[EM] Random and reproductible tie-breaks

[Raph Frank]

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.