean, then, with the A=0 & D=1
> renormalization, B’s renormalized value is the mean of all of the
> renormalized values.
>
> The position of the mean among the candidates doesn’t change with
> renormalization.
>
>
>
> On Sat, May 4, 2024 at 15:25 Michael Ossipoff <email9648742 at gmail.com>
> wrote:
>
>>
>>
>> On Sat, May 4, 2024 at 14:45 Kristofer Munsterhjelm <km_elmet at t-online.de>
>> wrote:
>>
>>>
>>> Yes, that's right. But consider a voter with the following utilities:
>>>
>>> A: 0.57
>>> B: 0.32
>>> C: 0.23
>>> D: 0.08
>>>
>>> Normalization to two steps fixes the highest value (0.57) to 1 and the
>>> lowest value (0.08) to 0 and rounds off the intermediate values after
>>> linearly scaling them.
>>
>>
>> Yes. So far, so good. But…
>>
>> This in essence says that a value is rounded off
>>> to 1 if it's greater than or equal to 0.325 (the midpoint between 0.08
>>> and 0.57)
>>
>>
>> What? You didn’t average the normalized values. You averaged two of the
>> values before normalization. The midrange isn’t usually the same as the
>> mean. You used the midrange as the mean.
>>
>> If you call the top value 1, & the bottom value 0,
>> then a rating’s new value is the number that’s the same % of the way from
>> 0 to 1 as the old number’s % from.08 to .57
>>
>> Average of those new values: .4475
>>
>> You still approve the best two.
>>
>>
>>
>> so the 0-1 normalized ballot is
>>>
>>> A: 1, B: 0, C: 0, D: 0
>>>
>>> On the other hand, the mean utility is 0.3. So the mean utility approval
>>> ballot is
>>>
>>> A: 1, B: 1, C: 0, D: 0.
>>>
>>> > 4 "dimensions" sounds like a lot.  What are the "strategy attempts" ?
>>> > How much and what information do the strategists have?  Are the
>>> > strategists confined to just trying to get their favourites elected,
>>> or
>>> > any candidate they prefer to the initial winner?
>>>
>>> The method works pretty much like this, for generating and testing a
>>> single election. (I've simplified the exact order that strategies are
>>> called upon, but this is in effect what happens.)
>>>
>>> ==== (Algorithm start) =====
>>>
>>> Draw candidate positions for each candidate (in this case, each is a
>>> point on a 4D normal distribution with mean 0 and variance 1).
>>> Draw voter positions for each voter, and create their honest ballots
>>> based on the distances between the voter and candidates.
>>> Pass the resulting ballots through the method to establish the honest
>>> outcome.
>>> If there's a tie, skip (because deciding what a strict improvement is
>>> when there's a honest tie is ambiguous). Otherwise let the winner be W.
>>>
>>> For each candidate X who is not the winner W:
>>>         For i = 1 to number of strategy attempts / number of candidates
>>>                 Set the strategic ballots to the honest ballots.
>>>
>>>                 For every voter who prefers X to W:
>>>                         Replace that voter's strategic ballot with a
>>>                         ballot according to a strategy that depends on
>>>                         i.
>>>
>>>                 Pass the modified strategic ballots through the method.
>>>                 If X is now a winner, the method is manipulable in
>>>                         this election. Return success.
>>>
>>> If we reach this point without any success, return failure; the method
>>> is (probably) not manipulable in this election.
>>>
>>> ==== (Algorithm end) =====
>>>
>>> The indexed strategies are
>>>         i=0: Compromising (raise X to unique top)
>>>         i=1: Burial (lower W to unique bottom)
>>>         i=2: Two-sided (do both at once)
>>>         i>2: Coalitional strategy
>>>
>>> The compromising, burial, and two-sided strategies modify the voters'
>>> otherwise honest ballots - for instance, compromising changes a
>>> strategist's ballot so that X is at unique top and the rest of the
>>> ballot is unchanged.
>>>
>>> The first time the coalitional strategy is called for a particular
>>> election, candidate to strategize for, and value of i, it chooses a
>>> random number of strategic ballots (between 1 and 3 inclusive). Each
>>> strategic voter then picks one of these ballots at random. This
>>> simulates strategies where every strategist ballot is equal, as well as
>>> ones where there are a few groups each with their own ballot type, thus
>>> covering more than JGA's simulations without becoming *too*
>>> computationally expensive.
>>>
>>> So with the setup for the stats that I gave, the full setup for a single
>>> method is like this:
>>>
>>> for j = 1 to 500k
>>>         Run the algorithm detailed above.
>>>         It returns one of three states: honest tie, success, or failure.
>>>         Increment the corresponding counter, call it TIES, SUCCESSES or
>>> FAILURES.
>>>
>>> manipulability = SUCCESSES/(500k - TIES)
>>>
>>>
>>> So to answer your questions:
>>>
>>> The strategists don't adapt their strategy to the information available
>>> to them, even though they strictly speaking have full information.
>>> However, they get to try over and over again until they win. If there is
>>> a full information strategy with not too many distinct ballots, then
>>> this random sampling will eventually find it, given a high enough
>>> strategy attempts value.
>>>
>>> For each non-winner X, everybody who prefers X to the current winner
>>> gets to have a go. So not just their favorites: anybody they all prefer
>>> to the current winner.
>>>
>>> >
>>> >>
>>> >> [2] The detailed stats suggest that pushover is a problem with
>>> Smith//DAC
>>> >
>>> > You don't have enough candidates for a sub-cycle, and so the method
>>> > can't fail mono-raise.  How can it have a Pushover problem?
>>>
>>> I did a bit more checking, and the full preference version doesn't have
>>> this high an "other strategy" count. Since I think it's unlikely that
>>> the version with truncation would have more pushover than the fully
>>> ranked one, I'm going to retract this; most likely it's just an artifact
>>> of the simulator's ballot reduction process that falsely attributes the
>>> strategy to the "other" category for cardinal methods.
>>>
>>> >
>>> >> - Margins-Sorted Approval, because I'm not sure how it works
>>> >
>>> > (I struggle to take this at face value.  Probably my promotion of MSA
>>> > has convinced you that it is the best method and you were concerned
>>> that
>>> > your simulation wouldn't do it justice.
>>>
>>> I'd like to believe both that I have enough scientific integrity not to
>>> do that, and that people know I have, too :-)
>>>
>>> Actually, I was planning on putting MSA at the same level as the other
>>> "I don't know enough about these or their dynamics" methods (double
>>> defeat Hare, MSMLV, and Max Strength Transitive Beatpath).
>>>
>>> > But our expert doing the
>>> > simulation claiming he can't understand the method isn't a good look
>>> for
>>> > its proposability.)
>>> >
>>> > Why didn't you simply ask me to explain it to you?
>>>
>>> I think it's the sorting phase that does it. My vague idea of how it
>>> works is that you essentially run a sorting algorithm on intermediate
>>> values, and that seems a little too complex to me. But I might just have
>>> got it wrong and then the initial impression of it as an intimidating
>>> method stuck.
>>>
>>> Ted Stern pointed me at the Electowiki article for MSA, which in turn
>>> led me to his Python implementation. I might port it if I have time, but
>>> I feel a bit exhausted after gathering all these stats. We'll see :-)
>>>
>>> > What happened to separate entries for BTR,  Woodall and Benham?
>>>
>>> They're in the other post. I didn't want to add them all to the post
>>> that was intended to focus on the new results. That's why I said "some
>>> for comparison" - the others are here:
>>>
>>>
>>> http://lists.electorama.com/pipermail/election-methods-electorama.com/2024-April/006029.html
>>>
>>> I could post all the stats - ordinal and cardinal methods' - in a
>>> summary post if you or other EM members would like.
>>>
>>> -km
>>> ----
>>> Election-Methods mailing list - see https://electorama.com/em for list
>>> info
>>>
>>
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