First and second order counts of Binomial STV example with seven seats.


Using the example of 10 voters prefering 21 candidates, suppose there are 7 vacancies instead of the 5, on the page, Further example and development of First and Second order Binomial STV. Also refer to this page for an explanation of how the count works. The present page mainly shows the arithmetic of the returning officer.

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


The quota becomes 10/(7+1) = 1.25.

First order count for seven seats.

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Consider the election of highest preferences.

A, O, K are elected with 2 votes each. They each have a surplus of 2 - 1.25 = .75. With reference to table 0, all the surpluses are split between two different second preference candidates, who each benefit by .75/2 = .375.

Candidates A and O both have keep values of 1.25/2. However, of the 10 voters, perm 1, with first preference for A, next prefers K. This improves K's keep value to 1.25/2.375. As K is already elected, we move on in perm 1 to the third preference E, who takes .375.

Perm 6 is the other voter who votes first for A. That voter's second preference goes to F, who takes the other .375 of A's surplus.

Perms 9 and 10 are the two voters whose first choice is K. Their respective second preferences are for H and B. Both these candidates already have a first preference each ( perms 5 and 2 respectively ). They now both have 1.375, which takes them over the 1.25 quota. They are elected with a surplus of .125 each.

H's surplus of .125 goes in different proportions to candidate I ( perm 5 ) at .125 x 1/1.375 = .091. And to candidate B ( perm 9 ) at .125 x .375/1.375 = .034. B is already elected but gets an improved keep value of 1.25/ (1.375 + .034) = 1.25/(1.409). As B has already passed the quota, the .034 passes next to J.
Perm 10 first prefers K then B, whose .125 surplus is passed on to E, who already has .375 ( from perm 1 ). E now has .5.

O's two voters next prefer G ( perm 3 ) and M ( perm 7 ). M gets .375.
G already has a first preference ( perm 4 ) and is elected with 1.375. G's .125 surplus is re-distributed as follows: Re perm 4, the next preference K is already elected, so on to D, who gets .125 x 1/1.375 = .091.
Perm 3 shows the rest of the surplus ( .125 x .375/1.375 = .034 ) goes to F, whose vote becomes .375 + .034 = .409.

First order Binomial STV table 1: Election. Quota = 10/(7+1) = 1.25

(0) Candidates (1) 1st preferences (2) A, K, O, elected.
Their surpluses transfered @ .75/2 = .375. B, G, H elected.
(3) B, G, H each have .125 surplus transfer'd. (4) Preference keep values.
A 2 1.25 1.25 1.25/2
B 1 1.375 1.25 1.25/1.409
C
D .091 1.25/.091
E .375 .5 1.25/.5
F .375 .409 1.25/.409
G 1 1.375 1.25 1.25/1.375
H 1 1.375 1.25 1.25/1.375
I .091 1.25/.091
J .034 1.25/.034
K 2 1.25 1.25 1.25/2.375
L 1 1 1 1.25/1
M .375 .375 1.25/.375
N
O 2 1.25 1.25 1.25/2
P
Q
R
S
T
U
Total vote: 10 10 10



The candidates A, K, O, H, B, G are elected with L closest to taking the seventh seat.

Consider the exclusion of least preferences on the 1.25 quota.

Reading for least preferences from right to left of table 0:

Candidate S has two least preferences and is excluded with a surplus of .75, which splits between R ( perm 6 ) and Q ( perm 10 ) at .375 each. Q also has one least preference ( perm 3 ). Q's 1.375 also passes the 1.25 exclusion quota. The remaining surplus of .125 passes from Q to U ( perm 3 ) at transfer value of .125 x 1/1.375 = .091. And passes from Q to O ( perm 10 ) at .125 x .375/1.375 = .034.

Candidate G also has 2 least preferences and is excluded. Perm 5 has G then J. Perm 7 has G then E. Both second preferences receive .375 each.
No further candidates reach an exclusion quota.

First order Binomial STV table 2: Exclusion. Quota = 1.25.

(0)Candidates (1) Last preferences. S and G excluded. (2) S and G's surpluses transfered @ .375 to next least prefered candidates. E and Q excluded. (3) E and Q's surpluses, @ .125 each, transfered. (4) Unpreference keep values.
A
B
C
D .034 1.25/.034
E 1 1.375 1.25 1.25/1.375
F 1 1 1 1.25/1
G 2 1.25 1.25 1.25/2
H 1 1 1 1.25/1
I
J .375 .375 1.25/.375
K
L
M 1 1 1 1.25/1
N
O .034 1.25/.034
P
Q 1 1.375 1.25 1.25/1.375
R .375 .375 1.25/.375
S 2 1.25 1.25 1.25/2
T .091 1.25/.091
U 1 1 1.091 1.25/1.091
Total: 10 10 10


A table of preference and unpreference keep values may now be collected.

Table 3: First order Binomial STV's averaged keep values from election and exclusion counts. Quota = 1.25.
(0) Candidates (1) Preference keep values (2) Unpreference keep values (3) Averaged keep values: Square root of (pref./unpref.) keep values
A 1.25/2
B 1.25/1.409
C
D 1.25/.091 1.25/.034 .302*
E 1.25/.5 1.25/.375 .866
F 1.25/.409 1.25/1 2.445
G 1.25/1.375 1.25/2 1.206
H 1.25/1.375 .727
I
J 1.25/.125 1.25/.375 1.732
K 1.25/2.375
L 1.25/1
M 1.25/1
N
O 1.25/2 1.25/.034 .13
P
Q 1.25/1.375
R 1.25/.375
S 1.25/2
T 1.25/.091
U 1.25/1.091

*D does not qualify for election despite the below-unity over-all keep value, because it only means that the preference keep value is relatively good compared to the exclusion keep value, but is itself not nearly good enough. To see that, one does not have to run a statistical deviation test, used on the 5-seat version of this election.

As one would expect from the previous use of this example with 5 vacancies, in this 7-seats count, K is the most prefered candidate, followed by A. O's slight vote of unpreference gives a more favorable over-all quota than the election quota alone. But this is still not as highly elective as K or A, who have no vote of unpreference at all, and whose over-all quota is only not stated because K and A's unpreference votes are indefinitely small, less even than O's.

This situation is peculiar to the excess of candidates over voters, so that the preference information of voters for candidates is very patchy and unbalanced. That isnt to say the count is erroneous, so far as it can go. But little, if any, significance can be read into such small numbers of voters for so many candidates.

B and H also achieve elective quotas, tho with smaller surpluses than K, A and O. B and H have no unpreferences. E also has an elective quota but with some small unpreference. This is a somewhat more respectable result than D's but also no-where near passing a deviation test of being within electing distance.
Candidate G does pass the election quota but not before also "passing" the exclusion quota. Having an over-all keep value just over unity, G just misses the over-all election threshold.

Thus E and even more so, D, reach an over-all election by being disliked even less than they are liked. If we object to such triumphs of the bland, we have to consider the case of G, who does receive the required support for election. Unfortunately, G also receives slightly more counter-support from those even quicker to reject him.

As I keep saying, one cannot read much into these miniscule figures. But in principle such things could happen on a large scale. So, with this in mind let's move on to the second order count.

Second order count for seven seats.

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The second order count requires four counts, which are qualifications, or "controls" ( to use the jargon of experiments ) on the above preference and unpreference counts.

Firstly, the preference-qualified preference count. This takes table 1 and re-distributes the 2 votes of most prefered candidate K ( with the best keep value ). B and H get one vote each, added to the one each that they already have. Like A and O, H and B now have 2 votes each. These four candidates are elected with a surplus of .75 each over the 1.25 quota.

As in table 1 of preference: Candidate A's two second preferences, getting .75/2 of the surplus each, are table 0 perms 1 and 6. Perm 1 goes to K. K's votes are re-distributed in this count, but the .375 surplus still improves K's keep value to 1.25/2.375, before the .375 surplus is passed on to E. Perm 6 gives F the other .375.

As in table 1 of preference: O's two voters next prefer G ( perm 3 ) and M ( perm 7 ). M gets .375.
G already has a first preference ( perm 4 ) and is elected with 1.375. G's .125 surplus is re-distributed as follows: Re perm 4, the next preference K is already elected, so on to D, who gets .125 x 1/1.375 = .091.
Perm 3 shows the rest of the surplus ( .125 x .375/1.375 = .034 ) goes to F, whose vote becomes .375 + .034 = .409.

Perms 9 and 10 are the two voters whose first choice is K. Their respective second preferences are for H and B. Both these candidates already have a first preference each ( perms 5 and 2 respectively ). They now both have 2, which takes them over the 1.25 quota. They are elected with a surplus of .75 each. H's surplus of .75 goes to candidate I ( perm 5 ) at .75 x 1/2 = .375. And to candidate B ( perm 9 ) at .75/2 = .375. B is already elected but gets an improved keep value of 1.25/ (2 + .375) = 1.25/(2.375). As B has already passed the quota, the .375 passes next to J.
Perm 10 first prefers K then B, whose .75 surplus is all passed on to E ( being the next preference to B in perm 2, as well ). E already has .375 ( from perm 1 ). E now has 1.125.

Table 4: Second order Binomial STV's Preference-qualified preference ( pp ) election. Quota = 1.25

(0) Candidates (1) 1st preferences. Most prefer'd, K has 2 votes re-distributed. (2) A, O, B, H elected.
Their .75 surpluses transfer'd.
(3) Elected G has .125 surplus transfer'd. (4) pp keep values.
A 2 1.25 1.25 1.25/2
B 1 + 1 1.25 1.25 1.25/2.375
C
D .091 1.25/.091
E .375+.75=1.125 1.25/1.125
F .375 .409 1.25/.409
G 1 1.375 1.25 1.25/1.375
H 1 + 1 1.25 1.25 1.25/2
I .375 .375 1.25/.375
J .375 .375 1.25/.375
K - - - 1.25/2.375
L 1 1 1 1.25/1
M .375 .375 1.25/.375
N
O 2 1.25 1.25 1.25/2
P
Q
R
S
T
U
Total vote: 10 10 10


From the preference-qualified preference election count, it is convenient to move to table 5 for the preference-qualified unpreference ( pu ) exclusion count. Candidate K this time has his least preferences re-distributed to qualify table 2. But K has no least preferences ( not until the ninth least preference ) So, table 5 is effectively table 2.

Table 6 is the unpreference-qualified unpreference ( uu ) exclusion count.

The most unprefered candidate has the lowest unpreference keep value. 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.

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 1.25, and a surplus of .75 goes half to U ( perm 3 ) and O ( perm 10 ).

G's surplus of .75 goes half to J ( perm 5 ) and half to E ( perm 7 ). E already has one unpreference vote ( perm 4 ). E is excluded with a surplus of 1.375 - 1.25 = .125. In perm 4, E's next preference, T, gets .125 x 1/1.375 = .091. In perm 7, from G, 2nd pref. E passes to D for .125 x .375/1.375 = .034.

Table 6: 2nd order Binomial STV's unpreference-qualified unpreference ( uu ) count. Exclusion Quota = 1.25.
(0)Candidates (1) Last preferences. Least prefer'd, S's 2 votes re-distributed. G's surplus transfered @ .375 each. (2) Q's surplus transfered @ .375 each. E's .125 surplus to T & D. (3) uu keep values.
A
B
C
D .034 1.25/.034
E 1 + .375 = 1.375 1.25 1.25/1.375
F 1 1 1.25/1
G 2 - .75 = 1.25 1.25 1.25/2
H 1 1 1.25/1
I
J + .375 .375 1.25/.375
K
L
M 1 1 1.25/1
N
O .375 1.25/.375
P
Q 1 + 1 = 2 1.25 1.25/2
R + 1 1 1.25/1
S - - 1.25/2
T .091 1.25/.091
U 1 1.375 1.25/1.375
Total: 10

Table 7 is the unpreference-qualified preference ( up ) election 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.


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 1.25/2 1.25/2 .625
B 1.25/2.375 = .526 1.25/1.409 = .887 .683
C
D 1.25/.091 = 13.736 1.25/.091 = 13.736 13.736 1.25/.034 = 36.765 1.25/.034 = 36.765 36.765 .027 .3709
E 1.25/1.125 = 1.111 1.25/.5 = 2.5 1.667 1.25/1.375 =.909 1.25/1.375 = .909 .909 1.1 1.834
F 1.25/.409 = 3.056 1.25/.409 = 3.056 3.056 1.25/1 1.25/1 1.25 .8 2.445
G 1.25/1.375 = .909 1.25/1.375 = .909 .909 1.25/2 1.25/2 .625 1.6 1.454
H 1.25/2 = .625 1.25/1.375 = .909 .754 1.25/1 1.25/1 1.25 .8 .603
I 1.25/.375 = 3.333 1.25/.091 = 13.736 6.766
J 1.25/.375 = 3.333 1.25/.034 = 36.765 11.07 1.25/.375 = 3.333 1.25/.375 = 3.333 3.333 .3 3.321
K 1.25/2.375 = .526 1.25/2.375 = .526 .526
L 1.25/1 1.25/1 1.25
M 1.25/.375 = 3.333 1.25/.375 = 3.333 3.333 1.25/1 1.25/1 1.25 .8 2.666
N
O 1.25/2 = .625 1.25/2 = .625 .625 1.25/.375 = 3.333 1.25/.034 = 36.765 122.538 .008 .005
P
Q 1.25/2 = .625 1.25/1.375 = .909 .568 1.76
R 1.25/1 1.25/.375 = 3.333 4.166 .24
S 1.25/2 = .625 1.25/2 = .625 .625 1.6
T 1.25/.091 = 13.736 1.25/.091 = 13.736 13.736 .073
U 1.25/1.375 = .909 1.25/1.091 = 1.146 1.041 .96


When I extended the definition of a keep value to deficit transfers as well as surplus transfers of votes, I also made the rule that a candidate's standing for an election guarantees at least one vote for the candidate, if only his own. That is because a minimum of one vote for a candidate ensures that a keep value cannot exceed the size of the quota. Only an indefinitely small fraction of a vote for a candidate will make his keep value indefinitely large.

Altho this example is taken from a real election, it was one which didnt give the candidates even a single preference each for themselves. And the quota itself is scarcely more than one vote. Consequently, the grossly misleading distortions in some of table 8's keep values.

However, it must be admitted that this cannot be blamed on the over-all "election" of candidate D with a final keep value of .3709. This merely owes to an inverted exclusion value more than cancelling out a poor election keep value. It doesnt need a statistical test to show that D is too far from any considerable positive support to merit re-assessment of his vote in terms of lack of unpopularity.

The test in question ( explained on the previous example page ) comes from taking the proportional representation of the number of seats. Here 7/8. This is taken as a lower bound to an average keep value of 1, with the upper bound as the inverse of 7/8. That is 8/7.

A measure, of a keep value probably being just a chance deviation from the average, is calculated. In the below-average direction, this is the square of the difference between the lower bound and the average. That is:

(7/8 - 1) = 1/64 = .016 ( to 3 decimal places ). Two or at most three times this deviation, 2 x .016 = .032, or 3 x .016 = .048 should be enough to decide how far below one, a candidate's keep value can be to escape being unelected by an unpreference quota enough less than one. Provided that the elected candidate has less than 1 - .032 = .968 or 1 - .048 = .952 at the out-side, he is safe from a poor unpreference keep value unelecting him.

(8/7 - 1 ) = 1/49 = .020 ( to 3 decimal places ). Two deviations makes this .04; three deviations, .06. A candidate with a preference keep value slightly more than one has a slightly better chance of qualifying for an over-all keep value of one or less, if he has an inverse unpreference keep value enough below one. But the election keep value must be no more than 1.04 or 1.06 at the most.

Almost no candidates come anywhere close these rather strict bounds. K, A, O, B, H, the elected candidates' election keep values are all far away from any chance of unelection. But G with .909 is only a moderate distance from an outlying unelective bound of .952. However, that is quite safe enough to prevent G being unelected by his inverse unpreference vote of 1.6, resulting in a disqualifying over-all keep value of 1.454. The statistical deviation measure spares G to take the sixth seat.

Whereas, the candidates who might have benefited from inverse exclusion keep values to elect them are all statisticly no-where near the election keep value of one. The runner-up for the seventh seat is candidate L with a keep value of 1.25. L is the seventh of the seven candidates to get at least one first preference from the 10 voters. It is quite usual in STV elections for the last seat to be taken by the candidate who has most votes towards the final quota, if there are no more transferable votes left.

Richard Lung.
December 2005.

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