[EM] Meek-based maths for equal ranked PR-STV
Raph Frank
raphfrk at gmail.com
Wed May 6 05:19:07 PDT 2009
Just expanding on concept from the other PR-STV thread. There is an
issue of how exactly to handle equal rankings.
One suggestion is made by Warren in voting matters
http://www.votingmatters.org.uk/ISSUE7/P5.HTM
A vote of
A=B>C
is converted into
0.5: A>B>C
0.5: B>A>C
This creates an effective keep factor that is different from the the
ordinary keep factor.
In the above example, the votes for A are
0.5*ka (from the first half vote)
0.5*(1-kb)*(ka) (from the 2nd half vote)
giving an effective keep factor of
ka_eff = 0.5*(ka + (1-kb)*ka)
ka_eff = 0.5*(ka + ka - kb*ka)
ka_eff = ka * ( 1 - kb/2 )
For 3 candidates, the result is (unless I messed up)
ka_eff = ka*[2 + (1-kb) + (1-kc) + 2*(1-kb)*(1-kc)]/6
The more candidates, the more complex the formula. I think the terms
are likely to be equal to 2^(N-1), as with N candidates you need to
multiply kx and ky for all combinations of x and y.
As, a possible simplified, but less accurate, method, you could use
the following.
Fairness rule 1: A=B>C should pass the same votes to C as A>B>C if A
and B have the same keep factors.
With multiple candidates (ka to kz) at the rank, the amount passed
onto the next rank is always
(1-ka)*(1-kb)*...*(1-kz)
no matter how you order a to z.
That means that there is always
Share = 1 - (1-ka)*(1-kb)*...*(1-kz)
available to share between all the candidates at the current rank.
Fairness rule 2: The vote weight is shared in proportion to k for each candidate
Thus the effective keep factors would be:
ka_eff = ka*Share/(sum(ka ... kz)
This gives
ka_eff = ka*Share/(ka+kb) for 2 candidates
and
ka_eff = ka*Share/(ka+kb+kc) for 3 candidates and so on
This seems reasonably fair and doesn't give an advantage to those who
equal rank.
The convergence theorem for Meek's method assumes that reducing ka by
30% will reduce A's votes by 30%. However, this is not the case with
this formula.
If ka is reduced by 30%, then all 3 parts of the formula change, so it
isn't linear.
However, this doesn't break the convergence rule. If ka is decreased
by 30% then,
ka will decrease by 30%
Share will increase (slightly)
sum(ka ... kz) will decrease (slightly)
Thus, ka*Share/(sum(ka ... kz) will decrease by (slightly) less then 30%.
This means that if a candidate has 15% above the quota and you divide
his ka by 1.15, then he will still be above the quota, but nearer than
15%.
Thus you can still use the convergence rule, i.e go through each
elected candidate and multiply his keep factor by (quota)/(total vote
held by candidate).
--Elimination score--
Assuming that it is acceptable to have a different election and
elimination score, then the above would be the method of working out
each candidate's election score.
Elimination score could be worked out using the same formula, but if
there is more than 1 hopeful candidate at a rank, then each one
receives the sum of all the hopefuls' score at that rank as their
elimination score. This has the same effect as multiplying the score
received by each hopeful candidate by the total number of hopeful
candidates at that rank.
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