There are two problems with this extended form of Condorcet. One is that, when two or more seats are being contested, it is not practicable for any but the smallest elections: 15 candidates contesting 5 seats would give rise to 5005 contests; 27 candidates standing for the 15 seats on the Council of the Electoral Reform Society would give rise to 13,037,895 contests. Confronted with the result sheet of such an election, the electorate would find it difficult to understand how the winning candidates won and, perhaps more importantly, how the losing candidates lost. The other problem is that there could be more than one set of n candidates (whether n>1 or n=1) which gain the equal greatest number of victories. We would have to provide some kind of tie-breaker.
I believe that we can achieve the effect of Condorcet for one or more seats without these practical difficulties. Indeed, David Hill has suggested one such scheme which selects sets of n candidates and tests each set against the other candidates one at a time. He admits that his scheme can elect a candidate other than the Condorcet winner in an election for one seat: I believe that the system propounded here will always elect the Condorcet winner, if there is one.
If, for some whole numbers K and L satisfying 0 < K <= L, more than K Droop quotas of voters put the same L candidates (not necessarily in the same order) as the top L candidates in their preference listings, then at least K of those L candidates should be elected.
A voter who puts those L candidates (in any order) as the top candidates in order of preference is said to be 'strongly committed' to that set of L candidates. We will refer to a set of candidates to whom one set of voters is strongly committed as a 'DPC set'.
Under any of the rules in current use, the elimination of candidates in an STV election makes votes available to other candidates in the DPC sets to which they belong. No candidate who has a quota at the relevant stage is eliminated, and, with insignificant exceptions, eliminations are made one at a time. This ensures that the result of an STV count is consistent with the Droop Proportionality Criterion. STV with Elimination by Electability Scores (STV(EES)) shares this characteristic.
On the other hand, the aim of STV(EES) is to identify those candidates who certainly should not be elected. It does so by taking account of all the preferences of every voter; in some circumstances, this feature will cause the system to fail on Woodall's second property. To identify candidates for elimination, it calculates 'electability scores' (see below) for the candidates: as new electability scores are calculated at successive stages, these form the basis for the elimination of candidates one by one until only sufficient are left to fill the available seats. These remaining candidates are elected.
STV(EES) differs in another way from conventional STV. As we are identifying candidates for elimination, not election, we do not have to use the Droop quota, and in fact its use can lead to perverse results. Instead, we calculate the 'threshold', which any of the other candidates must be able to attain in order to survive.
A stage of STV(EES) culminates in the withdrawal, either temporary or permanent, of a candidate. It consists of two substages: the first establishes the threshold of votes which a candidate must be capable of achieving in order to survive; the second is to test whether the candidates who start with less than the threshold can in fact achieve it. At the end of the second sub-stage, one of these candidates is withdrawn from contention. This withdrawal takes one of two forms: the candidate is either 'eliminated', which means that (s)he takes no further part in the count, and is treated from that point on as though (s)he had withdrawn before it started; or is 'temporarily excluded', which means that (s)he is withdrawn for the time being, but comes back in after the next elimination.
Before explaining how to calculate electability scores, we must define the 'retention factor', which Meek calls the 'proportion retained'. In a Meek count, a point will be reached when a candidate has more than the quota. Clearly, that candidate should get less of the incoming votes in the next iteration of the count than were credited this time; and in successive iterations, the proportion of each incoming vote that stays with that candidate will diminish. The tendency will be for each new total of votes credited to that candidate to be closer to the quota than the last. To achieve this, an incoming whole vote or fraction of a vote is multiplied by an amount m where 0<m<1; the result of this multiplication is the fraction of that vote which is credited to that candidate. This amount m is known as the retention factor. Retention factors start with a value of 1.0, and those for the candidates with more than the quota are re-calculated at every iteration; thus retention factors will diminish as the count progresses. The Droop quota is also re-calculated at every iteration on the basis of the votes credited to candidates, ignoring those which have become non-transferable.
In an STV(EES) election, the first sub-stage of each stage is the calculation of the threshold. It does this by calculating the mean of the votes of the n candidates who have the most votes. Surpluses over the mean are transferred, then a new mean is calculated. This process of distributing the votes, calculating the mean, and transferring surpluses is repeated until the first n candidates have the same number of votes. The top n candidates are then known collectively as the 'probables', and their common total of votes is the threshold (T). The value of T remains fixed throughout the second substage, which is the calculation of the contending candidates' electability scores. Let C be the contending candidate whose electability score we are calculating (the 'candidate under test'), and let all the contending candidates other than C have a common retention factor of c. C's own retention factor remains at 1.0. In successive iterations, c and the retention factors of the probables are recalculated until the votes credited to all the probables are equal to the threshold and C either has the threshold or has less than the threshold while no other contending candidate has any votes at all. At this point, c is declared to be C's electability score. The electability scores of the remaining contending candidates are calculated in like fashion. The smaller C's electability score, the greater the number of votes that have had to be transferred from contending candidates other than C in order to ensure that C and the probables get their thresholds.
If the votes credited to the candidate under test and the contending candidates have a collective total of less than T, this indicates that the probables had a Droop quota of votes each when that candidate's electability score was being calculated. In that case, that contending candidate is eliminated, and all the non-eliminated candidates are re- classified as contending. On the other hand, if all the contending candidates' electability scores are at least 0.0, the one with the highest electability score is temporarily excluded, and only the existing probables are re-classified as contending. The new set of contending candidates proceeds to the next stage.
Stage succeeds stage until there are only n candidates who have not been eliminated, and those final candidates are elected. Note that at any stage when there are only n+1 'active' candidates (ie, candidates who have not been eliminated or temporarily excluded), one of them is certain to be eliminated. We therefore know that candidates will be eliminated until only n active candidates survive; thus an STV(EES) election must come to an end.
STV(EES) aims to identify a set of n candidates which can score at least as many victories in Condorcet mini-elections as every other. This means, for every eliminated candidate X, that there must be no set of n candidates including X which can score more victories in Condorcet mini-elections than every set of n not including X. We know at any given stage that every probable is better supported at that stage than X, and that every temporarily excluded candidate was better supported than X at the time of their temporary exclusion. Any DPC set to which X belongs has more members than can be elected by the number of voters that support it, and every other member of that DPC set is better supported than X. We can therefore be confident, though not certain, that there is no set of n candidates including X that can score more victories in Condorcet mini-elections than every set of n not including X. We can, however, state with certainty that in a count for one seat, the Condorcet winner (if there is one) will win. This is because, by definition, the Condorcet winner will win a contest with any one other candidate: and since no candidate is eliminated unless n other candidates have a Droop quota of votes each, the Condorcet winner cannot be eliminated.
ABCDEF 3670 CBAEFD 3436 DEFABC 1936 EFDBCA 1039 FDECAB 1919 ===== 12000After sub-stage 1.1, A and C are probables, and the threshold (the number of votes held by both A and C when transfers are complete) is 3436. At sub-stage 1.2, electability scores are:
B 0.1319 D 0.4125 E 0.2860 F 0.3478
This means that if D, E, and F had a common retention factor of 0.1319, A, B, and C would have 3436 votes each when surpluses have been transferred; if B, E, and F had a common retention factor of 0.4125, A, C, and D would have 3436 votes each when surpluses have been transferred, and so on. As D has the largest electability score at this stage, we act on the presumption that D has a better chance of being elected than B, E, or F, and so we ensure by temporary exclusion that D does not run the risk of being eliminated at substage 1.2. Note that this presumption is like the presumption of innocence in a criminal trial: the process tests it and may very well overturn it.
At substage 2.1, effective votes are:
ABCEF 3670 CBAEF 3436 EFABC 1936 EFBCA 1039 FECAB 1919 ===== 12000Again, A and C are probables and the threshold is 3436. At 0.7608, E's electability score is higher than B's or F's, so E is temporarily excluded at substage 2.2. Effective votes are now:
ABCF 3670 CBAF 3436 FABC 1936 FBCA 1039 FCAB 1919 ===== 12000At substage 3.1, A and F are probables, and the threshold is 4016.9493, more than the current Droop quota. As neither B nor C can get that many votes if the other is temporarily withdrawn, we can eliminate both. D and E are now reclassified as contending, making effective votes:
ADEF 3670 AEFD 3436 DEFA 1936 EFDA 1039 FDEA 1919 ===== 12000At substage 4.1, A and D are probables, and the threshold is 3696.7554. At substage 4.2, E's electability score is 0.4741 and F's is 0.3385, so E is temporarily excluded. Active votes are now:
ADF 3670 AFD 3436 DFA 1936 FDA 1039 FDA 1919 ===== 12000At substage 5.1, A and F are probables, and the threshold is 4309.9757, more than the current Droop quota. D cannot get that many votes and therefore is eliminated. E now comes back in, and active votes are:
AEF 3670 AEF 3436 EFA 1936 EFA 1039 FEA 1919 ===== 12000At substage 6.1, the threshold is 5040.5, more than the current Droop quota, and A and E are probables. There is no prospect that F can attain the threshold, so we eliminate F. A and E are the only active candidates left, so they are elected.
If there are too few DPC sets with sufficient support to 'soak up' all n seats being contested, can the system still produce a reasonable outcome? Let there be 4 candidates contesting 2 seats with votes:
ABCD 41 BCDA 30 CDAB 25 DABC 24 === 120The results of an exhaustive Condorcet count are:
ABC AB ABD AD ACD AC BCD BCWe have a paradox in that AB wins the ABC contest, but AD wins the ABD contest and AC wins the ACD contest; there is also a four-way tie. As A starts with a quota of first preferences, A must be one of the winning candidates, but which of the other three should take the second seat?
Under STV(EES), A and B are probables, and the initial threshold is 35.5. At stage 1, the electability scores of C and D are respectively 0.5625 and 0.54, so C is temporarily excluded. At stage 2, A and D win the ABD contest, and B is eliminated. At stage 3, A, C, and D remain in the contest, so A and C are elected.
How can the elimination of B and D be justified? Part of the answer is that D was in only one winning set in the exhaustive Condorcet count, whereas the other candidates were in at least two. But is there any objective reason why B rather than C should be eliminated? Here we must confess that the system may be said to be perverse: 95 voters prefer B to C, but only 25 prefer C to B. In defence of this outcome, we can say that set AC is one of the joint Condorcet winners, so it meets the aim of STV(EES); and that when a tie is the result of a paradox, it will be arbitrary to some extent. But I would still have preferred AB to be the winning set in this case.
I submit that STV(EES) will in most cases (perhaps all) give a result that is compatible with an exhaustive Condorcet count: and that even if it does not, the result will still be defensible.
The suggested procedure is as follows: