Hello Raph,
maybe I was not clear. I mean an equiprobable but determined tie-breaker.
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.
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
For
A : 1234
B : 1234
C : 1234
we obtain 1234 = 411 x 3 +1. Thus A wins.
For
A : 6542
B : 6542
C : 6542
we obtain 6542 = 2180 x 3 +2. Thus B wins.
For
A : 2539
B : 2539
C : 2539
we obtain 2539 = 846 x 3 +1. Thus A wins.
Using total turnout is less random and easier to manipulate for fraud during
a recount.
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.
From: "Raph Frank" <[EMAIL PROTECTED]>
To: "Stéphane Rouillon" <[EMAIL PROTECTED]>
CC: [EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: Re: [EM] Random and reproductible tie-breaks
Date: Thu, 25 Sep 2008 11:25:24 +0100
On 9/25/08, Stéphane Rouillon <[EMAIL PROTECTED]> 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.
----
Election-Methods mailing list - see http://electorama.com/em for list info
