[EM] "Unmanipulable Majority" strategy criterion (newly amended version)
cbenhamau at yahoo.com.au
Wed Dec 3 12:34:00 PST 2008
Regarding my proposed Unmanipulable Majority criterion:
*If (assuming there are more than two candidates) the ballot
rules don't constrain voters to expressing fewer than three
preference-levels, and A wins being voted above B on more
than half the ballots, then it must not be possible to make B
the winner by altering any of the ballots on which B is voted
above A without raising their ranking or rating of B.*
To have any point a criterion must be met by some method.
It is met by my recently proposed SMD,TR method, which I introduced
as "3-slot SMD,FPP(w)":
*Voters fill out 3-slot ratings ballots, default rating is bottom-most
(indicating least preferred and not approved).
Interpreting top and middle rating as approval, disqualify all candidates
with an approval score lower than their maximum approval-opposition
(X's MAO score is the approval score of the most approved candidate on
ballots that don't approve X).
Elect the undisqualified candidate with the highest top-ratings score.*
Referring to the UM criterion: (a) if candidate A has a higher TR score than B
then the B>A strategists can only make B win by causing A to be disqualified.
But in this method it isn't possible to vote x above y without approving x, so
we know that just on the A>B ballots A has majority approval. It isn't possible
for a majority-approved candidate to be disqualified, and the strategists can't
cause A's approval to fall below majority-strength. And the criterion specifies
that none of the B>A voters who don't top-rate B can raise their rating of B to
increase B's TR score.
(b) if on the other hand B has a higher TR score than A but B is disqualified
there is nothing the B>A strategists can do to undisqualify B.
So SMD,TR meets the UM criterion.
B>A 101-95, B>C 87-20, A>C 102-20.
All Condorcet methods, plus MDD,X and MAMPO and ICA elect B.
B has a majority-strength pairwise win against A, but say 82 of the 93A change to
B>A 101-95, C>B 102-87, A>C 102-20
Approvals: A104, B101, C102
TR scores: A93, B87, C 20
Now MDD,A and MDD,TR and MAMPO and ICA and Schulze/RP/MinMax etc. using
WV or Margins elect A. So all those methods fail the UM criterion.
B>C 51-49, C>A 75-25, A>B 48-26
Schulze/RP/MM/River (WV) and Approval-Weighted Pairwise and DMC and MinMax(PO)
and MAMPO and IRV elect B.
Now say 4 of the 26C change to A>C (trying a Push-over strategy):
B>C 51-49, C>A 71-29, A>B 52-26
Now Schulze/RP/MM/River (WV) and AWP and DMC and MinMax(PO) and MAMPO
and IRV all elect C. Since B had/has a majority-strength pairwise win against C, all these
methods also fail Unmanipulable Majority. If scoring ballots were used and all voters score
their most preferred candidate 10 and any second-ranked candidate 5 and unranked candidates
zero, then this demonstration also works for IRNR so it also fails.
Who knew that such vaunted "monotonic" methods as WV and MinMax(PO) and MAMPO
were vulnerable to Push-over?!
B>A 51-49. Bucklin and MCA elect B, but if the 48 A>B voters truncate the winner changes
to A. So those methods also fail UM.
49: A9, B8, C0
24: B9, A0, C0
27: C9, B8, A0
Here Range/Average Ratings/Score/CR elects B and on more than half the ballots B is voted
above A, but if the 49 A9, B8, C0 voters change to A9, B0, C0 the winner changes to A.
So this method fails UM.
Here Borda elects B and B is voted above A on more than half the ballots, but if the 48
A>B>C>D ballots are changed to A>C>D>B the Borda winner changes to A, so
Borda fails UM.
This Unmanipulable Majority criterion is failed by all well known and currently advocated
methods, except 3-slot SMD,TR!
Given its other criterion compliances and simplicity, that is my favourite 3-slot s-w method
and my favourite Favourite Betrayal complying method.
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