[EM] Mono-switch-plump criterion

[C.Benham]

>  Mono-switch-plump:
>
> *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 
defeats.

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 
3-candidate cycle
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 
to 1,
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 
switching
any plumping ballots to 1 will only increase that distance.

I hope that (almost confused) waffle is not too confusing or opaque.

Chris Benham





  Mono-switch-plump:

*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 
Mono-raise.


25 A>B
26 B>C
23 C>A
22 C
04 A

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.

25 A>B
26 B>C
23 C>A
26 C

B>C  51-49       C>A 71-29       A>B 48-26

Top Preferences:  C45 > B26 > A25

Total Approval Chain Climbing  also fails.

25 A>B
06 A>C
32 B>C
27 C>A
08 C
02 B

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.

25 A>B
06 A>C
32 B>C
27 C>A
10 C

C>A>B>C,     Approvals C75 > A58 > B57