This note is a brief explanation of the quota system in a proportional representation election. It is intended as an adjunct to articles discussing issues arising from the implementation of proportional representation in parliaments.
If a parliament has N members, a “quota” is 1 + V/(N+1), where V is the total number of formal votes. In the counting process, as soon as a particular party or group (including groups of one) has enough votes for a quota, the person first on their candidate list is elected. Subsequent votes go toward their second candidate and so on until all votes have been counted and no more parties or groups have a quota.
In general, less than the requisite N members will have been elected at this stage, so proportional representation necessarily means preferential voting, unless it is agreed that after all full quotas have been allocated, the seats remaining are allocated according to the number of primary votes for the remaining candidates. However, this would seem contrary to purpose, since it would then be possible for someone to be elected with a small percentage of a quota if there were lots of minor parties and independents.
Of the groups without a quota, the one with the least votes is eliminated and their votes are allocated toward the remaining groups according to their second preferences, unless no second preference is allocated, in which case the vote is exhausted and is effectively informal. The elimination and preference distribution continues until N candidates are elected.
The big problem here is that say party A has 2.5 quotas, for example. We allocate 80% of their votes to the quota piles for candidates A1 & A2 on their list, then 20% of the votes to candidate A3, who now has 0.5 of a quota and must hope for a sufficient preference flow to be elected.
Suppose however that candidate A3 is eliminated in the preference count, so the preferences from their 0.5 of a quota will be allocated to other candidates. How do we know the preferences in the 20% of party A votes allocated to candidate A3 for their 0.5 quotas will have the same distribution as the preferences in the 80% of party A votes allocated to candidates A1 & A2 for their 2 quotas?
The strict answer is that we don’t. In NSW (and other jurisdictions), random sampling is used. This means that all of party A’s votes are collected and 20% of these randomly assigned to candidate A3. But, just like rolling a die, if we do it a lot and count the number of 6’s, the proportion will be close to 1/6 of the total rolls, but probably not exact. In fact, if we perform 60,000 rolls, we are expecting 10,000 6’s, but there is a 10% chance the actual outcome will be less than 9,883 and a 1% chance the actual outcome will be less than 9,787.
Those are significant differences from the expected or average outcome. If the rolls were votes instead, there’s a 10% chance we’ll miss out on more than 117 votes to which we’re entitled and a 1% chance we’ll miss out on more than 213 votes.
This is in essence one of Pauline Hanson’s two claims in support of a recount in the 2011 NSW Legislative Council election. It’s a fair point, although I’m pretty sure someone else must have noticed it on her behalf.
Prior to computers, a random sample was the only practical way to allocate a proportion of votes to preferences. However in modern elections, all votes, including preferences are scanned into a database. It is not difficult to work out all possible relevant preference combinations and allocate preferences to candidates exactly in proportion to those in the total voting pool. Any halfway decent undergraduate maths student could do it.
Democracy develops once there is the technology to support it. We now have the technology to do away with the sampling errors in the random allocation of preference votes. Let’s use it.
Let’s look at an example:
Suppose a parliament has 9 members and there are 10,000 voters, so a quota is 1 + 10000/(9+1) = 1001 votes ie. 10% plus 1 vote.
Suppose there are 4 parties who receive votes A: 4665, B: 3300, C: 1400, D: 635.
For ease of explanation, we’ll assume compulsory preferential voting, so all ballot papers have 1, 2, 3, 4 listed in some order against A, B, C, D. Optional preferential voting (you can write 1 or 1, 2 or 1, 2, 3 or 1, 2, 3, 4) follows the same principles of quotas, elimination and preference distribution, but the maths is slightly more complicated.
From the first round of counting, A has 4 whole quotas, B has 3 and C has 1, so we get 8 of our 9 members elected. Once these votes are allocated to the candidates on A, B & C’s lists, the remaining votes are A: 661, B: 297, C: 399, D: 635.
It’s B with the lowest total of those remaining, so all their candidates from 4th down are eliminated and we must distribute preferences from 297 of B’s 3300 votes amongst A, C and D. As discussed above, since the order of preferences can differ between ballot papers, how do we work out a fair way to choose which 297 of all B’s votes have their preferences allocated?
As I said above, in NSW, the 297 are selected at random. Prior to computers, this was really the only practical way to do it.
But suppose we have all B’s 3300 votes in a database. We count the second preferences and find 33 = 1% for A, 2970 = 90% for C and 297 = 9% for D.
Then of B’s 297 vote overhang, we should allocate 1% = 3 votes to A, 90% = 267 to C and 9% = 27 to D.
After allocation, we have A: 664, C: 666, D: 662.
D, who was ahead of B’s 4th candidate and C’s 2nd candidate, is now eliminated and their 662 votes distributed to A or C depending on who is preferred.
So, if most of these 662 voters prefer A to C, A’s 5th candidate will take the 9th seat in parliament and party A will have an outright majority. Otherwise, C’s 2nd candidate will be elected and party C will have the balance of power.
In this case, sampling 297 votes from 3300 at random could easily have given a couple of extra votes to D by pure chance and maybe only 1 or 2 votes to A instead of 3, thus altering the outcome of the election, since A would then have been eliminated and perhaps more of A’s preferences would have gone to D, giving D the final seat.
Hopefully you can see the point that although the preference distribution system is not perfect, being elected (or not) by the vagaries of the statistics of random sampling is a flaw we can now fix and therefore should.
Let’s modify the example slightly:
Suppose instead the primary votes were A: 4700, B: 2800, C: 1650, D: 850.
A has 4 whole quotas, B has 2 and C has 1, so we get 7 of our 9 members elected. Once these votes are allocated to the candidates on A, B & C’s lists, the remaining votes are A5: 696, B3: 798, C2: 649, D: 850, so now it is C’s candidates who are eliminated and the preferences of the 649 votes for C distributed.
By A5, I mean the 5th candidate on A’s list, since the first 4 have already been allocated a quota and elected. B3 is the 3rd candidate on B’s list and so on.
Suppose C’s preferences go to A: 254, B: 152, D: 243, giving totals of A5: 950, B3: 950, D: 1093.
Now D has a quota of 1001. Their first candidate is elected in 8th spot overall and the remaining 92 votes go to D’s 2nd candidate, who doesn’t have more than A5 or B3 and so is eliminated.
The 92 votes are distributed to A5 or B3 according to preferences. Both A and B need 51 of the 92 votes for a quota, so it is possible neither will achieve this. The outcome could be A: 48, B: 44 for example. This would see A5 elected since they have more votes than B3. A 46:46 split is also possible, which would necessitate a run off for the 9th seat.
The reason we make a quota 1001 instead of 1000 is that for anyone receiving 1001 votes, it is impossible for 9 other candidates to equal or pass them, so they must be in the top 9. If they only received 1000 votes, it is possible for 9 other candidates to draw with them, so they are not certainly in the top 9 until they receive the 1001st vote.
However, this second example shows that in a very tight race, it is possible for the final spots to be won by candidates who have not achieved a full quota (it could happen in the first example too, if D’s preferences split evenly between A and C).
Optional preferential voting:
This works in the same way as the examples above, except there is no requirement to allocate preferences to all candidates in order. In our examples, someone might vote AC or just A, instead of ACBD or ACDB.
In this system, it is often not possible to allocate the preferences of all votes for an eliminated candidate. If all the candidates chosen by a particular vote have already been eliminated, that vote “exhausts” and has the same practical effect as an informal one.
With optional preferential voting, there are usually sufficiently many exhausting votes that the last few candidates are elected without reaching a quota. This is not a serious problem per se, since it can happen with full preferential voting, as we saw above.
Let’s revisit our first example, but with optional preferential voting.
Suppose that in the first allocation of preferences ie. the 297 votes from B’s 4th candidate, we have 1% = 3 votes to A, 80% = 238 to C, 6% = 18 to D and 38 votes exhausting because they only chose B and didn’t preference any other candidate.
After allocation, we have A: 664, C: 637, D: 653.
Now it is C’s 2nd candidate who is eliminated. Their preferences are allocated to A or D or neither and exhaust. It is quite possible for A and D to receive only 100 preferences each if neither are popular amongst B & C voters, in which case A’s 5th candidate would be elected in 9th place, without even coming close to a quota.
This was the case with the final 4 candidates elected to the NSW Legislative Council in the recent poll.
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