Australian Senate Counting

A Bird's Guide

The Australian Senate uses a preferential and proportional election system to elect Senators from each Australian state and mainland territory. The technical name for this election system is the Single Transferable Vote.

Each state and territory holds their own election for the Australian Senate in which multiple candidates are elected to fill a number of available seats.

These elections are preferential—we rank parties and candidates from most to least preferred. They are also proportional in that the number of seats awarded to a party is in proportion to the number of votes they receive.

Australia is not the only country to use the Single Transferable Vote...

The Bird Republic—a fictional island off the coast of north-east Australia—recently adopted the Single Transferable Vote to elect candidates to the upper house of its parliament, the Aviary.

The Aviary is a bicameral parliament with two houses. Its upper house—the Nest—has five seats up for election.

Australian Senate voting and counting rules are used for Nest elections. Our reporters have provided this account of the republic's most recent Nest election.

The Ballot

Voters get the following ballot: one column per party, with each party's candidates listed in a fixed order.

A ballot with three boxes above a line, one for each party. Under the line, the candidates of each party are listed in order, with boxes next to each candidate.

This arrangement of boxes above a line, next to parties, and boxes below a line, next to candidates, is the way Australian Senate ballots are designed.

There are two ways to fill it out. The first is to number the boxes above the line and the second to number the boxes below the line.

Voting above the line

Let's say a voter numbered the boxes above the line as shown below.

Above the line voting for the Nest. The voter placed a 1 in the box for the Water-birds, a 2 in the box for the City-dwellers,
and a 3 in the box for the Raptors.

The voter prefers all candidates in the Water-birds party the most, then the City-dweller candidates, and then the Raptor candidates.

This ballot expresses the following preference ordering over the candidates.

How an above the line vote is interpreted. Description follows image.

As the Water-birds have been ranked first above the line, the voter's first four preferences go to the Water-birds candidates in the order they are listed: the Pelican (first), the Black Swan (second), the Brolga (third), and the Dusky Moorhen (fourth). The City-dwellers are ranked second, so their candidates take the next four preferences: the Magpie (fifth), the Kookaburra (sixth), the Sulfur-crested Cockatoo (seventh), and the Rainbow Lorikeet (eighth). The Raptors' candidates take the last two: the Powerful Owl (ninth) and the Wedge-tailed Eagle (tenth).

Voting below the line

Sometimes, a voter might like certain candidates in a party better than others. They can express these preferences by voting below the line.

Here is an example of a below the line vote in the Nest election.

Below the line voting for the Nest. Description follows image.

In the above ballot, the voter prefers the Rainbow Lorikeet from the City-dwellers the most out of all the candidates, and then the Black Swan from the Water-birds. The only way the voter can express these preferences is by numbering all the candidates below the line.

The voter has the following preference ordering over candidates, from most to least preferred: Rainbow Lorikeet; Black Swan; Magpie; Kookaburra; Powerful Owl; Pelican; Sulfur-crested Cockatoo; Wedge-tailed Eagle; Brolga; and then the Dusky Moorhen.

Election Day

On election day, 150 birds fill out their ballots for the Nest. All birds fill out their ballot correctly, and choose to vote above the line.

The table below shows every ranking that appeared on at least one of the ballots cast in this election (first column) and how many ballots were cast with each of these rankings (second column).

Preference ranking	Number of ballots
City-dwellers then Water-birds then Raptors	50
Water-birds then City-dwellers then Raptors	32
Water-birds then Raptors then City-dwellers	15
City-dwellers then Raptors then Water-birds	10
Raptors then City-dwellers then Water-birds	23
Raptors then Water-birds then City-dwellers	20

The Count

Counting the votes in a Single Transferable Vote election involves a number of steps.

The first key concept to understand is that of the quota. The quota for the five-seat Nest election is 26 votes.

What is a quota?

The quota of a Single Transferable Vote election is a threshold. It is the number of votes that a candidate needs to have to get a seat (i.e., to be one of the winners).

The quota can be calculated in different ways, but the most common approach, and the one we use for Australian Senate and Nest elections is shown below.

Formula used to compute the quota. Description follows image.

Divide the valid ballots by one more than the number of seats, round down, and add one. For the Nest: 150 divided by 6 equals 25, plus one—a quota of 26.

Now that we have the quota, the next step is to establish the initial tallies for each candidate.

When a voter has numbered candidates below the line, we put the ballot in the tally pile of the candidate who has been ranked first, with a number '1' in their box.

But how does this work when voters have ranked parties above the line? Who do these ballots belong to?

Let's look at one of the ballots that was cast in our election.

A ballot ranking the Water-birds first,
City-dwellers second and the Raptors third

At the start of counting, an above the line ballot belongs to the tally pile of the first listed candidate in the party that has been ranked first with a number 1 in its box.

This means the ballot above will initially be placed in the tally pile of the Pelican, as they are the first listed candidate in the most preferred party, the Water-birds.

Step One: Initial Tallies

Give each ballot containing a below the line vote to the candidate who is ranked first on the ballot, with a number '1' in their box.

Give each ballot containing an above the line vote to the first listed candidate in the party that has been ranked first on the ballot, with a number '1' in its box.

Each ballot is worth one vote, and the number of ballots in a candidate's tally pile is their initial tally.

Initial tallies of candidates in the Nest election.
Text-based explanation follows image.

Text-based explanation of candidate tallies

The 50 ballots that prefer the City-dwellers first, the Water-birds second, and the Raptors third, are placed in the Magpie's tally pile, along with the 10 ballots that prefer the City-dwellers first, the Raptors second, and the Water-birds third. As each ballot is worth one vote, the Magpie has 60 votes in their tally. The other City-dweller candidates, the Kookaburra, Sulfur-crested Cockatoo, and Rainbow Lorikeet, have no votes in their tally.

The Pelican, as the first listed Water-bird, is given the 32 ballots that rank the Water-birds first, the City-dwellers second, and the Raptors third, along with the 15 ballots that rank the Water-birds first, the Raptors second, and the City-dwellers third.

The Pelican has 47 votes in their tally, while the Black Swan, Brolga, and Dusky Moorhen have no votes in their tallies.

The Powerful Owl, as the first listed Raptor, is given the 23 ballots that rank the Raptors first, the City-dwellers second, and the Water-birds third, along with the 20 ballots that rank the Raptors first, the Water-birds second, and the City-dwellers third.

The Powerful Owl has 43 votes in their tally, while the Wedge-tailed Eagle has no votes in their tally.

Now that we have initial tallies for our candidates, we can explain the two key steps that we repeat throughout the rest of the counting process.

Step Two: Seating

Seat all candidates whose tallies are equal to or greater than the quota. These candidates are winners.

Adjust the value of the ballots in the tally pile of each winner, and give those ballots, with their new values, to the next highest ranked eligible candidate on the ballot.

A candidate is eligible to receive votes if they have not been seated or removed from the contest, and their tally is less than the quota.

What do we mean by adjusting the value of a ballot? And how do we work out who the next highest ranked candidate is? We will answer both questions shortly.

The first candidate in each of the three parties have tallies that are greater than the quota of 26 votes. According to Step Two, we give the first three of the five available seats to the Magpie, the Pelican, and the Powerful Owl.

We will seat these candidates in order of their tally, from highest to smallest.

Seating the Magpie, Pelican, and Owl

The Magpie has 60 votes in their tally, but only 26 of those are needed to get them elected.

What should we do with the other 34 votes? We are going to pass them on to their next highest ranked eligible candidate.

Who is the next highest ranked eligible candidate?

Let's look at one of the ballots in the Magpie's tally pile. It ranks the City-dwellers first, the Water-birds second and the Raptors third. It reads: Magpie, then Kookaburra, then Sulfur-crested Cockatoo, then Rainbow Lorikeet, and then the candidates of the other two parties.

We move down this list, starting just below the Magpie, until we find a candidate who is eligible—still in the contest, and without a quota. That candidate is the Kookaburra. If the Kookaburra had already been seated or removed, we would keep going down the list.

But how do we decide which votes to pass on?

We will pass all the ballots on, but will change how much they are worth so that the total value of the Magpie's tally pile becomes 34 votes.

When a candidate is seated, we work out the new value of each of the ballots in their tally pile.

The new value of each ballot is equal to the total number of votes in the candidate's
tally pile minus the quota all divided by the total number of ballots in the candidate's tally pile.

The Magpie has a tally of 60 votes, and 60 ballots in their tally pile. The new value of each of these 60 ballots is reduced to 0.5667. This value is called a transfer value. Under Australian Senate rules, transfer values are defined to four decimal places.

A formula stating that the new value of each of the Magpie's ballots is 60 minus 26 all
divided by 60 which equals 0.5667.

The Kookaburra is the next highest ranked candidate on all the Magpie's ballots. So, each of the Magpie's reduced-value ballots go to the Kookaburra.

The 60 ballots in the Magpie's tally pile,
now worth 34 votes in total, move to the Kookaburra. The Kookaburra's tally is 34 votes.

Now that we have stepped through what happens when a candidate is seated, let's look at what happens when the Pelican and Powerful Owl are seated.

Seating the Pelican

The Pelican has 47 votes in their tally—a surplus of 21 votes—so we adjust the value of their ballots.

Formula stating that the new value of each of the Pelican's ballots is
47 minus 26 all divided by 47 which equals 0.4468.

Each of the Pelican's ballots is now worth 0.4468 votes. There are two types of ballots in their pile, but the Black Swan is the next most preferred candidate on all of them.

The 47 ballots in the Pelican's tally pile move to the Black Swan, their total
value now 20 votes. The Black Swan's tally is 20 votes.

The Black Swan had no votes in their tally pile. Once the Pelican is seated, the Black Swan receives:

32 ballots ranking the Water-birds first, the City-dwellers second, and the Raptors third, valued at 14.2976 votes in total (32 × 0.4468 = 14.2976).
15 ballots ranking the Water-birds first, the Raptors second, and the City-dwellers third, valued at 6.702 votes in total (15 × 0.4468 = 6.702).

Together these are worth 20.9996 votes, which rounds down to 20—the Black Swan's new tally. Because transfer values are defined to four decimal places, and bundle totals are rounded down as they move between candidates, the Black Swan only gets 20 of the Pelican's 21 surplus votes.

Seating the Powerful Owl

The Powerful Owl has 43 votes in their tally—a surplus of 17 votes—so we adjust the value of their ballots.

Formula stating that the new value of each of the Owl's ballots is
43 minus 26 all divided by 43 which equals 0.3953.

Each of the Powerful Owl's ballots is now worth 0.3953 votes. There are two types of ballots in their pile, but the Wedge-tailed Eagle is the next most preferred candidate on all of them.

The 43 ballots in the Owl's tally pile move to the Wedge-tailed Eagle, their
total value now 16 votes. The Wedge-tailed Eagle's tally is 16 votes.

The Wedge-tailed Eagle had no votes in their tally pile. Once the Powerful Owl is seated, the Wedge-tailed Eagle receives:

23 ballots ranking the Raptors first, the City-dwellers second, and the Water-birds third, valued at 9.0919 votes in total (23 × 0.3953 = 9.0919).
20 ballots ranking the Raptors first, the Water-birds second, and the City-dwellers third, valued at 7.906 votes in total (20 × 0.3953 = 7.906).

Together these are worth 16.9979 votes, which rounds down to 16—the Wedge-tailed Eagle's new tally. Because of rounding, the Wedge-tailed Eagle gets only 16 of the Owl's 17 surplus votes.

Let's take a look at the candidates' tallies after these three seatings.

Tallies of each candidate after the Magpie, Pelican, and
Powerful Owl have been seated. Text-based description follows image.

Text-based description of candidate tallies

There are now 60 ballots in the Kookaburra's tally pile with a total value of 34 votes. The Kookaburra now owns the 50 ballots ranking the City-dwellers first, the Water-birds second and the Raptors third, and the 10 ballots ranking the City-dwellers first, the Raptors second, and the Water-birds third.

There are now 47 ballots in the Black Swan's tally pile with a total value of 20 votes. The Black Swan now owns the 32 ballots ranking the Water-birds first, the City-dwellers second, and the Raptors third, as well as the 15 ballots ranking the Water-birds first, the Raptors second, and the City-dwellers third.

There are now 43 ballots in the Wedge-tailed Eagle's tally pile with a total value of 16 votes. The Wedge-tailed Eagle now owns the 23 ballots ranking the Raptors first, the City-dwellers second and the Water-birds third, as well as the 20 ballots ranking the Raptors first, the Water-birds second and the City-dwellers third.

There are now two seats left to fill. When seating the Magpie, enough votes were transferred to the Kookaburra to give them more than 26 votes.

We now repeat Step Two until we either fill up the 5 seats, or there is no candidate with a quota in their tally pile.

Seating the Kookaburra

The Kookaburra has 34 votes and 60 ballots in their tally pile. Their surplus is 8 votes. We seat the Kookaburra and work out the new value for each of the ballots in their pile.

A formula stating that the new value of each of the ballots in the Kookaburra's
tally pile is 34 minus 26 all divided by 60 which equals 0.1333.

The new value of each ballot is the Kookaburra's surplus (8 votes) divided by the number of ballots in their pile (60).

All the ballots in the Kookaburra's tally pile have the Sulfur-crested Cockatoo as the next most preferred candidate.

Each time a candidate is seated, all their ballots move on to the next listed candidate in their party. This is because everyone voted above the line in this election.

The Kookaburra's ballots are given to the Sulfur-crested Cockatoo at a value of 0.1333 votes each. Again, due to rounding, the Cockatoo gets 7 votes from the Kookaburra and not the full surplus of 8 votes.

The 60 ballots in the Kookaburra's tally pile move to the Sulfur-crested
Cockatoo with a total value of 7 votes. The Sulfur-crested Cockatoo's tally is now 7 votes.

The Sulfur-crested Cockatoo had no votes in their tally pile. Once the Kookaburra is seated, the Sulfur-crested Cockatoo receives:

50 ballots ranking the City-dwellers first, the Water-birds second, and the Raptors third, valued at 6.665 votes in total (50 × 0.1333 = 6.665).
10 ballots ranking the City-dwellers first, the Raptors second, and the Water-birds third, valued at 1.333 votes in total (10 × 0.1333 = 1.333).

Together these are worth 7.998 votes, which rounds down to 7—the Sulfur-crested Cockatoo's new tally.

Let's see what the tallies of all of our candidates look like now.

Tallies after the Kookaburra is seated. Text-based description follows image.

Text-based description of candidate tallies

There are now 60 ballots in the Sulfur-crested Cockatoo's tally pile with a total value of 7 votes. The Sulfur-crested Cockatoo now owns the 50 ballots ranking the City-dwellers first, the Water-birds second and the Raptors third, and the 10 ballots ranking the City-dwellers first, the Raptors second, and the Water-birds third.

There are still 47 ballots in the Black Swan's tally pile with a total value of 20 votes. The Black Swan now owns the 32 ballots ranking the Water-birds first, the City-dwellers second, and the Raptors third, as well as the 15 ballots ranking the Water-birds first, the Raptors second, and the City-dwellers third.

The Magpie, Pelican, Powerful Owl and the Kookaburra have been seated. The Rainbow Lorikeet, Brolga, and Dusky Moorhen all have 0 votes with no ballots in their tally piles.

We now have one seat left to fill, but no candidates with a quota in their tally, apart from the ones that we have already seated!

We can now move on to Step Three of the counting process.

Step Three: Removing Candidates

If no candidate has a tally that is equal to or greater than a quota, remove the candidate with the smallest tally.

All ballots in this candidate's tally pile are given to their next most preferred eligible candidate at their current value.

There are three candidates with the smallest tally of 0 votes. How do we know which candidate should be removed?

Removing the Lorikeet, Brolga, and Moorhen

We repeat Step Three three times, removing the Rainbow Lorikeet, Brolga, and Dusky Moorhen. As these candidates have no votes, removing them doesn't change anyone else's tally.

So, whichever tie breaking rule is used, we get to the following situation.

Tallies after removing the Rainbow Lorikeet, Brolga, and Dusky Moorhen from the contest.
Text-based description follows image.

The Sulfur-crested Cockatoo, Black Swan and the Wedge-tailed Eagle are now the only candidates remaining, and we have one seat to fill. The tally piles of these candidates have not changed.

The Sulfur-crested Cockatoo still has 60 ballots in their tally pile with a total value of 7 votes, the Black Swan has 47 ballots in their pile with a total value of 20 votes, and the Wedge-tailed Eagle has 43 ballots in their pile with a total value of 16 votes.

As none of these remaining candidates has a quota, we can't perform Step Two. So, we need to keep repeating Step Three until either only one candidate is left, or one of the candidates gets a quota.

Removing the Sulfur-crested Cockatoo

The Sulfur-crested Cockatoo has the smallest tally at 7 votes, and so they will be removed from the contest.

Let's have a look at the ballots in the Sulfur-crested Cockatoo's tally pile.

Tally pile of the Sulfur-crested Cockatoo. The pile contains 50 ballots ranking
the City-dwellers first, Water-birds second, and Raptors third, valued at 0.1333 each.
The pile also contains 10 ballots ranking the City-dwellers first, Raptors second, and
Water-birds third, valued at 0.1333 each.

The next most preferred candidate on all the Sulfur-crested Cockatoo's ballots is the Rainbow Lorikeet. But the Rainbow Lorikeet is no longer in the contest! We need to work our way down the preference list until we find a candidate that is still in the contest, and has not already been seated.

For the 50 ballots ranking the Water-birds party second, the next most preferred candidate who is still around is the Black Swan. So, these ballots go to the Black Swan, retaining their value of 0.1333 each.

For the 10 ballots ranking the Raptors party second, the next most preferred candidate who is still around is the Wedge-tailed Eagle. So, these ballots go to the Wedge-tailed Eagle retaining their value of 0.1333 each.

The total number of votes that move from the Sulfur-crested Cockatoo to the Black Swan is
equal to 50 times 0.1333 all rounded down, which is 6 votes. The total number of votes that
move from the Sulfur-crested Cockatoo to the Wedge-tailed Eagle is 10 times 0.1333 all rounded down,
which is 1 vote.

Now let's look at the tally piles of the two remaining candidates, the Black Swan and the Wedge-tailed Eagle, after the Sulfur-crested Cockatoo was removed.

The Black Swan has a tally of 26 (20 + 6, exactly the quota) and the Wedge-tailed Eagle on 17 (16 + 1).

We now have a candidate that has reached the quota!

(Back to) Step Two: Seating the Black Swan

We elect the Black Swan to the last seat. As there are no more seats to fill, the counting process stops.

Summarising the Steps

To find the winners of a Single Transferable Vote election we repeat the steps of electing candidates when they have a quota, and removing candidates when no one has a quota.

We start with Step One: Initial Tallies, below, working out the starting tallies of all our candidates.