[EM] Random and reproductible tie-breaks

[Stéphane Rouillon]

Yes euclidian division means the same as remainder using integer division.

Yes you have necessarily the same counts if you have a tie (you can use 
previous counts like first preference as tie breaker, but I focused on the 
case of an exact tie). Then use the current count which is necessarily a tie 
to resolve the tie. Outside information gives more opportunity to
modify the result. Using the actual count, one cannot add just one ballot to 
win the tie-breaker, because it would mean there is no more tie. If, as you 
suggest, outside information can be of some consequence, it is easier to 
come out with a new count for the last party (13 votes instead of 14)
to change the outcome between the 22 000 tie. However this is detail.

The tie-breaker is clone-dependent because the method is clone dependent. 
Just take a clone-independent method! The tie-breaker is not the problem.

You are free to define the index as you which, prior from knowing the 
results!

Reversing digits of the turnout is a bad idea because it is highly not 
equiprobable. If I know between 7100 and 7700 people are going to vote, I 
want the 1 as index in order to win tie-breaks... XXX7 = 2 . YYY3 + 1.

>From: "Raph Frank" <raphfrk at gmail.com>
>To: "Stéphane Rouillon" <stephane.rouillon at sympatico.ca>
>CC: easmith at beatrice.rutgers.edu, election-methods at electorama.com
>Subject: Re: [EM] Random and reproductible tie-breaks
>Date: Thu, 25 Sep 2008 16:51:23 +0100
>
>On 9/25/08, Stéphane Rouillon <stephane.rouillon at sympatico.ca> wrote:
> >  maybe I was not clear. I mean an equiprobable but determined 
>tie-breaker.
>
>My question was what "... rest using euclidian divison ... " meant.
>
>I just wanted to confirm that it has the same meaning as "...
>remainder using integer division ... ".
>
> >
> >  First if we want to break a tie. Using your data,
> >
> > > A wins: 1234
> > > B wins: 6542
> > > C wins: 2539
> > >
> >  B wins and no need for any tie-breaker.
>
>Hmm, not necessarily.  In some election methods, you could have a tie,
>where the candidates have different first choice counts.
>
>Ofc, some ranked methods do use the first choice count as the tie
>breaker in that case.  However, it isn't clone independent.
>
>Lets say you had an approval election
>
>50: A approved
>50: B approved
>
>Result
>A: 50
>B: 50
>
>This is a tie and both get 50% chance of winning, if you add a clone
>of A, you get
>
>50: A1 + A2 approved
>50: B approved
>
>Result:
>A1: 50
>A2: 50
>B: 50
>
>This is also a tie, but the fair way to spilt the odds would be B:
>50%, A1: 25%, A2: 25%
>
> >
> >  However to determine a winner in case for triple equality
> >  I would proceed like that:
> >  1) establish an index for winner using alphabetical order (A is 1, B is 
>2
> > and C is 0)
> >  2) divide the tie value by the number of equally concerned candidates
>
>I would probably have indexed them A:0, B:1 and C:2, but that isn't a
>major issue.
>
>The effect is that that a single voter change in the total changes the
>result and so you get chaos and lack of predictability.
>
> >  Using total turnout is less random and easier to manipulate for fraud
> > during a recount.
>
>Well, I think that any total can be manipulated.
>
>By reversing the digits of the number, turnout becomes just as random
>as your proposal, e.g.
>
>Turnout = 1234
>
>becomes
>
>4321 / 9999 as the index.
>
> >  Tie-breakers have no need to be clone-independent. You cannot predict a 
>tie
> > will occur
> >  before the election and it would be ridiculous to strategize on this
> > hypothesis.
>
>In fairness, tie breaks are a low probability issue in any case.  If
>you are going to do them right, you might as well do them fully right.