<div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">If the candidate-lineup is so good that you do above-mean voting, then you’re indeed fortunate.</div><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">Maybe that was the assumed Approval strategy to which you were referring.</div><div dir="auto"><br></div><div dir="auto">Approval is particularly perfectly matched for an election with unacceptable candidates: </div><div dir="auto"><br></div><div dir="auto">Just approve (only) all of the Acceptables.</div><div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sat, May 4, 2024 at 19:27 Michael Ossipoff <<a href="mailto:email9648742@gmail.com">email9648742@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left-width:1px;border-left-style:solid;padding-left:1ex;border-left-color:rgb(204,204,204)"><div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">The position of the mean among the candidates doesn’t change with renormalization.</div></div><div><div dir="auto"><br></div><div dir="auto"><br></div><div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sat, May 4, 2024 at 15:25 Michael Ossipoff <<a href="mailto:email9648742@gmail.com" target="_blank">email9648742@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left-width:1px;border-left-style:solid;padding-left:1ex;border-left-color:rgb(204,204,204)"><div><br></div><div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Sat, May 4, 2024 at 14:45 Kristofer Munsterhjelm <<a href="mailto:km_elmet@t-online.de" target="_blank">km_elmet@t-online.de</a>> wrote:</div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left-width:1px;border-left-style:solid;padding-left:1ex;border-left-color:rgb(204,204,204)" dir="auto">
<br>
Yes, that's right. But consider a voter with the following utilities:<br>
<br>
A: 0.57<br>
B: 0.32<br>
C: 0.23<br>
D: 0.08<br>
<br>
Normalization to two steps fixes the highest value (0.57) to 1 and the <br>
lowest value (0.08) to 0 and rounds off the intermediate values after <br>
linearly scaling them. </blockquote><div dir="auto"><br></div><div dir="auto">Yes. So far, so good. But…</div><div dir="auto"><br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left-width:1px;border-left-style:solid;padding-left:1ex;border-left-color:rgb(204,204,204)" dir="auto">This in essence says that a value is rounded off <br>
to 1 if it's greater than or equal to 0.325 (the midpoint between 0.08 <br>
and 0.57)</blockquote><div dir="auto"><br></div><div dir="auto">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.</div><div dir="auto"><br></div><div dir="auto">If you call the top value 1, & the bottom value 0,</div><div dir="auto">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</div><div dir="auto"><br></div><div dir="auto">Average of those new values: .4475</div><div dir="auto"><br></div><div dir="auto">You still approve the best two.</div></div></div><div><div class="gmail_quote"><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto"><br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left-width:1px;border-left-style:solid;padding-left:1ex;border-left-color:rgb(204,204,204)" dir="auto">so the 0-1 normalized ballot is<br>
<br>
A: 1, B: 0, C: 0, D: 0<br>
<br>
On the other hand, the mean utility is 0.3. So the mean utility approval <br>
ballot is<br>
<br>
A: 1, B: 1, C: 0, D: 0.<br>
<br>
> 4 "dimensions" sounds like a lot. What are the "strategy attempts" ? <br>
> How much and what information do the strategists have? Are the <br>
> strategists confined to just trying to get their favourites elected, or <br>
> any candidate they prefer to the initial winner?<br>
<br>
The method works pretty much like this, for generating and testing a <br>
single election. (I've simplified the exact order that strategies are <br>
called upon, but this is in effect what happens.)<br>
<br>
==== (Algorithm start) =====<br>
<br>
Draw candidate positions for each candidate (in this case, each is a <br>
point on a 4D normal distribution with mean 0 and variance 1).<br>
Draw voter positions for each voter, and create their honest ballots <br>
based on the distances between the voter and candidates.<br>
Pass the resulting ballots through the method to establish the honest <br>
outcome.<br>
If there's a tie, skip (because deciding what a strict improvement is <br>
when there's a honest tie is ambiguous). Otherwise let the winner be W.<br>
<br>
For each candidate X who is not the winner W:<br>
For i = 1 to number of strategy attempts / number of candidates<br>
Set the strategic ballots to the honest ballots.<br>
<br>
For every voter who prefers X to W:<br>
Replace that voter's strategic ballot with a<br>
ballot according to a strategy that depends on<br>
i.<br>
<br>
Pass the modified strategic ballots through the method.<br>
If X is now a winner, the method is manipulable in<br>
this election. Return success.<br>
<br>
If we reach this point without any success, return failure; the method <br>
is (probably) not manipulable in this election.<br>
<br>
==== (Algorithm end) =====<br>
<br>
The indexed strategies are<br>
i=0: Compromising (raise X to unique top)<br>
i=1: Burial (lower W to unique bottom)<br>
i=2: Two-sided (do both at once)<br>
i>2: Coalitional strategy<br>
<br>
The compromising, burial, and two-sided strategies modify the voters' <br>
otherwise honest ballots - for instance, compromising changes a <br>
strategist's ballot so that X is at unique top and the rest of the <br>
ballot is unchanged.<br>
<br>
The first time the coalitional strategy is called for a particular <br>
election, candidate to strategize for, and value of i, it chooses a <br>
random number of strategic ballots (between 1 and 3 inclusive). Each <br>
strategic voter then picks one of these ballots at random. This <br>
simulates strategies where every strategist ballot is equal, as well as <br>
ones where there are a few groups each with their own ballot type, thus <br>
covering more than JGA's simulations without becoming *too* <br>
computationally expensive.<br>
<br>
So with the setup for the stats that I gave, the full setup for a single <br>
method is like this:<br>
<br>
for j = 1 to 500k<br>
Run the algorithm detailed above.<br>
It returns one of three states: honest tie, success, or failure.<br>
Increment the corresponding counter, call it TIES, SUCCESSES or FAILURES.<br>
<br>
manipulability = SUCCESSES/(500k - TIES)<br>
<br>
<br>
So to answer your questions:<br>
<br>
The strategists don't adapt their strategy to the information available <br>
to them, even though they strictly speaking have full information. <br>
However, they get to try over and over again until they win. If there is <br>
a full information strategy with not too many distinct ballots, then <br>
this random sampling will eventually find it, given a high enough <br>
strategy attempts value.<br>
<br>
For each non-winner X, everybody who prefers X to the current winner <br>
gets to have a go. So not just their favorites: anybody they all prefer <br>
to the current winner.<br>
<br>
> <br>
>><br>
>> [2] The detailed stats suggest that pushover is a problem with Smith//DAC<br>
> <br>
> You don't have enough candidates for a sub-cycle, and so the method <br>
> can't fail mono-raise. How can it have a Pushover problem?<br>
<br>
I did a bit more checking, and the full preference version doesn't have <br>
this high an "other strategy" count. Since I think it's unlikely that <br>
the version with truncation would have more pushover than the fully <br>
ranked one, I'm going to retract this; most likely it's just an artifact <br>
of the simulator's ballot reduction process that falsely attributes the <br>
strategy to the "other" category for cardinal methods.<br>
<br>
> <br>
>> - Margins-Sorted Approval, because I'm not sure how it works<br>
> <br>
> (I struggle to take this at face value. Probably my promotion of MSA <br>
> has convinced you that it is the best method and you were concerned that <br>
> your simulation wouldn't do it justice.<br>
<br>
I'd like to believe both that I have enough scientific integrity not to <br>
do that, and that people know I have, too :-)<br>
<br>
Actually, I was planning on putting MSA at the same level as the other <br>
"I don't know enough about these or their dynamics" methods (double <br>
defeat Hare, MSMLV, and Max Strength Transitive Beatpath).<br>
<br>
> But our expert doing the <br>
> simulation claiming he can't understand the method isn't a good look for <br>
> its proposability.)<br>
> <br>
> Why didn't you simply ask me to explain it to you?<br>
<br>
I think it's the sorting phase that does it. My vague idea of how it <br>
works is that you essentially run a sorting algorithm on intermediate <br>
values, and that seems a little too complex to me. But I might just have <br>
got it wrong and then the initial impression of it as an intimidating <br>
method stuck.<br>
<br>
Ted Stern pointed me at the Electowiki article for MSA, which in turn <br>
led me to his Python implementation. I might port it if I have time, but <br>
I feel a bit exhausted after gathering all these stats. We'll see :-)<br>
<br>
> What happened to separate entries for BTR, Woodall and Benham?<br>
<br>
They're in the other post. I didn't want to add them all to the post <br>
that was intended to focus on the new results. That's why I said "some <br>
for comparison" - the others are here:<br>
<br>
<a href="http://lists.electorama.com/pipermail/election-methods-electorama.com/2024-April/006029.html" rel="noreferrer" target="_blank">http://lists.electorama.com/pipermail/election-methods-electorama.com/2024-April/006029.html</a><br>
<br>
I could post all the stats - ordinal and cardinal methods' - in a <br>
summary post if you or other EM members would like.<br>
<br>
-km<br>
----<br>
Election-Methods mailing list - see <a href="https://electorama.com/em" rel="noreferrer" target="_blank">https://electorama.com/em</a> for list info<br>
</blockquote></div></div>
</blockquote></div></div>
</div>
</blockquote></div></div>