Seven seat example supplemented by Condorcet group-pairing.

This page follows from the previous page on a Binomial STV count for 7 seats, by way of its Condorcet group-pairing count supplement. But much a simpler example of this procedure is given all on the following page, Averaging Binomial and Condorcet counts of Keep-valued Transferable Voting.

The example of an election of 21 candidates by 10 votes, makes one wonder how the voters knew so exactly who they prefered. Moreover, the conventional STV count only takes into account a few of the first few preferences. Binomial STV also takes into account some last few preferences. Nevertheless, the bulk of preferences are not touched by the count and have no influence on the result. The problem with transferable voting is to make it more so, an "extra-transferable vote."

One way to combat this deficiency, if so it be, would be to limit the number of candidates per count. One may still take into account the whole order of preferences for all the candidates. But one could also allow an input from a system of lesser counts that reach into the virgin territories of middle preferences. For instance, the previous page is a Binomial STV count of 21 candidates for 7 seats. It would also be possible to hold a count of 14 candidates for 7 seats, by filtering out 7 of the candidates. This is just as valid an election as the one with the larger number of candidates. It still expresses the voters' preferences for the remaining 14 candidates. True, it loses the whole range of preference information, but we drew on what we could of that in the conventional STV or Binomial STV counts. Instead, it allows us to draw on previously untapped preferences.

A reduction of candidates for sub-counts must be done systematicly so no candidates out of the 21 are adversely affected. The 21 candidates were labeled A to U. We split them into three sub-count categories, A to G, H to N, O to U. There are three possible ways to combine the three groups so that we can hold three sub-elections of 7 out of 14 candidates.

They are A to G plus H to N, A to G plus O to U, and H to N plus O to U. From these three sub-counts of 14 candidates, all 21 candidates have two keep values each. These pairs of keep values can have their geometric means taken. These GM keep values for 7 out of 14 candidates can then suitably modify the keep values gained in an election of 7 out of 21 candidates.

The rule for sub-counts is an extension of Condorcet pairing. This takes place between individual candidates. But in this new system, the rule is extended to groups of candidates. The groupings are random and thus arbitrary. Perhaps the vote offers enough analysis for us to begin synthesis in the count.

Condorcet group-pairing may shuffle candidates enough to reduce possible bias from the arbitrary memberships of the candidate groups. The more shuffling, the less likely a bias. The rule for Condorcet pairing of candidate groups may be shown by the above example. 21 candidates contesting 7 seats may be divided into three groups of 7 candidates, which we'll call A, B and C ( not to be confused with the individual candidates so labeled ). Their possible combinations of 14 candidates are AB, AC, and BC. ( These correspond to below tables I, II, and III, respectively. )

Looking at these three groups, A, B, C, you notice that, say, A has two combinations with B and C. And that, say, B has one remaining combination with C. And that is the basis of the rule of Condorcet pairing: For 3 groups, there are 2 plus 1 equals 3 pairings.
Similarly, for 4 groups, A, B, C, D, there are 3 + 2 + 1 = 6 possible pairings.

On the previous page, I repeated the election of 21 candidates, as a contest for 7 seats, because the actual election for 5 seats is not convenient for Condorcet group-pairing. I would have had to spoil the true result by taking out, say, candidate C, whose preferences were all in the middle range. Then I could have split the 20 candidates into four groups of five candidates.

The significance of the extended rule of Condorcet pairing, to groups of more than one candidate, seems to be that unmanagable numbers of candidates can be reduced to sub-counts where there is only twice the number of candidates to representatives. Voters dont have too many candidates to choose from at once, in the sub-elections ( or primaries ) and their preferences can be explored in depth by the sub-count.

For this reason, the three 14-candidate sub-counts on this page are not a very good example, because the 14 preferences still prevents taking into account near enough the full range of preferences. Six sub-counts of 5 out of 10 candidates would have been better for that purpose.

The election between all the candidates at once has to be given proportionately more weight than the sub-elections between two-thirds of the candidates at a time. The votes in the sub-counts are ( geometricly ) weighted at 2/3 the votes in the general election for the total candidates. This is to make up for the fact that in the sub-counts only two-thirds the candidates share the same number of votes as in the general election: 14 instead of 21.

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

From the original table of 10 voters for 21 candidates, we filter three tables of 10 voters for 14 candidates.

Table I: 10 voters' preferences for candidates A to G and H to N.
1 A K E B F L I N J M D C G H
2 B E A C N F I L D K G J H M
3 G F C M A K B L E H J D I N
4 G K D M A F H L N B I C J E
5 H I B E M F A K N C L D J G
6 A F G I M C E H K D N B L J
7 M H B K A N I C J F L D E G
8 L G F A C M J E N K D B I H
9 K H B J A I E N C D L M G F
10 K B E N I A H C M F J L G D

Table II: 10 voters' preferences for candidates A to G and O to U.
1 A E B R F P T Q O D C S U G
2 B E R A O C F P D Q G S T U
3 O G F S C A B T R E P D U Q
4 G D P R Q A F S U B C O T E
5 B Q E O F R A P C U S T D G
6 A F G T C E U D O P Q B R S
7 O P B A Q C F R S U T D E G
8 G F O A C E S R D Q P B T U
9 B A T E O C D R Q G U P S F
10 B E T A C F P R U G D O Q S

Table III: 10 voters' preferences for candidates H to N and O to U.
1 K R L I N P T Q J M O S U H
2 R O N I P L K Q S T U J H M
3 O S M K T L R H P J I N U Q
4 K M P R Q H L N S U I J O T
5 H I Q O M R K N P L U S T J
6 I M T H K U N O P Q L J R S
7 O M H P K N Q I J L R S U T
8 L O M J N S K R Q P I T H U
9 K H J T I O N L M R Q U P S
10 K T N I H M P R U J L O Q S

We move on to the sub-counts. The first sub-count, shown in table 1, is based on the preferences shown in table I. The quota for all sub-counts is the same as for the count of 21 candidates, because it is still based on 10 voters for 7 representatives. Quota = 10/(7+1) = 1.25.

Drawing on table I, A, G, K are elected each with a surplus of .75 over the quota. Their surpluses transfer as follows. Their two voters each share the .75 surplus between two second preferences, each gaining .75/2 = .375.

The voter of perm. 1 for candidate A, has second preference for K, who is already elected and needs no more votes but can improve his keep value from 1.25/2 to 1.25/2.375. But third preference E gets the .375.
The voter at perm 6 also gave A first choice. The second preference goes to F, who gets the other .375 of A's surplus.

Perm 3 for G also next goes to F, who gets another .375, making .75 in all.
Perm 4 for G next goes to K, who gets another .375 improvement of keep value for 1.25/(2.375+.375) = 1.25/2.75. But the actual .375 votes goes to third preference D.

Perm 9 for K next goes to H, already with 1 vote, gets an extra .375. 1.375 elects H with a surplus of .125
Perm 10 for K next goes to B, already with 1 vote, gets extra .375, to elect with surplus of .125.

The transfer of these two .125 surpluses from H and B is as follows.
In perm 9, H's second preference transfers to third preference B, already elected. B's keep value is improved from 1.25/1.375 to 1.25/1.409. Fourth preference J gets the .034 vote.
Perm 5 is from first choice H to next choice I, who gets .091, the rest of H's .125 surplus.

Both B's next preferences in perm 10 and 2 go to E who gets all .125 surplus.

Table 1: Preference count of candidates A to G and H to N.
(0) Candidates. (1) First preferences. A, G, K elected. (2) .75 surplus transfers each from A, G, K. (3) .125 surplus transfers each from B, H. (4) Preference keep values.
A 2 1.25 1.25 1.25/2
B 1 1.375 1.25 1.25/1.409
D .375 .375 1.25/1.375
E .375 .5 1.25/.5
F .375 + .375 = .75 .75 1.25/.75
G 2 1.25 1.25 1.25/2
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 1.25/2.75
L 1 1 1 1.25/1
M 1 1 1 1.25/1
Total votes: 10 10 10

Table 2 of Unpreference is also based on table I with the preferences counted in reverse ( from right to left ).

Candidate G's 2 votes exclude with a surplus of .375 votes each, which go to the next preference, in perm 5, which is J, and to the next in perm 7, which is E.
J and E both already have one last preference each. Both are now excluded with a surplus each of .125.

.091 of J's surplus goes to L ( perm 6 ); .034 goes to D ( perm 5 ).
In perm 4, .091 of E's surplus worsens J's unpreference keep value from 1.25/1.375 to 1.25/1.466. As J is already excluded, the .091 is then passed on to C.
In perm 7, the next preference to E is D, who gets the .034 balance of E's surplus. That is D's second .034, bringing D's vote to 1.068.

Candidate H's two last preferences also produced two .375 surpluses. In perm 1, G was next prefered. G is already excluded and gets a worsened keep value of 1.25/2.375. The .375 passes on to C, whose vote, with the above-mentioned .091, becomes .466.
H's other last preference, in perm 8, next goes to I, who takes the other .375 surplus.

Table 2: Unpreference count of candidates A to G and H to N.
(0) Candidates. (1) Last preferences. G, H excluded. (2) .75 surplus transfers from G, H. E, J excluded. (3) .125 surplus transfers from E, J. (4) Unpreference keep values.
C .375 .466 1.25/.466
D 1 1 1.068 1.25/1.068
E 1 1.375 1.25 1.25/1.375
F 1 1 1 1.25/1
G 2 1.25 1.25 1.25/2.375
H 2 1.25 1.25 1.25/2
I .375 .375 1.25/.375
J 1 1.375 1.25 1.25/1.466
L .091 .091 1.25/.091
M 1 1 1 1.25/1
N 1 1 1 1.25/1
Total votes: 10 10 10

We now draw on table II for a different combination of sub-groups of candidates, A to G and O to U, to prefer, in table 3.

Candidate B is elected with surplus of 4 - 1.25 = 2.75. There are four next preferences to share 2.75/4 = .6875 each. E gets two of them ( perms 2 and 10 ). E also benefits from one of A's next preferences at .375 ( perm 1 ), more than electing E with total of 1.75.

Perm 5 for B next goes to Q, which .6875 is Q's lot.
Perm 9 for B next goes to elected candidate A, improving A's keep value to 1.25/2.6875, before the .6875 is passed on to T.

Candidate F benefits from 3 x .375 = 1.125 surplus votes from A ( perm 6 ); G ( perm 8 ) and O ( perm 3 ) where F is third preference, after elected G improves keep value to 1.25/2.375.

Other .375 surpluses go from G to D ( perm 4 ) and O to P ( perm 7 ).

This leaves E, which at 1.75, has to transfer .5 or 1/2 surplus. From perm 2, 1/2 x .6875/1.75 or about .1964 goes to R. Similarly, in perm 10, .1964 goes to T.
This leaves ( from perm 1 ) 1/2 x .375/1.75 = .107, which benefits elected B's keep value to 1.25/4.107, before passing on to fourth preference R, whose vote now totals .3034.

Table 3: Preference count of candidates A to G and O to U.
(0) Candidates. (1) First preferences. A, B, G, O elected. Transfer of surpluses: B @ .6875; A, G, O @ .375. E's surplus transfer of .5. Preference keep values.
A 2 1.25 1.25 1.25/2.6875
B 4 1.25 1.25 1.25/4.107
D .375 .375 1.25/.375
E 1.75 1.25 1.25/1.75
F 1.125 1.125 1.25/1.125
G 2 1.25 1.25 1.25/2
O 2 1.25 1.25 1.25/2
P .375 .375 1.25/.375
Q .6875 .6875 1.25/.6875
R .107+.1964=.3034 1.25/.3034
T .6875 .6875+.1964~.884 1.25/.884
Total votes: 10 10 10 ( 3 dec. places )

Table 4 draws on table II in reverse to count unpreference.

Elected candidate G's surplus per next unpreference is (3-1.25)/3 = .5833. Perm 1 passes this to excluded U, worsening U's keep value to 1.25/2.5833. Excluded S is the third unpreference. ( In fact there are several such successive preferences for already excluded candidates, in this count. ) S's keep value would only be changed if Extra-transferable Voting's use of reduced numbers of candidates in sub-counts on occasion filtered out the first of two already elected or excluded candidates At least, the E-t V system has some potential for registering otherwise passed-over information such as this.
At any rate, .5833 is passed on to 4th pref. C.
Perms 5 and 7 pass on .5833 to T and E, respectively.

T also gains 2 x .375 unpreferences after U ( perms 2 and 8 ) excluding T on .75 + .5833 = 1.3333. This leaves a surplus of .0833. Of this, .375/1.3333 = .02343 goes to excluded S ( perm 2) worsening S's unpreference keep value to 1.25/2.0234. Next come excluded G and, as it turns out, excluded Q, so that next unpreference D finally takes the .0234.
The other .0234 surplus of T ( perm 8 ) goes to B.
This leaves T's surplus balance of .03644 to go to D ( perm 5 ).

One of S's two last preferences next goes to R ( perm 6 ) for .375 surplus.
Q is excluded from 1 unpreference, in perm 3, plus .375 surplus from S ( perm 10 ).
Q's .125 surplus splits to .0909 passing to excluded U ( perm 3 ) whose unpreference keep value worsens again from 1.25/2.5833 to 1.25/2.6742, before the .0909 rests with D.
The .0341 balance of Q's surplus ( perm 10 ) goes to O.

E has one last preference in perm 4 and .5833 as surplus from G ( perm 7 ). This excludes E with a surplus of .3333. In perm 4, the part of this surplus transferable is .3333 x 1/1.5833 = .2105. Excluded T is next least prefered, getting worse keep value, 1.25/(1.3333+.2105) = 1.25/1.5438. Then O is in line for the .2105 to add to .0341 above.
The surplus balance of .1228 goes to D ( perm 7 ). This makes four tiny surpluses to add up for D.

Table 4: unpreference count of candidates A to G and O to U.
(0) Candidates. Last preferences. G, S, U excluded. Surpluses / next pref.: G's @ .5833; S & U's @ .375. Surpluses / next pref.: E's @ .2105 & .1228 = .3333; Q's @ .0909 & .0341 = .125; T's @ (2x.02343) & .03644 = .8333. Unpreference keep values.
B .0234 1.25/.0234
C .5833 .5833 1.25/.5833
D .1228+.0909+.0234+.0364=.2735 1.25/.2735
E 1 1.5833 1.25 1.25/1.5833
F 1 1 1 1.25/1
G 3 1.25 1.25 1.25/3
O .2105+.0341=.2446 1.25/.2446
Q 1 1.375 1.25 1.25/1.375
R .375 .375 1.25/.375
S 2 1.25 1.25 1.25/2.034
T 1.3333 1.25 1.25/1.5438
U 2 1.25 1.25 1.25/2.6742
Total votes: 10 10 10

We now draw on table III of the third combination of candidates to make table 5.

K's surplus of 4-1.25= 2.75 is divided between four next preferences @ .6875. Two of those next preferences H and R already have one first preference each and so are both elected with a surplus of 1.6875-1.25= .4375. This surplus is proportionately shared: .4375 x 1/1.6875 = .2593 goes from H to I in perm 5. ( This gives I, already with 1 vote in perm 2, a small .0093 surplus. ) The rest, .4375 x .6875/1.6875 = .1782 goes from H to J in perm 9.

Likewise, from R in perm 2, .2593 is passed to elected O improving O's keep value to 1.25/2.2593, before N gains .2593 of R's surplus.
The other .1782 of R's surplus goes to L ( perm 1 ).

Finally, I's small .0093 surplus is split so: .0093 x 1/1.2593 = .0074 goes to M ( perm 6 ) who already has a surplus each from K, .6875 and from O, .375.
The balance of I's surplus, .0093 x .2593/1.2593 = .0019 goes to Q.

Table 5: Preference count of candidates H to N and O to U.
(0) Candidates. (1) 1st preferences. K & O elected. (2) Surplus transfers: K's @ 2.75/4 = .6875; O's @ .375. (3) Surplus transfers: H's & R's both @ .2593 & .1782. Transfer I's .0093 surplus. (4) Preference keep values.
H 1 1.6875 1.25 1.25/1.6875
I 1 1 1.2593-.0093=1.25 1.25/1.2593
J .1782 1.25/.1782
K 4 1.25 1.25 1.25/4
L 1 1 1.1782 1.25/1.1782
M .6875+.375=1.0625 1.0625+.0074=1.0699 1.25/1.0699
N .2593 1.25/.2593
O 2 1.25 1.25 1.25/2.2593
Q +.0019 1.25/.0019
R 1 1.6875 1.25 1.25/1.6875
S .375 .375 1.25/.375
T .6875 .6875 1.25/.6875
Total votes: 10 10 10

Table III is now used from right to left for a count from the least prefered candidates in table 6.

As a result of transfering S's surplus of .5833 to each next unpreference, Q is also excluded, already having 1 last preference, in perm 3. There 1/1.5833, of the .3333 surplus, equaling .2105 next goes to elected U. U has 1 last pref. ( perm 8 ) and .375 surplus from T ( perm 7 ). So U's keep value becomes 1.25/1.5855. The .2105 surplus is passed on to N.
The .1228 balance of Q's surplus goes to O ( perm 10 ).

N also gains .034 out of the .125 of U's surplus ( perm 3 ).
H gets the .091 balance ( perm 8 ).

Table 6: Unpreference count of candidates H to N and O to U.
(0) Candidates. Last preferences. Surplus unprefs.: S's @ 1.75/3; T's @ .75/2. Surplus unprefs.: Q's @ .2105 & .1228; U's @ .091 & .034. Unpreference keep values.
H 1 1 1.091 1.25/1.091
J 1 1 1 1.25/1
M 1 1 1 1.25/1
N .2105+.034=.2445 1.25/.2445
O .375 .4978 1.25/.4978
P .5833 .5833 1.25/.5833
Q 1 1.5833 1.25 1.25/1.5833
R .5833 .5833 1.25/.5833
S 3 1.25 1.25 1.25/3
T 2 1.25 1.25 1.25/2
U 1 1.375 1.25 1.25/1.5855
Total votes: 10 10 10

We now combine all the keep values from the Condorcet group-pairing counts into table 7.

Table 7: Condorcet group-pairing geometric mean preference and unpreference keep values.
(0) Candidates. (1) Multiplied pref. keep values from tables 1, 3, 5. (2) Geometric mean pref. keep values: Square root of col. (1). (3) Multiplied unpref. keep values from tables 2, 4, 6. (4) Geometric mean unpref. keep values: square root of col. 3. (5) Effectively elective keep values: inverse of col. (4).
A 1.25/2 x 1.25/2.6875 .5392
B 1.25/1.409 x 1.25/4.214 .513 ---- 1.25/.0234
C 1.25/.466 x 1.25/.5833 2.3976 .4171
D 1.25/.375 x 1.25/.375 3.3333 1.25/1.068 x 1.25/.2735 2.3128 .4324
E 1.25/.5 x 1.25/1.75 1.3363 1.25/1.375 x 1.25/1.5833 .8472 1.1804
F 1.25/.75 x 1.25/1.125 1.3608 1.25/1 x 1.25/1 1.25 .8
G 1.25/2 x 1.25/2 .625 1.25/2 x 1.25/3 .5103 1.9596
H 1.25/1.375 x 1.25/1.6875 .8206 1.25/2 x 1.25/1.091 .8462 1.1818
I 1.25/.091 x 1.25/1.2593 3.6925 1.25/.375 ----
J 1.25/.034 x 1.25/.1782 16.0589 1.25/1.375 x 1.25/1.25/1 1.4658 .6822
K 1.25/2.75 x 1.25/4 .3769
L 1.25/1 x 1.25/1.1782 1.1516 1.25/.091 ----
M 1.25/1 x 1.25/1.0699 1.2085 1.25/1 x 1.25/1 1.25 .8
N ---- 1.25/.2593 1.25/1 x 1.25/.2445 2.528 .3956
O 1.25/2 x 1.25/2 .625 1.25/.2446 x 1.25/.4978 3.5822 .2792
P 1.25/.375 --- ---- 1.25/.5833
Q 1.25/.6875 x 1.25/.0019 34.5857 1.25/1.375 x 1.25/1.5833 .8472 1.1804
R 1.25/.3034 x 1.25/1.6875 1.7469 1.25/.375 x 1.25/.5833 2.6729 .3741
S ---- 1.25/.375 1.25/2.0234 x 1.25/3 .5074 1.9708
T 1.25/.884 x 1.25/.6875 1.6034 1.25/1.5438 x 1.25/2 .7114 1.4057
U 1.25/2.0909 x 1.25/1.375 .7342 1.3565

From table 7, candidates K, B, A, O and G, H are elected with L as the runner-up. This was also the case for the election of all 21 candidates at once, tho the elected were not in quite the same order.

Looking at permutation table 0 of all 21 candidates, apart from the one first preference, L does not get another preference listing till the seventh preference of row 1. The 14 candidates sub-set table III, advances this to a third preference but it is not enough to do much for L's candidature. In principle, the Condorcet count supplement could advance a candidature. But equally if the in-depth preferences are not there, as in L's case, then the candidate's cause will deservedly not prosper.

The excluded candidates are S, G, T, U, H, Q. But H still would win an over-all election ( multiplying .8206 ( col. (2) ) by 1.1818 ( col. (5) ) which equals .9698, just under the required keep value of unity.

G would not win an over-all election but with a preference keep value of .625 is beyond statistical doubt safe from being eligible for unelection ( as previously explained ).

Altho the result seems reasonably reliable, in its own right, we have not taken into account the original election of 7 representatives from 21 candidates all at once. This information shouldnt be left out. That count has to be combined with the above sub-counts to get a more fully representative result than either result on its own.
So, on to table 8.

Table 8: Combined general and ( Condorcet group-pairing ) primaried, election and exclusion keep values.
(0) (1) General election of 21 candids: preference keep values (from previous web page). (2) Primaried elections of 14 candids.: GM pref. keep values (from table 7 col. 2). (3) Over-all weighted* geometric mean preference keep values. (4) General exclusion of 21 candids.: unpref. keep values. (5) Primaried exclusions of 14 candids.: GM unpref. keep values (from table 7 col. 4). (6) Over-all weighted* geometric mean unpref. keep values. (7) Effective election inverse of (6) (8) Final over-all GM keep value: square root of (3) x (7).
A .625 .5392 .58914
B .8872 .513 .71259
C 2.3976
D 13.7363 3.3333 7.796 36.76 2.3128 12.15935 .08224 .8007
E 2.5 1.3363 1.9459 3.3333 .8472 1.92716 .5189 1.0049
F 3.0562 1.3608 1.99647 1.25 1.25 1.25 .8 1.597
G .9091 .625 .78255 .625 .5103 .5763 1.7352 1.1653
H .9091 .8206 .8726 .8462
I 3.6925
J 10 16.0589 12.0865 3.3333 1.4658 2.11267 .47333 2.3918
K .5263 .3769 .4605
L 1.25 1.1516 1.20967
M 1.2085 1.25 1.25 1.25 .8
N 2.528
O .625 .625 .625 36.76 3.5822 14.4849 .069 .2077
Q 34.5857 .90909 .8472 .8838 1.13148
R 1.7469 3.3333 2.6729 3.05145 .32771
S .625 .5074 .57498 1.7392
T 1.6034 13.7363 .7114 4.203 .23792
U 1.1457 .7372 .96048 1.04115

*Weighted geometric mean.

In table 8, columns 3 and 6, the over-all keep values for preference and unpreference were calculated, in the same way, by proportional weighting determined by the ratio of candidates in the general election ( 21 ) and the ( Condorcet group-pairing ) primaried elections ( 14 ). The primaried elections had their own geometric mean keep values, derived in table 7. But here the geometric mean was not weighted because all the sub-counts shared the vote between the same number of 14 candidates.

Taking a geometric mean of a general election with 21 candidates and the sub-count out-come for 14 candidates, the keep values for the general election were harder to obtain, because the votes had to go round more candidates. Therefore the general election keep values should be given proportionately more weight in deriving a geometric keep value out of the 21 candidates election and 14 candidates sub-routine.

We are dealing with geometric weights not arithmetic weights. That means it is no good weighting the general election votes by three times to the primaried election votes multiplied by two. I did try it and it gave wild results. The proper weighting is geometric. That means the votes won by candidates in the general an primaried elections must be to powers not multiples.

What are the correct powers? Well, remember the unweighted geometric mean takes the square root of two keep values to give an over-all keep value. The numerators of the multiplied keep values are the same, namely the quota, in this ridiculous example, 1.25. So, the numerator stays 1.25. This leaves the two denominators as the votes a candidate won in the general and primaried elections. For the unweighted geometric mean taking the square root of both, the respective votes are both to the power of one half.

Tho these respective votes won by a candidate differ, the multiplication of two values, both to the power of a half, add up to a value which is of the order of unity. Because, one adds powers to derive value, X to a power, multiplied by X to another power. Moreover, we want a final keep value result that is of the power order of unity, a standard unitary value. That is one keep value, not a keep value and a bit or a bit of a keep value.

Therefore, the weighting of candidates' general election votes, X, and primaried votes, Y, should be powers that add up to one and do so in the ratio of three to two, respectively. The answer is X to the power of three-fifths or .6, and Y to the power of two-fifths or .4. These two values for each candidate are multiplied and divided into the quota, 1.25. That is how, in table 8, the weighted keep values for preference and unpreference are calculated, in columns 3 and 6, respectively.


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The following sections were written before the idea of supplementing Binomial STV with Condorcet pairing, including group pairing, and weighted averaging keep-value averages of both types of count. The advantage of this kind of averaged average count is that it allows more analysis of the preference information that is given, rather than requiring the voters to make very full returns of preferences, which previous count systems might not make so much use of.
Moreover, keep-valued Condorcet ( group ) pairing brings quantitative count of primaries to one election, without making the voters have to vote in a series of primary elections. On the other hand, voters might prefer more primaries of fewer candidates to choose from.

Too many preferences?

A reversible or binomial STV gives the voters greater power and with that comes greater responsibility. Indeed, it is this greater power which is a measure of the improved efficiency of the system, that justifies the introduction of this reform. But, as always when there is a prospect of greater power to the people, detractors can point to a greater prospect for its irresponsible use.

With traditional STV, voters have only to prefer the candidates they care anything about. Voters know that not extending preferences, to candidates they dont like or actively dislike, will do nothing to help any of them. With a reversible or binomial STV, if voters extend their preferences right down to the candidate they least like, they know they will certainly be helping to exclude candidates in the order they least like them, as well as STV's usual advantage of electing candidates in the order that the voters most like them.

This should encourage voters to extend their preferences in the ballot. For this reason, it would not be legitimate to re-count elections held under traditional STV, to say some candidates should or should not have been elected. STV voters did not have as strong an incentive to maximally extend their preferences. Therefore, after-the-fact re-counts of STV elections by a "two-ways" STV need not be legitimate re-assessments of the results.

STV has been disliked by partisans because it allows voters a unifying escape, from their prison of party, by prefering candidates across party lines. It might be too hasty to suppose that a reformed STV's encouragement, to express dislikes, incites an oppositional feeling of two nations. Encouraging voters to think about degrees of dislike, as well as degrees of liking, obliges one to think about the relative merits of opponents. Instead of just bundling together 'the other side' as a faceless horde of enemies, one is obliged to recognise opponents also as individuals deserving personal consideration.

Critics may say this is asking too much of voters. But it should not be asking too much of people to consider their diametric opponents as human beings. In any case, that is no criticism of the rightness of a new method. And it is perhaps enough that some voters more or less respond to the chance to more fully express their views to greater effect.

Resumed preferences in large constituencies.

One problem is the large numbers of candidates in multi-member constituencies. The original STV of Hare's system treated the country as a constituency. John Stuart Mill's promotion of the system didnt envisage the voters having to prefer hundreds of voters. He knew perfectly well that national figures could draw votes from all over the country, and that well-known regional or local characters could get enough local support to achieve the quota. And this didnt require the voters to prefer all that many candidates before all the seats to parliament were taken.

Then how is a reversible or binomial STV to function in such large constituencies, as those in Australia, for example? Australia, except Tasmania, obliges voters to either state every preference for scores of candidates or just put an X by a party list. Not surprisingly, nearly everybody abdicates such an unrealistic statement of preference between so many candidates they cannot know all that well or reliably assess.

There is a way a reversible STV system could work in large constituencies without compulsory ballot of every preference. The ballot paper should explain that you can order your favorite candidates, first, second, third etc. Also, when you have run out of highest preferences, you may state your lowest preferences.

For instance, the ballot paper may say there are 37 candidates standing in this constituency. If you have a least prefered candidate, you may number him 37, and a next least prefered candidate may be numbered 36, and so on. In other words, a resumed preference or resumed order of choice could be allowed with STV.

The provision of a count from least prefered, as well as from most prefered, implies that voters need not express preferences for candidates they neither much like nor dislike. That is to say, the voters may make a Resumed Preference order that gives power of election and exclusion but does not compel a complete order of preference of such length that the voters cannot possibly know and assess all the candidates.
A reversible STV with Resumed Preferences means the public is not obliged to prefer all or none of a herd of candidates using their large numbers to run in safety from public discernment.

Primaries to make quota achievement more likely.

Quotas of election and exclusion are more likely to be achieved if there are not too many candidates chasing too few seats. For instance, only five candidates for three seats is a ratio that makes it unlikely no candidate will have a quota. Suppose 32 voters. The quota is 32/( 3 + 1) = 8. Five candidates would have to be nearly in a "dead heat" for none of them to achieve a quota. That is something like three candidates having six votes each and the other two having seven votes each.

This would be a state close to equilibrium of choice. No stored "energy" in the form of surplus votes over the quota is available for the electoral energy transfer known as transferable voting.

But voting tends to follow probability distributions, with different levels of support for candidates, more or less about an average vote that identifies candidates of middling popularity. So, it is possible to set higher ratios of seats to candidates, to improve the probability of quota attainments. Also, there is a second chance of quota attainment, with an exclusion quota as well as the election quota.

To make the greater accuracy of quota measurements more probable, primaries might be needed to increase the ratio of seats to candidates. Tho, STV, in multi-member constituencies, already does vital work, in this respect, by allowing rankings of candidates in the same party, as well as other parties.

Reversible STV systems belong to the evolution of voting method towards greater efficiency, in its innovatory use of proportional counting, to further replace plurality counting. Voting is not just an axiomatic system, which is somehow irrational if it does not meet all the reasonable demands of social choice theorists. Voting, as it exists in the real world, is more like, say a thermo-dynamic engine with limits to its efficiency. The theorists are actually measuring those limits. For them to expect all their reasonable conditions to be met -- what amounts to an ideal system -- is like expecting a perpetual motion machine.

Richard Lung.
3 January 2006.

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