[EM] Multiwinner participation rule. Reply to M. Schulze

[Craig Carey]

At 01:14 14.12.99 , Markus Schulze wrote:
>Dear Craig,
>
>Craig Carey wrote (13 Dec 1999): ...
>> -------------------------------------------------------------------------
>> Definition: (Q2), "multiwinner participation axiom" rule. (13-Dec-99):
>>
>> (For All V)(For All p)(For All t) [
>>    (0<=t)(#(p.W(V))<=#(p.W(V+tp))) .=>
>>       (Satisf(W(V),p) <= Satisf(W(V+tp),p))]]
>> -------------------------------------------------------------------------
>> (Satisf(W,p) <= Satisf(X,p)) = (For All j=1..length(p)).[ G(W,X,p,j) ]
>> G(W,X,p,j) = ((W.trunc(p,j-1) = X.trunc(p,j-1)) => (W.{p[j]} <= X.{p[j])})
...
>> Write a n:A+B+CDEF to mean that the paper (ABCDEF) has weight n and that
>>  {A,B} = ({A,B,C,D,E,F}.Winners).
>>
>> Let Z(p) = [#(p.W(V)), #(p.W(V+tp)); Satisf(W(V),p), Satisf(W(V+tp),p)]
>> The rule fails the particular alteration if this is false:
>>   (z1<=z2)=>(z3<=z4).
>>
>> (0:AB)--(9:AB+): Z(AB)=(0,1; 0,1/4), Allowed
>> (0:AB+)--(9:AB): Z(AB)=(1,0; _,_), Allowed
>>         Allow the adding of AB to causes B to start losing
>>
>> (0:ABC+DE+FG)--(9:ABC+DE+FG+): Z(ABCDEFG)=(2,3;5/32,5/32+1/128), Allowed
>> (0:ABC+DE+FG+)--(9:ABC+DE+FG): Z(ABCDEFG)=(3,2; _, _), Allowed
>> (0:A+BCD)--(9:AB+C+D+):
>>   Z(AB)  =(1,1; 1/2, 1/4), Not allowed
>>   Z(ABCD)=(1,3; 1/2, 1/4+1/8+1/16), Not allowed

I didn't find a problem where specifying and failing to specify that last
 preference makes a difference.

...
>Example (1 seat; 21 voters; Alternative Voting):
>
> 7 voters vote A > B > C > D.
> 6 voters vote B > A > C > D.
> 5 voters vote C > B > A > D.
> 3 voters vote D > C > B > A.
>
> Candidate A is elected. But if the three DCBA voters had
> truncated their votes and had voted only for candidate D,
> then candidate B would have been elected.

There doesn't seem to be a problem having B win there.

The example you gave involves remarking of papers which is certainly
 is not permitted when your/my "participation axiom" is being applied.

Anyway, the alteration given is:                (DCB {A+)--(D {B+);
  that alteration is fully permitted by (P1).
This alteration would not be permitted by (P1): (DCB {A+)--(DC {B+).
That B must lose if the paper is marked (DC) seems OK to me, since
under my (P2) [unannounced] axiom:
   3 (D {B+)
; is identical to:
   1 (DA {B+)
   1 (DB+)      <- and this could explain why (D {B+) can have B win.
   1 (DC {B+)

Perhaps you would tell me where you got your "participation axiom"
 definition from. I recall that Mr Ron Holzman, in Israel, published a
 participation axiom definition. I can't recall anything about the
 definition particularly. (He has a website currently. It has no
 mention of voting. There was a bit on citations of graph theory
 publications by Mr Holzman.)

>
>Could you -please- explain why Alternative Voting meets your
>multiwinner participation axiom so that Alternative Voting is
>not "too defective to be used in practice" (20 Oct 1999)?

I do not know if the AV is passed or failed by the participation
 axiom. It will be passed in the 3 candidates case. An interesting
 question could be: assuming (P1) and/or the "participation axiom",
 deduce the values of quotas, for a AV method that was modified so
 that it used quotas (for losers). Some easy inductive argument
 maybe, or a computer simulation done in a day if it takes more
 than a few hours to solve the problem. Lord Alexander wasn't
 impressed with the method, and nor was Winston Churchill.


>
>Markus Schulze


I read Mr Cachpole's  response. The connection between the
 "regularity" definition and the "probability" idea is nearly no
 connection at all. You just happen to be have them in mind at the same
 time, plus they were written about at the same time. You did not prove
 a connection and none exists. It is a weak form of rebuttal: to connect
 on failing idea to another when they are in truth separate.

If AV and FPTP and Condorcet and Borda and IFPP and would have
 candidate A win, then what are the probabilities for candidate A
 winning in your method: 0.96?. Just how small do those numbers (that
 ought be 1) get?.
 

Craig Carey
Auckland, 14 Dec 1999