[EM] Final Conclusion on strong FBC!!!!!!

[Forest Simmons]

Alex, how do you generalize to more than three candidates?  Is it still
Top 2 or does it become Median Rank and Above?

Or could you make it "Above Lowest Rank" or "Non-truncated."

It seems to me that any of these would satisfy the FBC, and all are
equally valid generalizations of the three candidate Top 2 method.

If fact, giving one point to each of the non-truncated candidates would be
instrumentally equivalent to Approval, and so would satisfy the usual FBC,
while the additional expressivity would make it satisfy the Strong FBC in
the same sense that Top 2 does (as well as Joe Weintein's Five Slot
Approval).

Am I missing something?

Forest

On Thu, 2 Jan 2003, Alex Small wrote:

> Yesterday I posted to the list a proof of the Gibbard-Satterthwaite (GS)
> Theorem.  I said that if my methods are valid then I can reach conclusions
> on strong FBC.  I won't post my proof yet, because it is lengthy and I
> want to streamline it.  However, I will give the conclusion and sketch the
> forthcoming proof.
>
> In elections with 3 candidates, there is only 1 way to remove any
> incentive to insincerely rank another candidate ahead of your true
> favorite:  "Top 2 Voting", AKA "Negative Voting" AKA "Anti-Plurality
> Voting".  Each voter ranks the candidate, and a single point is assigned
> to each voter's first and second choices.  The candidate with the most
> points wins.  All other ranked methods fail strong FBC (if all voters and
> all candidates are treated equally, so that Random Ballot is excluded, as
> well as methods that only elect a certain candidate in a very tiny set of
> cases.)
>
> Let us first recall the fundamental requirement of strong FBC, that a
> voter either cannot improve the outcome by strategic voting, or that
> swapping second and third choices will have the same effect as insincerely
> ranking somebody ahead of your favorite.
>
> I began by looking at the boundary between the region where A wins and the
> region where B wins in the 5D space of electorates.  Although there are 6
> factions of voters, a basic symmetry requirement on swapping A and B in
> the rankings means that I only need to examine the strategic incentives
> (or lack thereof) for 3 factions.  Another symmetry requirement enables us
> to draw conclusions on the A-C and B-C boundaries after examining the A-B
> boundary.
>
> Since each faction can have either no incentive to vote strategically, or
> an incentive to betray #2, there are 2^3 = 8 cases to consider:
>
> The case where all 3 factions have no incentive to vote strategically
> violates the GS Theorem.  It places such severe restrictions on the
> normals to boundaries that we encounter Condorcet's paradox.
>
> Likewise, in 6 other cases, severe restrictions are placed on the normals
> to the boundaries.  The normals are not the ones you would get from
> Condorcet, but there are electorates which fall into a cyclic region:  One
> restriction says A>B, another says B>C, and a third says C>A.  Since the
> severe restrictions on the normals to boundaries exclude the possibility
> of an auxillary procedure in cyclic regions, strong FBC leads to a
> paradox.
>
> The only case that does not produce a paradox is "top 2 voting."  If we're
> near the boundary where A and B are tied, the factions A>B>C and B>A>C
> have an incentive to betray #2.  The factions C>A>B, C>B>A, B>C>A, and
> A>C>B have no incentive to vote insincerely.  Although the incentives for
> each faction place severe constraints on the normals to boundaries, the
> normals to the boundaries are not linearly independent, and it can be
> shown that cycles are then forbidden.
>
> (It is possible to exclude cycles when the normals are linearly
> independent, but rather than having a single rule "A beats B if
> <condition>" you have "A beats B if <condition 1> and <condition 2> or
> <condition 3>..."  Examples of ranked methods like that are Condorcet,
> IRV, and Bucklin.)
>
>
> Assuming there are no holes in the reasoning, there's still one weakness:
> The reasoning employed is specific to the case of 3 candidates.  The case
> of 4 or more candidates is very difficult, because there are many more
> cases to examine.  It would be nice if there were some sort of inductive
> argument:  "If for N candidates one cannot satisfy strong FBC while
> distinguishing between #1 and #2, then for N+1 candidates the same holds."
>
> Anybody have an idea for that?
>
>
>
>
>
> Alex
>
>
> ----
> For more information about this list (subscribe, unsubscribe, FAQ, etc),
> please see http://www.eskimo.com/~robla/em
>
>

----
For more information about this list (subscribe, unsubscribe, FAQ, etc), 
please see http://www.eskimo.com/~robla/em