In part 5, we presented the weighted inclusive Gregory method. In that part and all previous parts, during the distribution of preferences we have ‘skipped over’ any candidates who have already been elected, as we said that to give extra votes to any of those candidates would be a waste.

While it is true that to leave a candidate with more votes than the quota, while there are vacancies still to fill, would be wasteful, there is another option we could have taken, rather than skipping those preferences completely.

## Transfers to prior winners

Let's use again the example from parts 4 and 5, where the ballots cast were:

21 D1 > D2 > F > C1 > C2
19 D2 > D1 > F > C1 > C2
40 C1 > C2 > F > D1 > D2
6 C2 > C1 > F > D1 > D2
7 F > D1 > D2 > C1 > C2
6 F > D1 > C1 > C2 > D2
1 F > D1 (and no further preferences)
100 Total

In this part, we will present the standings using the ‘fractions of a vote’ model of the Gregory method. After the exclusion of Fish Owner, the standings were as previously discussed:

Dog Owner 1 35.000 Elected
Dog Owner 2 19.000
Cat Owner 1 26.000 Elected
Cat Owner 2 20.000
Fish Owner 0.000 Excluded
Total 100.000
Quota 26.000

Dog Owner 1 now has a surplus of 9.000 votes. Let's suppose, as in the weighted inclusive Gregory method, we will distribute this by examining all of Dog Owner 1's 35.000 votes. For most votes, it is clear who the next preference should be:

• For the 21 votes ‘D1 > D2 > F > C1 > C2’, the next preference is Dog Owner 2.
• For the 7 votes ‘F > D1 > D2 > C1 > C2’, the next preference is Dog Owner 2 as well.
• For the 1 vote ‘F > D1’, the ballot paper is non-transferable and should be set aside as exhausted.

However, for the 6 votes ‘F > D1 > C1 > C2 > D2’, the next preference after Dog Owner 2 is actually Cat Owner 1 – who has already been elected. In parts 4 and 5, we skipped this preference and proceeded directly to Cat Owner 2. What would happen, however, if we were to transfer these votes to Cat Owner 1, despite the fact that they have been elected already?

Applying the weighted inclusive Gregory methodology, we would obtain the following standings:

Dog Owner 1 26.000 Elected
Dog Owner 2 26.196 Elected
Cat Owner 1 27.542 Elected
Cat Owner 2 20.000
Fish Owner 0.000 Excluded
Exhausted 0.257
Lost to Rounding 0.005
Total 100.000
Quota 26.000

In this case, Dog Owner 2 still reaches the quota and is elected, and so the count is complete. However, suppose that this were not the case. Cat Owner 1 has now been pushed over the quota, again, and now has a surplus, again. This, one supposes, would need to be distributed. We will come to this shortly.

## Quota reduction

Before we deal with Cat Owner 1's new surplus, though, consider this: In this situation, 0.257 votes have been exhausted, and 0.005 votes have been lost to rounding. These, therefore, cannot contribute to any candidate.

Recall from part 2 that we defined the quota in this election as ‘What is the smallest number of votes, such that no more than 3 candidates can have that many?’. We calculated our quota of 26 based on having 100.000 total votes. However, now that some votes have been exhausted or lost to rounding, it seems only reasonable we should take this into account.

In our election at this stage, we have 100.000 − 0.257 − 0.005 = 99.738 votes remaining for the candidates to compete for. Applying the definition of the Droop quota, $Q = \left\lfloor\frac{V}{S + 1}\right\rfloor + 1$, using this new figure, we obtain $Q = \left\lfloor\frac{99.738}{3 + 1}\right\rfloor + 1 = \lfloor 24.9345\rfloor + 1 = 25$. We find that at most 3 candidates can have 25 votes, so 26 is actually now an overestimate!

In fact, now that we are counting to 3 decimal places using the Gregory method, a better bound would actually be 24.935, as this is a smaller number than 25, and yet still at most 3 candidates can have 24.935 votes. (Using this logic, our original quota should really have been calculated as 25.001.)

In any case, all 3 elected candidates actually have more votes than they need. Had there still been more vacancies to fill, it seems reasonable that we should also allow these surplus votes to be distributed.

## Dealing with recursion

One imagines that distributing the new surplus of one of these elected candidates could add more votes to another one of the elected candidates. Distributing that new surplus could create yet new surpluses and so on. Similarly, distributing one surplus could result in more votes exhausting, which might reduce the quota, creating new surpluses which need to be distributed, and so on ad infinitum.

Clearly, this recursive process could not be finalised by hand. We would need a new approach, leveraging the power of computers.

## Reimagining STV

In order to tackle this problem – of how to allow ballot papers to be transferred to already elected candidates – we need to reconceptualise the STV process. Specifically, we must move beyond the model of surpluses being distributed in stages. In Hare's original method, surpluses were not distributed in stages – rather, the process of distributing ballot papers ensured that surplus ballots were distributed as and when they arose (by declaring a candidate elected as soon as they reached the quota, and not giving them any further ballots). Can we come up with an equivalent model which uses the Gregory method rather than random transfers?

Let us assign to each candidate a keep value (kv), which will represent what fraction of each vote a candidate retains when the vote is distributed to that candidate. The remainder, 1 − kv, is then free to be distributed to the next preference.

For each excluded candidate, the keep value will be 0: an excluded candidate cannot retain any votes. Likewise, for each continuing candidate (candidate who is neither elected nor excluded), the keep value will be 1: a continuing candidate receives the full value of the vote. So far, this is essentially the same as the STV systems described so far.

The key innovation now is that an elected candidate's keep value can be a fraction between 0 and 1. The fraction of the vote corresponding to the keep value will be retained by the candidate, while the remainder will be distributed to the next preference. This models the surplus distribution process of the Gregory method – where some fraction of the vote remains with the elected candidate, and the remainder is distributed to next preferences as the surplus.

To explain how this all comes together, let's work through an example using the same ballots from parts 4 and 5.

## The first stage

At the start of the count, no candidates are elected or excluded, so every candidate has a keep value of 1.

We distribute the ballot papers by proceeding down the preferences, and allocating votes to the preferred candidates according to their keep value. For example, for the 21 votes ‘D1 > D2 > F > C1 > C2’, the first preference is for Dog Owner 1, who has a keep value of 1, so we allocate the entire vote to Dog Owner 1.

Naturally, since all candidates have keep value 1 at this stage, this simply equates to allocating every vote to its first-preference candidate:

Dog Owner 1 21.000
Dog Owner 2 19.000
Cat Owner 1 40.000 Elected (kv = 1.000000000)
Cat Owner 2 6.000
Fish Owner 14.000
Total 100.000
Quota 25.001

100.000 votes have been allocated to the candidates, and so the quota (as previously discussed, now calculated to 3 decimal places) is 25.001. Cat Owner 1 meets the quota and is declared elected.

## Distribution of surpluses with 1 elected candidate

Cat Owner 1 now has a surplus which must be distributed. In this system, we distribute surpluses not by fiddling with individual ballot papers, but simply by adjusting the elected candidates' keep values. As previously discussed, the distribution of one candidate's surplus may affect another candidate's surplus and/or reduce the quota, so the calculation of the correct keep values generally requires a computer.1 Suffice to say, however, that there is always a particular set of keep values which can be calculated, which distributes all elected candidates' surpluses and is consistent with any reduced quota.

It is also worth noting that, even though the calculation of keep values requires a computer, the computer's calculation can easily be verified by hand, which we will do now. In this situation, the correct keep value for Cat Owner 1 is 0.625025000.2

We now imagine distributing the ballot papers again from scratch, but using this new keep value for Cat Owner 1. Looking at the 40 votes ‘C1 > C2 > F > D1 > D2’:

• 0.625025000 of each vote will be retained by Cat Owner 1, for a total of 40 × 0.625025000 = 25.001 votes
• The remainder, 0.374975000 of each vote, will be distributed to the next preference, Cat Owner 2, for a total of 40 × 0.374975000 = 14.999 votes

All of the other votes continue to be distributed to their first preferences:

Dog Owner 1 21.000
Dog Owner 2 19.000
Cat Owner 1 25.001 Elected (kv = 0.625025000)
Cat Owner 2 20.999
Fish Owner 14.000
Total 100.000
Quota 25.001

We can see, now, that Cat Owner 1's surplus has been distributed, and they have been left with exactly a quota of votes.

## Exclusion of candidate

All surpluses have now been distributed, and only 1 candidate has been elected, so as usual, we now exclude the candidate with the fewest votes (Fish Owner).

Recall from earlier that, in this system, exclusion is achieved by setting the candidate's keep value to 0. We now, again, imagine distributing the ballot papers from scratch, but with Fish Owner 1's keep value set to 0.

For example, for the 7 votes ‘F > D1 > D2 > C1 > C2’, the first preference, Fish Owner, has a keep value of 0, so Fish Owner is distributed none of those votes, and the entire value flows to the next preference, Dog Owner 1.

The standings then are:

Dog Owner 1 35.000 Elected (kv = 1.000000000)
Dog Owner 2 19.000
Cat Owner 1 25.001 Elected (kv = 0.625025000)
Cat Owner 2 20.999
Fish Owner 0.000 Excluded
Total 100.000
Quota 25.001

Dog Owner 1 now reaches the quota and is elected.

## Distribution of surpluses with multiple elected candidates

Dog Owner 1 now has a surplus which must be distributed. As discussed earlier, in a system which permits ballot papers to be transferred to candidates who have already been elected, this will cause some votes to be transferred to Cat Owner 1, and will reduce the quota as ballot papers exhaust – both of which will produce further surpluses which must be distributed.

Since this process cannot be easily completed by hand, we use a computer to calculate the correct keep values that will fully distribute the surpluses of Dog Owner 1 and Cat Owner 1.1 These are 0.712257142 for Dog Owner 1, and 0.597438692 for Cat Owner 1.

As mentioned previously, however, it is very much possible to verify by hand that these are the correct keep values. We will, yet again, imagine distributing the ballot papers from scratch, but using the new keep values.

Take, for example, the 6 votes ‘F > D1 > C1 > C2 > D2’:

• The first preference, Fish Owner, has a keep value of 0, so will receive no votes.

• The second preference, Dog Owner 1, has a keep value of 0.712257142, so will receive 0.712257142 of each vote, for a total of 6 × 0.712257142 = 4.273542852. 0.287742858 of each vote remains to be transferred to next preferences, for a total of 6 × 0.287742858 = 1.726457148.

• The third preference, Cat Owner 1, has a keep value of 0.597438692, so will receive 0.597438692 of each remaining vote, for a total of 1.726457148 × 0.597438692 = 1.031452300. 0.402561308 of each remaining vote remains to be transferred to next preferences, for a total of 1.726457148 × 0.402561308 = 0.695004847.

• The fourth preference, Cat Owner 2, has a keep value of 1, so will receive the entire 0.695004847 remaining votes.

We can proceed analogously to work out how much of each of the votes should be credited to each candidate:

Votes credited to Votes Preferences Dog Owner 1 Dog Owner 2 Cat Owner 1 Cat Owner 2 Fish Owner 21 D1 > D2 > F > C1 > C2 14.957399982 6.042600018 19 D2 > D1 > F > C1 > C2 19.000000000 40 C1 > C2 > F > D1 > D2 23.897547680 16.102452320 6 C2 > C1 > F > D1 > D2 6.000000000 7 F > D1 > D2 > C1 > C2 4.985799994 2.014200006 0.000000000 6 F > D1 > C1 > C2 > D2 4.273542852 1.031452300 0.695004847 0.000000000 1 F > D1 0.712257142 0.000000000 0.287742858

Adding up the number of votes to credit to each candidate, and rounding to 3 decimal places:

Dog Owner 1 24.929 Elected (kv = 0.712257142)
Dog Owner 2 27.057 Elected (kv = 1.000000000)
Cat Owner 1 24.929 Elected (kv = 0.597438692)
Cat Owner 2 22.797
Fish Owner 0.000 Excluded
Exhausted 0.288
Total 100.000
Quota 24.929

We can see that 0.288 votes have been exhausted, leaving 99.712 votes available for the candidates to compete for. Recalculating the quota, $\frac{99.712}{3+1}$ = 24.928, and so the quota (being the smallest number of votes such that at most 3 candidates can have that many) is 24.929. We can confirm that the 2 previously elected candidates, Dog Owner 1 and Cat Owner 1, have had their surpluses fully distributed, and are left with exactly a quota of votes. This confirms that the keep values were computed correctly!

Dog Owner 2 now meets the quota, and is elected to the final vacancy, completing the count.

The method presented above is the Meek method of counting STV elections, and is used in New Zealand for some local government elections.

## Benefits of the Meek method

Compared with the Gregory method and other methods previously presented, the Meek method has a number of distinct advantages.

Firstly, because transfers can be made to already elected candidates, the Meek method treats all voters more equally. Whether a preference for a candidate is counted does not depend on whether the preference is reached (somewhat arbitrarily) before or after the candidate reaches the quota.3

As a corollary, whenever a candidate is excluded under the Meek method, the count proceeds as if that candidate had never stood. Contrast this with the situation under previously described STV methods, where the distribution of an excluded candidate's preferences will depend on which candidates had previously been elected.

This problem, which is solved by the Meek method, is illustrated particularly clearly by exclusive Gregory and random transfer STV systems: In those systems, if no candidate reaches the quota on first preferences, and ballot papers from an excluded candidate allow a candidate to be elected, only the excluded candidate's votes (the last parcel) will contribute to the surplus distribution – even though, had the excluded candidate not run in the election, all the votes of both the excluded and elected candidate would contribute to the surplus distribution.

Secondly, the Meek method allows the quota to be continually recalculated during the count, to account for ballot papers which are exhausted or lost to rounding. This avoids the situation where the quota is set too high, and the final candidate must be elected with less than a quota. Fixing the quota too high becomes unfair to candidates elected early in the count, as votes are wasted on them which they did not, in fact, actually require to secure election. As shown in the example, the Meek method allows the quota to be recalculated, and for surpluses to be redistributed from previously elected candidates, such that each candidate faces the same quota for election.

For these reasons, the Proportional Representation Society of Australia (Victoria–Tasmania branch) recognises the Meek method as superior to the Gregory methods previously presented. The Meek method, alongside WIGM, is regarded by the PRSA as one of the options in a gold standard STV system.

2. Because no ballot papers exhaust, and there is only 1 elected candidate, in this case the keep value can actually be calculated by hand: $\frac{25.001}{40}$ = 0.625025.