[EM] Random and reproductible tie-breaks

[Raph Frank]

On 9/25/08, Stéphane Rouillon <stephane.rouillon at sympatico.ca> wrote:
> Hello Allen,
>
>  simply using the number of ballots involved in the tie is enough. Compare
> its rest using euclidian divison by the number of involved candidates to the
> alphabetical rank of the candidates.

Just to clarify, you mean divide the number of ballots by the number
of candidates involved in the tie and use the remainder (rest?) to
decide who wins (candidates sorted alphabetically)?

>  Simple, effective and greatly equiprobable. It works for winner selection
> as for elemination rounds.

However, it isn't clone independent.  Random ballot is used to achieve that.

What about reversing the digits in the number of ballots and using
that as an index.

For example, assuming a tie between A, B and C

Check each ballot to see who would win if that ballot is picked

A wins: 1234
B wins: 6542
C wins: 2539
Turnout: 10315

Reverse turnout: 51301
Max possible: 99999 (with 5 digits)

Rank used is ( Reverse / Max possible) * ( turnout - 1 )

= 51301/99999 * (10315 - 1) = 5291 (rounded to nearest)

A wins for 0 - 1233
B wins for 1234 - 7775
C wins for 7776 - 10314

Thus in this case B wins.  If the turnout had been 10319, then C would
have won as the rank would have been 9417.

It might require some tweaking to make it so that it gets the
probabilities exactly right.

Ofc, the process assumes that the number of ballots is reasonably
solid.  I am not sure that a tie could occur in a real election that
was solid over multiple recounts.  I think it would come down to which
ballots the courts includes/excluded rather than the official tie
breaking rule.

If the tie breaking rule needs multiple rounds, the turnout could be
squared and the result used for the 2nd round.  Aternatively, a
different base could be used to express the numbers.