[EM] Random and reproductible tie-breaks

Raph Frank raphfrk at gmail.com
Thu Sep 25 11:27:44 PDT 2008


On Thu, Sep 25, 2008 at 7:10 PM, Stéphane Rouillon
<stephane.rouillon at sympatico.ca> wrote:
> 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.

Approval is clone-independent, but the ties wouldn't be.

However, I do accept your point about the just taking the count that
resulted in the tie would mean that everyone is equal, as it is a tie.

Also, if there was a tie, then odds are that there is only 2 people
involved, so it doesn't really matter about clones.

My suggestion went back to all the ballots to simulate a random ballot
process, so the number of ballots which gives each member of the tie a
win, is not necessarily equal.

>
> 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.

I seem your point, if the first digit is 7, then the results would be

2 in tie: 1
3 in tie: all possible
4 in tie: 1 or 3
5 in tie: 2

Hmm, maybe the 2 most significant digits should be excluded, but that
gives less bits.



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