Hello,
I deduce from an off-list enquiry that, in my recent posts, I may not have made the exact rules of this method completely clear.
I shall now attempt to spell it out very clearly:
The method is a plain ranked-ballot method, that allows truncation and can accomodate equal non-last preferences.
It is a Condorcet-completion method. The first step is to eliminate all non-members of the Schwartz set. Thereafter the eliminated
candidates are ignored, that is the count proceeds as if though they hadn't stood.
(The "Schwartz set" is the smallest possible non-empty set of candidates with none of the members having any pairwise losses to
any non-member candidates. If there are no pairwise ties, it is the same as the Smith set.)
Next the ballots are "symetrically completed". That means that each ballot that doesn't strictly rank all the candidates is replaced by
a number of ballots, one for each alternative way of strictly ordering the candidates that were not strictly ordered on the original ballot,
and the total weight of these new ballots will sum to one.
(To take a simple example, if non-last equal-ranking is allowed and there are three candidates ABC, then a A=B>C ballot will be
"replaced" by two ballots, each with a weight of 1/2, one A>B>C and one B>A>C).
Lastly, on these "completed" ballots, the "Weighted Median Approval" (WMA) method is used to pick the winner, thus:
The ballots are converted to "Approval" ballots, with each ballot wholly "approving" or "not approving" each candidate.
(There are no partial approvals).
Each candidate is considered to have a "weight" that is equal to the number of ("symetrically completed") ballots on which it is ranked
in first-place (of course disregarding eliminated candidates).
Each ballot approves the candidate it ranks first. Is the "weight" of this candidate less than half the total weight of all the candidates?
If yes, then the ballot also approves the candidate ranked second. Is the combined weight of these two candidates less than half
the total weight of all the candidates. If no, then this ballot approves no other candidate. If yes, then this ballot also approves the
candidate it ranks third, and so on, until each ballot has approved at least "half" the candidates, measured "by weight".
The candidate with the highest (thus derived) "approval" score wins. If there is tie, elect the tied candidare with the higher "weight".
In practice there will often be easy shortcuts. For example, if there are three candidates in the Schwartz set, then (without bothering
with the "symetric completion") every ballot will approve the two candidates it ranks non-last, and those that only rank one candidate
will approve that candidate and half-approve the other two.
The justification for the method is that unlike Winning Votes, it meets the Sincere Expectation Criterion, which I now regard as
essential. It is a weaker form of No Zero-Information Strategy, and it says that with every possible valid by the other voters being
equally likely, then regarding the categories:(1) the voter's favourite candidates, (2) the voter's first and second favourites, etc. down
to all the candidates the voter ranks non-last; then there should be no insincere vote that gives the voter a better chance in all the categories than voting sincerely. The sincere vote must do better in at least one, or as well in all of them.
This standard is met by Margins, but unlike Margins, Schwartz // SC-WMA meets Steve Eppley's "Minimal Defense" and
"Truncation Resistance" defensive strategy criteria.
(I've been wrong before. In a later post I might get around to justifying these claims.)
The cost of this is that it fails his "Immunity from Majority Complaints" criterion, and (in my view fairly benignly) it can fail
Mono-raise (aka monotonicity) when there are more than three candidates in the Schwartz set.
It may be more vulnerable to uncountered Burying than Winning Votes. It gives more Bucklin-like results which are probably
(with sincere voting) higher "social utility".
Here, without the ratings, is an example James G-A gave to illustrate the need for his "Weighted Pairwise" method.
26: B>D>K 22:B>K>D 26:D>K>B 01:D>B>K 21:K>D>B 04>K>B>D
100 ballots. B>D 52-48, D>K 53-47, K>B 51-49. All the candidates are in the Schwartz set.
All the ballots give a full "strict" ranking (no equal preferenes), so there is no "symetric completing" to do.
Each ballot simply approves the two candidates they rank non-last, to give these scores:
B:53, D:74, K:73.
D has the highest score and so wins. Ranked Pairs, Beat Path, MinMax etc. all elect B
Chris Benham
Sincere Expectation Standard Given that a voter has no knowledge about how others will vote, a sincere vote must be at least as likely as any insincere vote to give results that are in some way better in the eyes of the voter.
Or expressed as a more rigid criterion: ----- Sincere Expectation Criterion (SEC) Consider a voter with a preference order between the possible outcomes of the election. Let us call his sincere ballot, X. Now, assuming that every possible legal ballot is equally likely for every other voter, there must be some justification for the vote X over any other way to fill out the ballot, which I will call Y. This justification is given by the following comparisons:
The probability of X electing one of the voter's first choices vs. the probability of Y electing one of these choices
The probability of X electing one of the voter's first or second choices vs. the probability of Y electing one of these.
The probability of X electing one of the voter's first, second or third choices vs. the probability of Y electing one of these. ... And so on through all the voter's choices
X must either do better in one of these comparisons than Y, or equal in all. Otherwise the sincere vote can not be justified. ---- In other words, there must be some justification for voting sincerely even if the voter does not know how any one else is voting.
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