The Single Transferable Vote: A Model Election
Our sample election uses a hypothetical 5-member constituency of 48 voters. In this case there are seven candidates
from two political parties, A and B.
Stage I - Counting the Votes
The Returning Officer sorts and counts the ballot papers according to the candidates marked as first
preference. Here we are concerned with 48 electors, and 48 votes. First preferences
for each candidate are represented by the number "1" on the ballot paper, and
are indicated in the diagram below by appropriate letters. Imagine the voter lining up to
stand beside their favourite candidate.
When the votes have been totalled, the Returning Officer calculates the quota needed for election, that is,
the smallest number of votes a candidate needs in order to be elected.
To calculate the quota, the Returning Officer divides the total number of valid votes (in this case 48) by one
more than the number of members to be elected (in this case 5+1=6) to make a total (in
this case 8.00). This total is exact to two places of decimals; otherwise it would then be
rounded up to the next highest 0.01.
In this case, the quota is 8 votes exactly.
Single Transferable Vote Election: Stage I
| Candidates |
Party |
Votes |
Totals |
Results |
| Erin |
A |
EEE |
3 |
-- |
| Lance |
B |
LLLL LLLL |
8 |
Elected |
| Peter |
A |
PPPP PPPP PPPP PPPP |
16 |
Elected |
| Rodney |
B |
RRRR |
4 |
-- |
| Shaw |
A |
SSS |
3 |
-- |
| Todd |
B |
TTTT TT |
6 |
-- |
| Vann |
A |
VVVV VVVV |
8 |
Elected |
| -- |
-- |
-- |
48 |
-- |
Three candidates, Peter, Lance and Vann have already obtained the quota of votes, and are therefore declared elected at Stage I.
Stage II - Transfer of Peter's surplus votes
Peter has 16 votes, but only needs 8 to be elected. His surplus, the equivalent of 8 votes, must be transferred, i.e. he does
not require eight-sixteenths of his votes. All sixteen of "Peter" first
preference ballot papers are examined and transferred, each at a value of 8/16th, or 0.5
of a vote, to the next available preference. In the case of a second preference for Lance
or Vann, (already elected) the paper is passed to the third preference (or fourth if
second and third preferences are for both).
If we suppose that 10 of the 16 papers show a next available preference for Shaw, and 6 for Erin, then Shaw
receives 5 (10 votes x 0.5) votes, and Erin receives 3 (6 votes x 0.5) votes from the
surplus of 8. The transfer of this surplus constitutes Stage II of the Count. Shaw now has
8 votes, equal to the quota, and is elected.
Single Transferable Vote Election: Stage II
| Candidates |
Party |
Votes |
Totals |
Results |
| Erin |
A |
EEE PPP |
6 |
-- |
| Lance |
B |
LLLL LLLL |
8 |
Elected |
| Peter |
A |
PPPP PPPP |
8 |
Elected |
| Rodney |
B |
RRRR |
4 |
-- |
| Shaw |
A |
SSS PPPP P |
8 |
Elected |
| Todd |
B |
TTTT TT |
6 |
-- |
| Vann |
A |
VVVV VVVV |
8 |
Elected |
| -- |
-- |
-- |
48 |
-- |
Stage III - Exclusion of Rodney
As there are no further surpluses to be distributed, the candidate who now has fewest votes, Rodney, is excluded, and all his
votes are transferred to the next available preferences of all those who voted for him. We
may suppose that all four of Rodney's votes pass to Todd. Any with preferences for Lance
or other elected candidates are passed to the next preference not already elected.
Exclusion of this candidate constitutes stage III of the count.
Single Transferable Vote Election: Stage III
| Candidates |
Party |
Votes |
Totals |
Results |
| Erin |
A |
EEE PPP |
6 |
Runner-up |
| Lance |
B |
LLLL LLLL |
8 |
Elected |
| Peter |
A |
PPPP PPPP |
8 |
Elected |
| Rodney |
B |
-- |
-- |
-- |
| Shaw |
A |
SSS PPPP P |
8 |
Elected |
| Todd |
B |
TTTT TT RRRR |
10 |
Elected |
| Vann |
A |
VVVV VVVV |
8 |
Elected |
| -- |
-- |
-- |
48 |
-- |
Lance and Todd, on attaining the quota, are declared elected.
As five candidates now have the
quota, the count is complete. Erin is runner-up.
Apart from the 3 votes which Erin
obtained in Stage I, 45 of the 48 electors have been effective in helping to elect a
candidate. Party A with 30 votes has 3 seats, Party B with 18 votes has 2 seats, giving
proportional representation to the main streams of opinion.
Summary
The successive stages
of the model election just discussed are exhibited by the Returning Officer on a Result
Sheet.
| Candidates |
Party |
Stage I |
Stage II |
Stage III |
|
| Erin |
A |
3 |
+3 |
6 |
|
6 |
Runner-up |
| Lance |
B |
8 |
|
8 |
|
8 |
Elected |
| Peter |
A |
16 |
-8 |
8 |
|
8 |
Elected |
| Rodney |
B |
4 |
|
4 |
-4 |
-- |
|
| Shaw |
A |
3 |
+5 |
8 |
|
8 |
Elected |
| Todd |
B |
6 |
|
6 |
+4 |
10 |
Elected |
| Vann |
A |
8 |
|
8 |
|
8 |
Elected |
| Total |
|
48 |
|
48 |
|
48 |
|
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