[EM] Mono-switch-plump criterion
cbenham at adam.com.au
Sat May 10 12:30:47 PDT 2014
> *The probability of candidate X winning must not be reduced if one or
> more ballots that
> plump for any not-X are replaced by an equal number of ballots that
> plump for X.*
Previously I showed that this is failed by the following methods:
Schulze (aka Beatpath), Ranked Pairs, River, MinMax (all equivalent with
3 candidates) if they use Winning Votes to weigh pairwise defeats.
IRV and the Condorcet methods based on IRV (such as Benham and Woodall)
Total Approval Chain Climbing.
I claim that it is met by Margins, any positional method, IBIFA,
Bucklin and Bucklin-like methods like Median Ratings and MCA and MTA.
And also it is met by MMLV(erw)M. To support that claim I'll just
talk about the Margins Sort version with 3 candidates.
Plumping ballots for any X always contribute to X's score and
switching plumping ballots to X might get rid of one of X's pairwise
If X has no pairwise defeats then that will always be still the case
after switching some plumping ballots to X and so X will still win. X can't
be a winner with all pairwise defeats so we are only concerned about the
case when X has just one (and so will the other 2 candidates).
Say we designate the candidate with the highest score 1, the
second-highest 2 and and the lowest 3. The algorithm in this
situation elects 1 unless 2 both pairwise beats 1 and has a score that
is closer to 1's than to 3's.
If winning candidate X is in position 2 then the effect of plumping
ballots being switched from 1 to 2 will be to just make 2 still closer
and the effect of plumping ballots being switched from 3 to 2 will have
the same effect (and make 3 further away).
If winning candidate X is 1 and pairwise beats 2 and loses to 3, then
the only hope of making 1 lose is to switch some plumping ballots from
2 to 1 sufficient for 2 and 3 to change places but that won't work
because then 2 and 3 will be adjacent candidates that are out of pairwise
order and will be much closer together score-wise than the other such
pair and they'll be switched back to give the final order 1>2>3.
And if X is 1 and losing to 2 then it means that 1's distance
(scorewise) from 2 is such that 2 and 3 are switched in the order, and
any plumping ballots to 1 will only increase that distance.
I hope that (almost confused) waffle is not too confusing or opaque.
*The probability of candidate X winning must not be reduced if one or
more ballots that
plump for any not-X are replaced by an equal number of ballots that
plump for X.*
Mono-raise is the traditional monotonicity criterion, but I don't see
why anyone would
see failure of Mono-switch-plump as less embarrassing than failing
B>C 51-45 C>A 71-29 A>B 52-26
Top Preferences: C45 > A29 > B26
When there are three candidates the MinMax , Beatpath (aka Schulze),
Ranked Pairs and River algorithms
are all equivalent. When they use Winning Votes as the measure of defeat
strength they all elect C.
IRV (aka the Alternative Vote) and Benham (and Woodall) also elect C.
But if we replace the 4A ballots
with 4C ballots the winner with all these methods changes from C to B.
B>C 51-49 C>A 71-29 A>B 48-26
Top Preferences: C45 > B26 > A25
Total Approval Chain Climbing also fails.
C>A>B>C, Approvals C73 > B59 > A58
TACC elects C, but if the 2B ballots are changed to 2C, then the
winner changes to A.
C>A>B>C, Approvals C75 > A58 > B57
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