[EM] Meek and Newland-Britton

[Chris Benham]

David,
You wrote (Sat.Sept.11):

>With regard to the example:
>
>A 61
>A>B 60
>C>D 50
>D>C 9
>
>for 2 seats, I'd be interested in knowing who list members think should win 
>and why.
>
CB: I haven't yet made up my mind what the best ranked-ballot PR method 
is yet, but if the ballots are symetrically completed,
and then those ballots that ranked A first are all  fractioanlly 
devalued by an amount that sums to a Droop quota (180/3 = 60),
then the winners are A and C.

Multiplying all the numbers by six, the symetrically completed ballots are:

61: A>B>C>D
61: A>B>D>C
61: A>C>B>D
61: A>C>D>B
61: A>D>B>C
61: A>D>C>B
180:A>B>C>D
180:A>B>D>C
150:C>D>A>B
150:C>D>B>A
27: D>C>A>B
27: D>C>B>A

With the numbers multiplied  again , this time by 121 (726 times the 
starting numbers), my numbers for the last seat look like:

14701:B>C
14701:B>D
03721:C>B
40021:C>D
03721:D>B
10255:D>C

C has more than half  the first prefernces, and so wins.

In a Dec.1994 something Woodall wrote, I've seen an arguement that says 
that in multi-winner STV it is better to handle truncation by
reducing the size of the quota (as ballots "exhaust") rather than 
symetrically complete them.

First election, for two seats:
40:A>B
02:B>A
12:C>D
06:D>C
The quota is 20, so  A and  B are neatly elected.

Second election, this time for three seats:
40:A>B
02:B>A
12:C>D
06:D>C
180:E
The quota is now 60, so E is easily elected.  If the ballots are 
symetrically completed, the other winners are A and C.
The arguement goes that compared to the first election the only changes 
are that there is an extra seat and a lot of new voters
who have expressed no desire to do anything other than have E fill that 
seat, and therfore it is absurd  that C should now be
elected instead of  B.

Chris Benham