# Counting methods for transferable individual votes

There are many different counting methods for the process of transferable individual votes .

## Voting

Each voter ranks candidates in the order of their preferences.

 1 Andrea 3 Bernd 2 Christian 4th Doris

## Determination of the quota

The quota is the number of votes a candidate needs to be elected. Since it is an absolute limit and not a percentage, it is sometimes called a hurdle . It is calculated from the number of valid votes cast and seats to be filled. Two different calculation methods are common, one according to Hare and one according to Droop.

With the Hare quota, there is a high probability that at least one candidate will be elected with less than a full quota, even if each voter indicates a preference for all candidates.

When each voter gives a full list of their preferences, the droop quota guarantees that each elected candidate will meet the quota and not just be elected because he is the last remaining candidate after weaker candidates are eliminated. Therefore, the droop rate is preferred. It is the smallest number that ensures that no other candidate can achieve the quota as soon as as many candidates each have a quota of votes as there are seats to be filled.

A candidate's excess votes are carried over to other candidates based on the next available preferences. With Meek's method, the quota has to be recalculated during the counting.

### Hare quota

When Thomas Hare devised his version of the transferable individual voting, he envisioned the use of a simple quota, which has therefore become known as the Hare quota:

${\ displaystyle {\ text {Quote}} _ {\ text {Hare}} = {\ frac {\ text {votes}} {\ text {seats}}}}$

It requires that, in the end, all votes cast are evenly divided between the elected candidates. The only difference between the votes received for each candidate would be based on the distribution of voters between constituencies - Hare's original proposal only included a single statewide constituency - and the number of incomplete votes; H. of people who did not include all candidates in their ranking, which means that some candidates would be chosen as the last remaining with less than the quota.

In the special case of only one seat to be awarded, all preferences must be evaluated by eliminating the weakest candidate in each round and distributing his votes until only one remains.

### Droop rate

Nowadays the droop rate is mostly used. It is usually stated as follows:

${\ displaystyle {\ text {Quote}} _ {\ text {Droop}} = \ left ({\ frac {\ text {votes}} {{\ text {seats}} + 1}} \ right) +1}$

In contrast to the Hare quota, this does not require that all preferences contribute to the choice of a candidate. It is enough that enough votes are gathered so that no other candidate still in the running can win. As a result, votes for almost an entire quota are not allocated, but it is said that this quota simplifies voting and that counting these votes would not change the final result.

In the special case of only one seat to be allocated, 50% + 1 vote is sufficient.

## Counting the votes

Allocation ( process A ): The first preferences are counted. If one or more candidates receive more votes than the quota, they will be declared elected. After a candidate is elected, they cannot receive any further votes (but see a modernization below).

The surplus votes of the successful candidate will be allocated to the candidates who have received the next higher place on the voting slip of the elected candidate. There are various methods of transferring the excess voices (see below).

Process A is repeated until there are no more candidates who have reached the quota.

Elimination ( process B ): The candidate with the lowest support is eliminated, his votes are transferred to those candidates who were next highest on the eliminated ballot papers. After a candidate is eliminated, they cannot receive any further votes.

At the end of each iteration of process B, if candidates have now been chosen, process A begins again. This continues until all candidates are either chosen or eliminated. Process B cannot be resumed while there is still an elected candidate whose excess votes have yet to be transferred.

### Reallocation of the surplus

The votes that an elected candidate receives over and above the quota form a surplus . In order to ensure that as few votes as possible are wasted, most of these are transferred to the remaining, unelected candidates. This happens according to the next specified preference.

There are several methods of deciding which of the candidate's votes to transfer. Some are usually only applied to the starting surplus when an unelected candidate first crosses the quota. Others are also applied to subsequent surpluses if an elected candidate receives additional transferred votes.

#### Random selection

Some of the methods of distributing the surplus rely on choosing votes at random. The guarantee of randomness is done in different ways. In many cases, all of the voices to be considered are simply mixed by hand.

In Cambridge , one constituency is counted at a time; this creates an artificial sequence of voices. To prevent all votes to be broadcast from being selected from the same constituency, every th ballot is selected, with the fraction to be selected being . ${\ displaystyle n}$${\ displaystyle {\ frac {1} {n}}}$

### Transfer only the initial excess

Suppose Andrea has 190 votes at a certain point in the count and the quota is 200. Now Andrea receives 30 votes from Bernd transferred (after Bernd is either elected or eliminated). That gives Andrea a total of 220 votes. H. a surplus of 20 votes is to be carried over. But which 20 votes should be transmitted?

#### Hare method

From the 30 votes that were transmitted by Bernd, 20 votes will be drawn at random. Each of these 20 votes will carry over to the next available preference listed after Andrea on the ballot. This method is the easiest to use when counting paper ballots by hand; it was Thomas Hare's original proposal in 1857. It is used in the Republic of Ireland (except in Senate elections).

However, there is no guarantee that the later preferences of the 30 votes Bernd transmitted would be similar to the other 190 votes Andrea received first. Hence, this method is potentially unfair and opens up the possibility of tactical choice. In addition, exhausted ballots are excluded. So if more than 10 of the 30 votes after Andrea have no further preference, then 20 votes cannot be selected for transmission, so some votes will have to be wasted.

#### Cincinnati method

20 votes are drawn at random from all 220 votes. This procedure is used in Cambridge . (There every 11th vote (220-200) / 220 = 1/11) would be drawn for transmission. This method is more likely to be representative than Hare's method and less prone to too many exhausted voices. However, there is still an element of chance. In the event of a tight election, this can be decisive in determining who wins. In addition, if the votes are recounted, it must be ensured that exactly the same random selection is used. (That is, the recount is only there to check for errors in the original count, not to be used to re-randomize.)

If a candidate exceeds the quota with the initial preferences alone, there is no difference between Hare's method and the Cincinnati method, because all of the candidate's votes are in the “last received batch” from which the Hare surplus is drawn.

#### Clarke Method

All 220 ballots are split into separate piles according to the next specified preference. An equal proportion of votes is drawn from each batch by random selection and transferred to the relevant candidate.

${\ displaystyle {\ text {proportion}} = {\ frac {\ text {excess}} {({\ text {total number}} - {\ text {exhausted}})}}}$

If we assume in the example that no further preference is given on 40 ballots after Andrea, the proportion is . For example, if 54 of Andreas give 220 votes as the next preference Bernd, 90 Christian and 36 Doris, then 6, 10 and 4 votes are drawn from the corresponding three stacks and transferred to the corresponding candidates. ${\ displaystyle {\ frac {20} {220-40}} = {\ frac {1} {9}}}$

This method is used in Australia. In Australia, non-integer votes are rounded down: With a division of 52: 88: 40 (i.e. 5.8: 9.8: 4.4), the transmission would be in the ratio 5: 9: 4, so that only 18 instead of 20 Votes are broadcast. The number of such “lost” votes is always smaller than the number of remaining candidates; in practice this is a very small part as the number of votes is much larger than the number of candidates.

The Clarke Method reduces but does not eliminate the problem of chance of the Cincinnati Method.

#### Gregory Method / Senate Rules

Another method is known as both Senate Rules (after its use for most seats in Irish Senate elections ) and the Gregory Method (after its inventor, JB Gregory ). This eliminates any chance and practically applies Clarke's method to all subsequent transmissions. Instead of transferring part of the votes at full value , all votes are transferred at part of their value . The proportion involved is the same as in the Clarke method; H. in the example . ${\ displaystyle {\ frac {1} {9}}}$

It should be noted that part of the total of 220 votes for Andrea can already be made up of proportional votes from previous broadcasts; z. B. Bernd had perhaps been elected with 250 votes, 150 with Andrea as the next preference, so that the transfer of 30 votes was actually 150 votes with a value of each . In that case, those 150 votes would now be retransmitted with a combined share value of . In practice, the transmitted value of a vote would normally not be given as a common fraction , but as a decimal fraction with two or three decimal places. To make it easier to count the votes, you can give the original votes a numerical value of 100 or 1000 so that you can then work with whole numbers . ${\ displaystyle {\ frac {1} {5}}}$${\ displaystyle {\ frac {1} {5}} \ times {\ frac {1} {9}} = {\ frac {1} {45}}}$

Computing combined fractions is labor intensive. As a result, this method is only used in Ireland for the Senate , where only about 1500 councilors are eligible to vote. In Northern Ireland, however, this method has been used in all STV elections since 1973; up to 7 transmissions take place (in Local Councils with 8 seats) and up to 700,000 votes are counted (in the European elections with 3 seats).

### Transfer of later surpluses

All of the above methods only apply to the transfer of an initial surplus if an initially unelected candidate exceeds the quota of votes for the first time during the count. A similar question arises where votes are to be transferred and the preference given next is for an already elected candidate. The usual practice has always been to ignore this preference and instead transfer the votes to the next unelected (and not eliminated) candidate. This boils down to Hare's method and suffers from all of the inadequacies of that method.

In principle it would be possible to use one of the other methods. Let us assume that Andrea, who has already been elected, receives 20 transmissions from the newly elected Bernd in addition to the quota of 200 previously retained: With the Cincinnati method, one would mix all 220 votes from Andrea and randomly select 20 votes for transmission. The problem here is that some of these 20 voices contain retransmissions from Andrea to Bernd and thus create a recursion. This is unclean. In the case of the Senate rules, it would be an infinite recursion, since with every step all votes are transferred and ever smaller fractions are created.

#### Meek method

In 1969, Brian L. Meek designed a counting method based on the Gregory method (Senate rules). As there, fractions of votes are transmitted. However, candidates who have already been elected are not skipped during the further counting, but can receive additional votes through broadcasts. The resulting new surplus must, however, be redistributed so that each elected candidate ultimately does not have more votes than the quota. This is done by reducing the fraction that a candidate keeps from each vote received (and also from each fraction of votes received). This fraction to be retained is called the retained value. It is further reduced each time the candidate receives votes again. The retained value of the candidates is reduced each time so that they only have a whole quota of votes.

Since an infinite recursion is created through the transmission and retransmission of fractions of votes (which are smaller in each subsequent round), Meek's method uses an iterative approximation process that continues the transmissions until the percentage of votes to be transmitted is vanishingly small, e.g. B. a millionth. This procedure would be extremely time-consuming in the case of manual counting, since all ballot papers have to be re-evaluated and transmissions have to be carried out in each pass.

Therefore, the Meek method requires that the votes be counted by computer, even if the voting is done on paper. However, once it has been calculated what proportion of each vote each candidate will keep, the results can also be checked using paper ballots. (You can think of it as a system of equations with many variables. Calculating the variables is time-consuming. But once they have been calculated, the correctness can be checked relatively easily by inserting the variables into the equations.)

A candidate who has not yet reached the quota but is still in the race always has the retention value 1. His votes are counted in full so far. If the candidate exceeds the quota in the further course of the count, the retained value drops to less than 1.

If a candidate has too few votes and is therefore deleted, their keep value is set to 0. Accordingly, the respective vote is fully carried over to the next preference. If a candidate is deleted, the count will be as if they had never run (except that no other previously deleted candidate cannot be revived).

In general, the part of the vote that is not retained is carried over.

${\ displaystyle {\ text {Transfer value}} = 1 - {\ text {Retained value}}}$

If a candidate is elected , all votes cast for him are weighted with the retention value, and the remaining amount of the vote value is proportionally passed on to the following preferences, each of which only retains the portion that corresponds to its retention value and the remainder again passes on to the next preference.

If votes cannot be transferred because no further preferences have been specified, the quota will be recalculated. Meek's method is the only one where the quota is changed in the middle of the process. The rate is calculated as follows

${\ displaystyle {\ text {Quote}} _ {\ text {Meek}} = {\ frac {{\ text {valid votes}} - {\ text {no longer transferable votes}}} {{\ text {seats} } +1}}}$

It is therefore a variation of the droop rate. The recalculation has the consequence that the retention value (the "weight") of each candidate is also changed.

This process is repeated until the vote value of all elected candidates is close to the quota (within a very small range, i.e. between 0.99999 and 1.00001 of a quota).

With the Meek method, all surpluses are transferred at the same time instead of in a specific order. The surpluses come with the corresponding share from all ballots, not just from those received in the previous transfer.

The Meek method is considered to be the STV procedure that best realizes the principles of the STV. It is currently used in some local elections in New Zealand.

#### Warren method

Warren's method is similar to Meek's method. However, the retained fractions of votes are not multiplied, but added.

## example

Assuming we carry out an STV election using the droop rate, we have 2 seats to be filled and 4 candidates: Andrea, Bernd, Christian and Doris. Let us further assume that 57 voters participate, who indicate the following order of preference on their ballot papers, the many other possible orders do not occur.

 16 votes 24 votes 17 votes 1. Andrea Andrea Doris 2. Bernd Christian Andrea 3. Christian Bernd Bernd 4th Doris Doris Christian

The droop rate and thus the hurdle is:

${\ displaystyle \ left ({\ frac {57} {2 + 1}} \ right) + 1 = 20}$

The hare rate would be:

${\ displaystyle \ left ({\ frac {57} {2}} \ right) = 28 {\ frac {1} {2}}}$

In the first round Andrea receives 40 votes and Doris 17. Andrea is elected and has 20 excess votes. These surplus votes are distributed proportionally to their secondary preferences. 12 of the votes to be transferred go to Christian and 8 to Bernd.

${\ displaystyle {\ frac {16} {40}} \ times 20 = 8}$
${\ displaystyle {\ frac {24} {40}} \ times 20 = 12}$

Since none of the remaining candidates reaches the quota, Bernd, the candidate with the fewest votes, is eliminated. All of his votes have Christian as the next-placed preference and are now transferred to Christian. Christian thus receives 20 votes and is elected; he occupies the second seat.

Round 1 round 2 Round 3 Andrea 40 20th 20th Elected in round 1 0 8th 0 Eliminated in round 2 0 12 20th Elected in round 3 17th 17th 17th Beat in round 3 57 57 57

## See also

• CPO-STV (Comparison of Pairs of Outcomes by the Single Transferable Vote)