Further example and development of First and Second order Binomial STV counts.

Voters preferences or, conversely, unpreferences.

For more explanation of how Binomial STV works, see previous page on the subject.

The following example is of an actual contest that took place between 21 candidates for 5 seats. There are only 10 voters, from which one could not infer much of a voting pattern. The count of such a small number is mostly useful as an exercise in a new procedure. The 10 voters all expressed different permutations of choice for all 21 candidates. There are factorial twenty one possible permutations of choice, which is a big number! Had even two voters shared the same order of 21 choices, one would suspect that they had agreed a common slate, before casting their preference votes.

Table 0 shows the 10 voters' orders of 21 choices. Following the convention of languages that read from left to right, the order of choice reads from left to right. For instance, in permutation row 1, the first choice is A, second choice K, and so on, till the last choice, number 21, which is H. All 10 voters express this on their ballot papers by numbering their orders of choice from 1, 2, 3, ... to 21.

The count by Binomial Single Transferable Vote also makes use of the reverse order of choices, reading right to left, from last preference to next last, 21, 20, 19, ... Conventional STV, needing but a single election table ( as shown in the section at the end ), is usually simpler than Binomial STV, using at least an election ( or preference ) quota table and an "exclusion ( or unpreference ) quota" table. But Binomial STV makes more use of the voters' preferential information, given in table 0, and therefore has more pretensions to accuracy.

I repeat, larger numbers of voters, than the below example, are really needed to draw realistic conclusions about voters' preferences for candidates. Very small numbers of voters tend not to have any significant pattern of distribution, which a highly rational system of election like STV, and even more so, Binomial STV, depends on, to ensure large proportions of the vote are transferable, for clear-cut result.

Binomial STV starts like conventional STV with an election quota and the transfer of surplus votes over the quota to help elect next prefered candidates. When this source of election runs out, conventional STV merely excludes the candidate, unfortunate enough to be currently last past the post.

The numbers 1 to 10 in the first column of table 0 have nothing to do with the voters' order of choice. They are there for the benefit of the returning officer, in a manual count, to show which number permutation row directs any transfer of votes.

Binomial STV table 0: 10 voters' permutations on 21 candidates.
1 A K E B R F L I N P T Q J M O D C S U G H
2 B E R A O C N F I P L D K Q G S T U J H M
3 O G F S C M A K B T L R E H P J D I N U Q
4 G K D M P R Q A F H L N S U B I C J O T E
5 H I B Q E O M F R A K N P C L U S T D J G
6 A F G I M T C E H K U D N O P Q B L J R S
7 O M H P B K A N Q I C J F L R S U T D E G
8 L G F O A C M J E N S K R D Q P B I T H U
9 K H B J A T I E O N C D L M R Q G U P S F
10 K B E T N I A H C M F P R U J L G D O Q S

First order Binomial STV example.

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We start with a first order Binomial STV count, which consists of an election count and an exclusion count. These counts are expressed in keep values assigned the candidates. The exclusion keep values are inverted to make them effective election keep values. These can then be multiplied by the election keep values, which when their square roots are taken, produce average elective keep values for the candidates, determining the winners. ( There are different kinds of average to best represent different distributions of data and the average used here is called the geometric mean. )

( If the first order binomial STV count is not completely decisive, it may be necessary to embark on a second order binomial count. )

The returning officer starts the manual count of first order Binomial STV by counting the number of votes and counting the number of first preferences, shown in the first column of letters for candidates, on the left of table 0. Three candidates, A, K, O each have 2 votes, which are enough to elect them, being over the elective quota of 10/( 5 + 1 ) = 1.667 votes ( to three decimal places ).

The surplus vote is transfered at a transfer value of: (2-1.667)/2 = .167 to three decimal places, according to the next preferences of the six voters, who gave two votes each to A, K and O. In this case, all six second preference candidates each receive .167 of a vote. This is recorded in table 1, col. 2.

For reference, the returning officer keeps track of these transfers by noting the number of each permutation row in which a next preference gains a surplus vote. So, the six transfers of .167 votes are as follows. On permutation row one ( perm 1 ) elected candidate A, has already elected K for second preference, so the surplus transfers to third preference, E. But the fact that K was the second preference of that voter is important, because it increases the size of K's transferable vote from 2 to 2.167 votes. This improves K's keep value, which is the quota divided by K's total transferable vote, to 1.667/2.167.

A candidate, whose transferable vote is the same as the quota has a keep value of 1, which means the candidate is just elected with no surplus vote to spare. The larger the candidate's transferable vote than the quota, the smaller the keep value. In this election, K has the smallest elective keep value, which means the best elected candidate.
( As it happens, the exclusion count does not modify this result: K stays in the lead for the over-all result. )

The other vote transfers are: On perm 6, A to F. On perm 9, K to H. On perm 10, K to B. On perm 3, O to G. On perm 7, O to M. The numbers of votes for the candidates, after these transfers, is recorded in table 1, column 2.

In column 3, as there are no more votes to transfer, the keep values are calculated from the quota divided by the highest votes each candidate received. Three candidates, with keep values, of one or less, are elected out of the five vacancies. But an exclusion vote, significiantly higher than the quota, might be enough to unelect a candidate. That is unless one devised a qualifying rule, such as that a candidate, receiving a quota of election, cannot be unelected. On my previous page on Binomial STV: first and second order counts, there was a limited qualifying rule.

Binomial STV table 1: Election. Quota = 10/(5+1) = 1.667 ( to 3 dec. places ).
(0) Candidates (1) 1st preferences (2) A, K, O, elected.
Their surpluses transfered @ .167.
(3) Preference keep values.
A 2 1.667 1.667/2 = .833
B 1 1.167 1.667/1.167 = 1.428
E .167 1.667/.167 = 9.982
F .167 1.667/.167 = 9.982
G 1 1.167 1.667/1.167 = 1.428
H 1 1.167 1.667/1.167 = 1.428
K 2 1.667 1.667/2.167 = .769
L 1 1 1.667/1 = 1.667
M .167 1.667/.167 = 9.982
O 2 1.667 1.667/2 = .833
Total vote: 10 10

Table 2, for the exclusion count, follows table 0 from right to left, for next least prefered candidates. Candidates S and G, each with 2 last preferences, are over the exclusion quota, which ( like the election quota ) is 1.667. S and G's surpluses of least preference are transfered to next least prefered candidates. The transfer values are again .167. Hence, on perm 6, S transfers to R. On perm 10, S transfers to Q. On perm 5, G transfers to J. On perm 7, G transfers to E.

But no further candidate reaches an exclusion quota. Then, the exclusion keep values are calculated.

Binomial STV table 2: Exclusion. Quota = 1.667.
(0)Candidates (1) Last preferences. S and G over exclusion quota. (2) S and G's surpluses transfered @ .167 to next least prefered candidates (3) Unpreference keep values.
E 1 1.167 1.667/1.167 = 1.428
F 1 1 1.667/1 = 1.667
G 2 1.667 1.667/2 = .833
H 1 1 1.667/1 = 1.667
J .167 1.667/.167 = 9.982
M 1 1 1.667/1 = 1.667
Q 1 1.167 1.667/1.167 = 1.428
R .167 1.667/.167 = 9.982
S 2 1.667 1.667/2 = .833
U 1 1 1.667/1 = 1.667
Total: 10 10

The keep values from the election table 1 and the exclusion table 2 are now combined in table 3 for the over-all keep values of the candidates. These are calculated by multiplying the elective keep values by the inverted exclusion keep values. The inversion of the exclusion keep values makes them effectively elective keep values. Then multiplying the two sets of values and taking their square roots produces ( geometric mean ) average keep values, which aim to be more representative of the candidates' support.

Table 3: Over-all keep values for 1st order binomial STV.
(0) Candidates. (1) Elective keep values. (2) Exclusive keep values inverted. (3) Over-all keep values: square root of col. (1) x col. (2).
A 1.667/2 = .833
B 1.667/1.167 = 1.428
E 1.667/.167 = 9.982 1/(1.667/1.167) = .7 Sq. rt. of (1.167/.167) = 2.643
F 1.667/.167 = 9.982 1/(1.667/1) = 1/1.667 2.447
G 1.667/1.167 = 1.428 1/(1.667/2) = 1.2 1.309
H 1.667/1.167 = 1.428 1/(1.667/1) = 1/1.667 .926
K 1.667/2.167 = .769
L 1.667/1 = 1.667
M 1.667/.167 = 9.982
O 1.667/2 = .833

The over-all keep values from table 3 show K as the most prefered candidate, with A and O also elected. Candidate G reached an exclusion quota with a slight surplus. When this was inverted for an effective elective quota of 1.2, this was enough to increase G's over-all elective keep value from the elective keep value. This meant that candidates H and B with the same elective keep values as G, do slightly better over-all, as H's unpreference was small enough to help effectively elect him. Whereas B had no unpreference and, therefore must be presumed to have some elective keep value better than H or G.

H and B qualify for the two remaining vacancies out of five. Not far behind, candidate G might demand a second order count under the Binomial STV system.

I mentioned above, the possibility of qualifying rules that restrict the power of the inverted exclusion keep values to effectively elect candidates. A previously used rule, on my Binomial STV web page, sets bounds of influence on the election keep values of candidates.

Suppose the elective keep value of one is made the average keep value, and the lower bound of influence is set according to the proportional representation. If there are five vacancies or seats in a multi-member constituency, giving a PR of five-sixths, then the keep values lower bound is 5/6. The upper bound can be found if we know both the average and the lower bound.

The average used is the geometric mean, which reflects that it is progressively harder for a candidate to get more and more votes that produce a smaller and smaller keep value. The geometric mean equals the square root of the lower times the upper bounds. That is one equals the square root of ( five-sixths times the upper bound ). So, the upper bound equals one times six-fifths. That is 1.2.

If this were the qualification, it would make the result less decisive, because the two nearest contenders, H and B, have election ( or preference ) keep values of 1.428. They are outside the 1.2 upper bound on qualifying for effective election from an inverted exclusion keep value, if it is sufficiently below one.
In any case, H and B might qualify, for the two final seats, as runners-up. You cannot really expect better from 10 votes spread between 21 candidates for 5 seats. And some relaxation of the qualifying bounds is a possibility.

Having described the first order Binomial STV procedure, the results must be taken with caution, especially for a statistical interpretation of a mere 10 votes. Statistics measures averages of distributions and the variations from the average in the distribution. But to fill out a distribution such as the binomial distribution would require at least 30 voters.

Second order Binomial STV count

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As explained in the previous page, Binomial STV, the second order Binomial STV count is more complicated than the first order count. The first order count, as we have seen above, merely does a quota count of transferably voted preferences. Then another count of transferably voted unpreferences, the exclusion count takes table zero's preferences in reverse order ( from right to left ).

The second order count involves not two but four counts. The former two counts of the first order are both qualified in two ways, to make these four counts. The count of preferences has the votes of the most prefered candidate re-distributed. I call this a preference-qualified preference count ( symbolised pp ). Likewise, the count of unpreferences can have the votes of the most unprefered candidate re-distributed. This is the unpreference-qualified unpreference count ( symbolised uu ).

Moreover, another count can be run with the most prefered ( quota-achieving ) candidate's least preferences re-distributed to their next least preferences, to qualify an unpreference count. That is a preference-qualified unpreference count ( symbolised pu ). Conversely, a count can be run with the least prefered candidate's greatest preferences re-distributed to their next greatest preferences, to qualify a preference count. This is an unpreference-qualified preference count ( symbolised up ).

The four symbols, pp + up + pu + uu, constitute a second order ( non-commutative ) binomial expansion of p plus u. "Second order" means here the square of ( p + u ). And "non-commutative" means that the count tables for "up" and "pu" are not the same. ( So the expansion terms, 2pu or 2up, would be wrong in this context. )
The first order binomial STV is simply a preference count plus an unpreference count or ( p + u ).

Commencing the second order count, with the preference-qualified preference count: From table 1 of the preference count, table 4 below re-distributes the votes of the most prefered candidate, who is K, with the lowest election keep value. K's voters' two second preferences give candidates H ( table 0 perm 9 ) and B ( perm 10 ) an extra vote each. They are both elected with a surplus of .333 or one-third of a vote.

In both cases ( perms 2 and 10 ) B's next preference is E, who gets all of B's .333 surplus.

Candidate H's next preferences go to I ( from H's one vote in perm 5 ) and to B ( from the one vote K passes-on in perm 9 ). Thus I and B receive 1 vote each of H's transferable 2 votes, whose surplus transfer value is .333. In short, I and B share 1/2 of .333 surplus, or about .167 each.
B is already elected and passes on the .167 to J ( perm 9 ). But B's keep value improves to 1.667/2.167.

The candidates benefiting from the re-distribution of K's votes have their keep values correspondingly improved.

Table 4: Second order Binomial STV's preference-qualified preference count ( pp ). Quota = 1.667.

(0) Candidates (1) 1st preferences. K, lowest keep-valued candidate's votes transfer'd. (2) A, K, O, elected.
Their surpluses transfered @ .167. H, B elected. Their surpluses transfer'd @ .333.
(3) pp keep values.
A 2 1.667 1.667/2 = .833
B 1 + 1 = 2 1.667 1.667/2.167 = .769
E .167 + .333 = .5 1.667/.5 = 3.333
F .167 1.667/.167 = 9.982
G 1 1.167 1.667/1.167 = 1.428
H 1 + 1 1.667 1.667/2 = .833
I .167 1.667/.167 = 9.982.
J .167 1.667/.167 = 9.982.
K - - 1.667/2.167 = .769
L 1 1 1.667/1 = 1.667
M .167 1.667/.167 = 9.982
O 2 1.667 1.667/2 = .833
Total vote: 10 10

It is convenient now to move to a preference-qualified unpreference count, table 5. This is based on the unpreference table 2. This time, the most prefered candidate K is excluded to have his least preferences re-distributed. But K has no least preferences. ( Indeed, only one voter has unprefered K by the ninth least preference. ) Therefore, there is a zero transfer of least preference votes from K, who contributes no preference-qualification of the unpreference count. In effect, table 5, the preference-qualified unpreference ( pu ) count is unchanged from table 2, the unpreference ( u ) count: For table 5, see table 2.

For table 6, we unpreference-qualify unpreference table 2. Only two candidates have election keep values, as well as exclusion keep values. But this may not remain the case even with only 10 votes to go round 21 candidates. For large scale elections, most probably all candidates would have election and exclusion keep values -- or, more accurately, preference and unpreference keep values, since not all candidates' keep values will signify that they have reached a quota of election or exclusion.

The most unprefered candidate has the lowest unpreference keep value. That is two candidates, S and G both at .833, which being below one, signify exclusions. For reason discussed on the "Binomial STV" page, it is not advisable to re-distribute the votes of two candidates. ( At least, no more than a quota of votes should be re-distributed in a qualified count ). A choice has to be made between S and G. Altho they are equally unprefered, G is more prefered with an elective 2 first preferences to none for S. This may be a less arbitrary reason for choosing S, than a random choice.

Very close decisions of this sort might have an unwarranted influence on an election result. But an averaging of counts, as with Binomial STV, may reduce the effect of any anomaly that occurs in a single count. And conventional STV is still made up of single count systems.

S's two votes go to R and Q as next least prefered, respectively permutation rows 6 and 10 in table 0. Thus Q, with 2 votes, passes the exclusion quota, of 1.667, and a surplus of .333 goes half to U ( perm 3 ) and O ( perm 10 ).

Table 6 column 3 shows the unpreference-qualified unpreference keep values as a result of the re-distributions of votes.

Table 6: 2nd order binomial STV's unpreference-qualified unpreference ( uu ) count. Exclusion. Quota = 1.667.
(0)Candidates (1) Last preferences. Least prefer'd, S's 2 votes re-distributed. G's surplus transfered @ .167 each to 2 next least prefered candidates. (2) Q's surplus transfered @ .167 each to 2 next least prefered. (3) uu keep values.
E 1 + .167 = 1.167 1.167 1.667/1.167 = 1.428
F 1 1 1.667/1 = 1.667
G 2 - .333 = 1.667 1.667 1.667/2 = .833
H 1 1 1.667/1 = 1.667
J + .167 .167 1.667/.167 = 9.982
M 1 1 1.667/1 = 1.667
O .167 1.667/.167 = 9.982
Q 1 + 1 = 2 1.667 1.667/2 = .833
R + 1 1 1.667/1 = 1.667
S - - 1.667/2 = .833
U 1 1.167 1.667/1.167 = 1.428
Total: 10

Table 7 is the unpreference-qualified preference ( up ) count. We redistribute the highest preferences of least prefered candidate S. But S has none. And so table 7 is the same as table 1. ( Even had we chosen G as least prefered, there would have been little change: In table 0, perm 4, there is one first preference for G, which would go to K as the second preference. K is already elected, indeed, the most prefered candidate but this transfer would serve to further improve K's leading elective keep value. K's ( up ) keep value would be 1.667/( 2.167 + 1 ) = 1.667/ 3.167 = .526. As K would not need this further surplus from G, the vote would be passed on to D, who is not otherwise in contention. D would have ( up ) keep value of 1.667/1 = 1.667. )

Now we have got all the keep values we need to average them for a second order binomial STV count. In table 8, the pp and up keep values are multiplied because they are both ( qualified ) preference keep values. The square root of the multiple is taken to derive its average ( called the geometric mean ). The uu and pu keep values are multiplied because they are both ( qualified ) unpreference keep values. ( It just happens with this very small number of 10 voters, that the preference-qualification is nil on the unpreference count. ) The multiple, of the qualified unpreference keep values, also has its square root taken, and is then inverted to make it effectively elective, instead of exclusive.

The two square-rooted multiples are in turn made a multiple, of each other, and its square root taken, to derive the average of the two averages. This gives the over-all keep values, as shown in the last column of table 8.

Table 8: Second order Binomial STV over-all keep values.
(0) Candidates (1) pp (table 4) (2) up (table 7=1.) (3) Square root of (pp x up) (4) uu (table 6) (5) pu (table 5=2) (6) Square root of (uu x pu) (7) Inverse of col. 6. (8) Over-all keep values: Square root of { col. (3) x col. (7)}.
A .833 .833 .833
B .769 1.428 1.048
E 3.333 9.982 5.768 1.428 1.428 1.428 .7 4.038
F 9.982 9.982 9.982 1.667 1.667 1.667 .6 2.447
G 1.428 1.428 1.428 .833 .833 .833 1.2 1.309
H .833 1.428 1.091 1.667 1.667 1.667 .6 .809
I 9.982
J 9.982 9.982 9.982 9.982 .1
K .769 .769 .769
L 1.667
M 9.982 9.982 9.982 1.667 1.667 1.667 .6 2.447
O .833 .833 .833 9.982
Q .833 1.428 1.091 .917
R 1.667 9.982 4.079 .245
S .833 .833 .833 1.2
U 1.428 1.667 1.543 .648

The problem with table 8 is that 10 voters for 21 candidates cannot furnish data for all the candidates to have keep values in all the counts. Normally, there are more voters than candidates and this deficiency shouldnt be a problem. But here we have to take the peculiar circumstances into account to show a reasonable result. Firstly, we note at the top of the table, candidate A has an elective keep value of .833. But A registers no unpreference. That means the exclusion keep value is 1.667 divided by an indefinitely small number approaching zero. That equals some large number for A's unpreference keep value. Invert that for an effective election keep value and you have an indefinitely small number. This tiny fraction multiplied by .833 would only serve to diminish A's over-all elective keep value to indefinitely less than .833. Therefore, we can say that no evidence of an exclusive keep value serves rather to confirm A's election.

The same reasoning can be applied to candidate B with a more interesting result. For, B's elective keep value of 1.048 is not quite below the elective keep value of 1. But given an indefinitely small fraction for the effectively elective keep value, it can be confidently stated that B has an over-all elective keep value of less than one, and is therefore elected.

K has the lowest elective keep value and with some indefinitely small effectively elective keep value is the over-all most prefered candidate.
All the keep value data is available for H, who is elected on an over-all keep value of .809.
Candidate O has an elective keep value of .833. The partial data on exclusion quotas indicates that the effectively elective quota would strengthen O's position as an elected candidate.

Therefore, candidates K, A, O, B and H are elected by second order binomial STV, confirming the slightly less decisive result of the first order count, which candidate G might have called in question.

As for the above first order count of this example, we review the result in terms of possible qualifying bounds on the exclusion keep values. The lower bound is 5/6 or .833. The upper bound is 6/5 or 1.2.

Take the lower bound. Only K is below it and therefore couldnt be adversely affected by an exclusion quota below one and therefore an effective election quota above one. In any case K does not have such an adverse function of the unpreference keep values.

A and O both have elective keep values of .833, which is exactly the lower bound. But neither have adverse exclusion keep values and so are also safely elected.

B and H have election keep values just above one, which therefore just fail to elect them. But they are both well within the upper bound of 1.2, which permits them relief from the exclusion keep values, in so far as they fail to exclude them on a quota of unpreference votes. H has an unexclusive unpreference keep value, that permits H to have an over-all elective keep value less than one. B's position is better with a slightly lower election keep value and no recorded unpreference keep value, which is some ( indefinite ) improvement on H's slight unpreference vote.

Thus, the second order Binomial STV, with or without the given qualifying bounds, elects the five candidates, K, A, O, B, H, in that order.

The traditional STV manual count, given in the last section, would have elected G rather than H or B. The difference in result is perfectly reasonable. It simply boils down to the fact that Binomial STV takes into account unpreference as well as preference. And G, tho quite well prefered, was more unprefered.

Measuring probability of dispersion about the geometric mean.

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At the end of the previous section, I set qualifying bounds, about a geometric mean of one, in accordance with the proportional representation of the election given by the number of seats. Consider these bounds for the most limited situation of one vacancy. Even if a candidate gets all the votes, say 100 votes, his keep value is only the quota of 50 votes divided by the 100, or a keep value of one half. This is the lower bound. A candidate receiving just the quota has the geometric mean keep value of one. And a candidate receiving all 100 votes as least preferences has an exclusion keep value of 1/2 and an inverted exclusion keep value of 2/1 = 2.

A keep value of one means a candidate has just been elected. It is the typical value for marginal candidates. In a single member constituency, the first round is a marginal contest with candidates vying for more or less half the votes. That is the expectation of two such candidates is more or less 50 votes out of 100. The ( Droop ) quota is votes divided by ( number of seats plus one ) or 100/(1+1) = 50. Then, on average, the expected keep value is the quota of fifty divided by an expected vote of about fifty, or about one.

In a two member constituency, the second round is the marginal contest. A candidate has usually been elected with a third of the votes, probably with a surplus considerably more than 34 out of a hundred votes, say. That most prefered candidate's consequent keep value is somewhat less than one. Then there may be two other front-runners getting more or less another third of the votes each, to decide which takes the second seat, probably with just about the 34 required votes out of the hundred, and a consequent keep value of about one, our geometric mean.
And so on for any number of seats per constituency.

In general, consider lower and upper bounds from one representative to any number of representatives, R, per constituency. That is geometric mean lower and upper bounds are R/(R+1) and (R+1)/R, respectively. These bounds may mark the limits of dispersion about that mean of unity.

A foremost purpose of statistics is the summarising, of distributions or collections of data, in terms of an average item and the extent of the dispersion of other items about that typical or most representative item.

For instance, the average known as the arithmetic mean has a complementary measure of dispersion known as the standard deviation. I dont remember any reference to a dispersion measure for the geometric mean. But by comparing the relation of the standard deviation to the arithmetic mean, it seems straight-forward to infer what calculating dispersion about the geometric mean must involve.

We need only consider the simplest case involving an average of two end-points or items on the outer bounds. The arithmetic mean is the sum of the items, divided by their number, which is two in this simplest case. To find the standard deviation, find the difference of each item from the arithmetic mean. Square each difference. Then add them, divide by the number of items ( just two again in this case ). This total is called the variance. Its square root is the standard deviation.

The biggest proportion of items in a binomial distribution, about one third either way from an arithmetic mean, is given by plus or minus one standard deviation from that average. Two standard deviations, both ways, covers about 95 per cent of the items in the distribution. Three standard deviations both ways covers all but about one in every thousand items of the distribution.

Let's hazard this line of thought on a dispersion round the geometric mean. The deviations above and below the arithmetic mean tend to be similar or symmetric. This geometric distribution is asymmetric. I assume this means in practise that calculation of geometric dispersion takes separately items above and below the geometric mean. So, take the difference of items from the geometric mean item. Square the difference of each item. Multiply these squared differences of items above the mean. And multiply separately items below the mean. Take the root of the above-average items' multiple for dispersion above the geometric mean. And take the root of below-average items' multiple for dispersion below the geometric mean. The order of the root depends on the number of items.

For just two items, below or above the mean, the square root is taken. ( If three items were involved, a cubic root would be taken, if four items, a quartic, and so on. ) But of course, in our example, there is just one above-average item, the upper bound, and one below-average item, the lower bound.

We dont multiply the upper by the lower bound, because we want distinct upper and lower dispersions ( not symmetric deviations ) about a geometric mean. As there is only one item each above and below the mean, there is no other such item to multiply with. A second such item could have been multiplied and the square root ( the root of two ) of the multiple taken. But in the case of one item, in effect its "linear" root ( the root of one ) is taken, which is to say it is left unchanged.

For five representatives giving a PR of 5/6, the lower and upper bounds about a geometric mean keep value of 1, are 5/6 and 6/5. To calculate a geometric deviation, subtract the bounds from the mean. That is 6/5 - 1 = 1/5. And 5/6 - 1 = -1/6. It doesnt matter about the negative, because the next step is to square the differences. That is 1/25 and 1/36. (But see postscript below.)

The previous paragraph explained that, as one item each, above and below the geometric mean, 1/25 = .04 and 1/36 = .028 are the respective deviations. However, they are only one deviation. Two deviations, may cover probable ranges of electability or unelectability of candidates with above average or below average keep values, respectively.

So, 2 x .04 = .08 marks out a probable limit above unity for a candidate's electability. Under this rule, a candidate with a higher keep value than 1.08 might be deemed probably too far above the elective keep value to be given the benefit of the doubt, and assisted to an over-all elective keep value, with a possible good showing by the candidate's inverted unpreference keep value.

This statistical upper bound of 1.08 would vindicate candidate B at 1.048 and only just exclude candidate H at 1.091. H would easily be within the three deviations of .04 x 3 = .12. That would be an above-average bound of 1.12.

Coming now to the dispersion below the mean keep value of one, two deviations below average are 2 x .028 = .056. This would mean that a candidate with less than 1 - .056 = .944 for a keep value might be deemed securely elected and not qualify for being unelected by an unpreference vote above the exclusion quota.

The three candidates K, A. O all have keep values well below .944 and so put beyond reasonable doubt as to their election, which means they wouldnt qualify for possible unelection, had they surpassed the exclusion quota. They hadnt, anyway.

If this statistical analysis is justified, then my simplistic range of qualifying bounds, R/(R+1) and (R+1)/R may be too generous, giving too much power to the inverse exclusion keep values, say, for practical political elections. Tho, a symmetry of election to inverse exclusion may be theoreticly sound and of some possible application.

(Postscript, 26 february 2007: Ive had second thoughts. I dont believe that considerations of a deviation from the geometric mean, here require the step of squaring the differences of the bounds from the mean. Squaring fractions makes the differences smaller. The effect of that is to restrict the range of influence within which an inverse exclusion count may help determine which candidates are returned. But the restriction may be too great, after all. I think one can just take the magnitude of the difference between the mean and a bound.)

Traditional STV version of above example.

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Table 9 shows how a traditional count of the single transferable vote would typicly take place, ignoring slight variations in the rules over the years and over the globe.

Table 9: STV election. Quota = 10/(5+1) = 1.667.
Candidates (1) First preferences. A, K, O elected. (2) Transfer surpluses of A, K, O all at value .166 (3) C, D, I, J, N, P, Q, R, S, T, U; E, F, M excluded. (4) L excluded. G elected. (5) Transfer of G's surplus @ .285
A 2 1.667 1.667 1.667 1.667
B 1 1.166 1.333 1.333 1.665
C 0
D 0
E .166 0
F .166 0
G 1 1.166 1.333 2.333 1.667
H 1 1.166 1.333 1.333 1.665
I 0
J 0
K 2 1.667 1.667 1.667 1.667
L 1 1 1 0
M .166 0
N 0
O 2 1.667 1.667 1.667 1.667
P 0
Q 0
R 0
S 0
T 0
U 0
Total votes: 10 10 10 10 9.998

The conventional STV election table shows A, K, O and G elected, with a tie. B or H, would have to be elected randomly to the fifth vacancy. I'm told that using a computer count by Meek's method on this example, H beats B. I was also told by the same expert that candidates A, K, O, G, H always win against any other candidate who may form a sextet of candidates competing for five seats. This way of thinking is based on the avowed philosophy that a candidate must be elected, if achieving an elective quota, no matter how much other voters may dislike him.

Binomial STV has rather a different slant on the matter. An exclusion quota is not of itself a veto in any one count. Neither is one count's election quota for a candidate complete assurance of election if that is an above average result compared to the other counts.
What matters is the over-all result that averages more than one count. Most simply, a candidate may be elected on an election count but be effectively unelected on an exclusion count. It is the average of these two counts that finally counts - or indeed, the average of two averaged counts in second order Binomial STV. On the other hand, that means that a large enough popularity can out-weigh even a quota of unpopularity.

Admittedly, too great a variation of a candidate's keep value from the elective value of unity may make it statisticly unrealistic to suppose that a safely elected candidate or a too little supported candidate can have those decisions reversed to, respectively, unelect or elect him.

Foot-note on the keep value of candidates without any votes:

The keep value of a candidate with zero votes is infinity ( ∞ ). The transfer value is ( zero minus the quota ) divided by zero. Zero goes into a number an infinite number of times. In this case, the numerator is a negative number, so the transfer value is minus infinity. The keep value equals one minus minus infinity, which equals plus infinity. Or, 1 - -∞ = ∞. This leads to the interesting convention that infinity minus infinity equals one: ∞ - ∞ = 1. This defies the common sense that infinity minus infinity should equal zero.

While infinities may not be unmanagable in an averaging count, like Binomial STV, it is simpler to do without them and they are not really needed or even justifiable. Every candidate might be deemed to have cast a vote of confidence in their candidature, just by standing. Every candidate should have at least one vote, if only their own. In public elections of any size, this is always the case. The problem of zero votes for a candidate could only crop up in small committee elections, where the candidates have been left out from the voting.

One vote is enough to reduce the candidate's keep value from infinity to the mere size of the quota. That is the transfer value for a candidate with one vote is: ( 1 - the quota )/1. Therefore, the keep value is 1 - ( 1 - the quota ), which equals the quota. This limit on the size of the keep values, in Binomial STV, makes for a neater comparison between the candidates' keep values, without rather meaningless infinities obtruding.

Richard Lung.
13 december 2005;
minor modification, 2 January 2006; 26 february 2007.

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