Tuesday, 02 June 2015 14:31

Single transferable vote (STV) – an example

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Hello last week I was involved in a conference, where I had to vote via procedure called single transferable vote, for the first time in my life. Well the procedure is over a 100 years old and now days is broadly used in Australia, New Zeeland, Ireland and Malta, but being new for me I decided to learn something more about how it works. In Finland there is almost no literature about, so I developed the example bellow on the basis of what I’ve learned from the net. Verbally that’s an iterative voting process for multiple choices, where the excessive and the lacklustre votes are redistributed according to the ordered preferences of the voters. The qualifying quota is defined on the basis of the voters and the seats.

Let’s assume we have 4 applicants for 2 seats and 24 voters. Then the threshold (quota) , calculated by the formulae proposed by Newland and Britton in 1973 (there are also other options), would be V/(s+1)=24/(2+1) =8 votes, where “V” is the number of voters and “s” the number of seats. In a real vote you are not requested to arrange all candidates, but let’s see in how many ways we could arrange the 4 candidates assuming all possible ways, that’s 4!=4*3*2*1=24 or the number of permutations (arranged order) of the four options. Well for those of you who will perceive it better visually hear I made a table:

 ballot

I

II

II

IV

1

a

b

c

d

2

a

b

d

c

3

a

c

b

d

4

a

c

d

b

5

a

d

b

c

6

a

d

c

b

7

b

a

c

d

8

b

a

d

c

9

b

c

a

d

10

b

c

d

a

11

b

d

a

c

12

b

d

c

a

13

c

a

b

d

14

c

a

d

b

15

c

b

a

d

16

c

b

d

a

17

c

d

a

b

18

c

d

b

a

19

d

c

a

b

20

d

c

b

a

21

d

a

c

b

22

d

a

b

c

23

d

b

c

a

24

d

b

a

c

As I said that’s the full number of possible options, which in real elections will or may not be reached (some of the 24 options may not be chosen, some of the voters could put less than 4 options in the ballot paper, some of the ballots could be invalid or so). Well let’s assume that our 24 voters made the following 4 preferences:

Raw/ballot

I

II

II

IV

Voters

4

a

c

d

b

5

10

b

c

d

a

3

12

b

d

c

a

4

16

c

b

d

a

3

17

c

d

a

b

3

22

d

a

b

c

6

 

 

 

 

Sum

24

Then if we arrange the choices according to their first preference we get:

Candidates

Voters

 

b

7

3 and 4

c

6

3 and 3

d

6

 

a

5

 

Or our favourite is “b” with 7 first choices, but none of the candidates has passed the threshold of 8. Well in such case we drop the candidate with least votes off, in our case that’s “a”, and redistribute the votes among the other candidates. We also drop the 5 voters, who put “a” as their first choice, hence we have to recalculate the quota: 19/(3+1)=4,75. That way “b” automatically qualifies with 2,25 excessive votes (7-4,75), but let’s see who comes next since “c” and “d” have 6 votes each.” The rule for weighting the count of the transferable votes is: transferable votes minus quota equals surplus, and the surplus is divided by the transferable votes. (This is Gregory's method or the Senatorial Rules, named after the use of STV in various Commonwealth senates)” (http://www.voting.ukscientists.com/stvcount.html ). So since the votes of both “c” and “d” is 6 it’s obvious that whatever equal number or fraction we add to the base, the equality will remain, but I will do it just to show the principle of redistribution. So with the recalculated quota (4,75) our transferable votes (5) give a surplus of 1/4 and transferable weight would be (¼)/5, which is 1/20 or 0,05. Therefore the new values for “c” and “d” will be 6,05 both (that’s if 1 vote is assumed redistributed) and we have to eliminate again one of them, but which one. Well in such case we could simply chose them randomly by a lot or if accepted as a general rule we could follow the order of the initially eliminated permutation a>c>d>b, where the 5 voters ranked “c” second and “d” third or arithmetically following that we could redistribute to “c” 3 votes (6+(3/20)) =6,15 and to “d” 2 votes (6+(2/20))=6,1 hence final winners are “b” and “c”. There are several different methods for redistributing the votes and you could read about them here:

http://en.wikipedia.org/wiki/Counting_single_transferable_votes and here a few more good links:

http://www.utu.fi/fi/yksikot/soc/yksikot/pcrc/huippu/julktoiminta/tyopaperit/Documents/A%20new%20method_42.pdf  and

http://www.hawaii.edu/uhmfs/documents/2010_11/20110216_reso_votingmethodscounting.pdf

Kind regards

Rosti

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