On the previous page, the Retransferable Vote was treated as four possible ways to conduct a transferable vote. This came about after I realised that, in developing a Reversible STV, there were two ways that a transferable vote could be counted with its preferences in reverse, either considered in relation to the normal preferential count or regardless of it.
This looked like a progression of the proportional count. Firstly, there was traditional STV with one election quota count. Secondly, there was Reversible STV ( rSTV ) with an election and an exclusion quota count. Thirdly, there was the Retransferable Vote ( RV ) with a preference election count qualified by an unpreference exclusion count and an unpreference election count qualified by a preference exclusion count. In other words, there was STV with one quota count, rSTV with two quota counts and RV with four quota counts.
Whenever one comes across combinations like the four combinations of RV, one thinks of the possibility of more elaborate combinations, such as those offered by the binomial theorem as it is expanded using higher powers. The bi in binomial signifies two factors, in this case, preference and unpreference. Thus, the binomial theorem expanded to the power of two produces two, to the power of two, equals four combinations.
Moreover, the binomial theorem at zero power, or two, to the power of zero, equals one. This corresponds to STV's one quota count. The binomial theorem, at two, to the power of one, corresponds to the two quota counts of rSTV. Ive updated this count, which was not keepvalue weighted, to a first order binomial STV count, in the next section.
We should seek a binomial theorem expansion of the STV count, because it forms a series of closer approximations to a measure of the voters' choice.
This treatment of the count improves on Reversible STV and its contingent exclusion procedure.
Table 0: Voters' permutations.
Votes per permutation.  Permutations.  Permutation row number  

2  D  E  A  C  B  1 
5  A  C  E  D  B  2 
6  E  B  C  D  A  3 
1  C  D  E  B  A  4 
3  C  D  A  B  E  5 
5  C  A  D  E  B  6 
3  C  B  E  A  D  7 
7  B  E  A  D  C  8 
32 votes total. 
The returning officer starts table 1 for preference election, by adding up, from table 0, how many first preferences go to each candidate. Candidate C has more votes than the quota and is declared elected in stage 1. The value of C's surplus is proportionly shared between all C's voters' next preferences, which each have a transfer value, calculated as (12  8)/12 = 4/12 = 1/3 of a vote.
Table 0 shows 12 first preference voters for C fall into four different
permutations. 1 voter, for perm 4, and three voters, for perm 5, both make D
their second choice. So D gets 4 transferable votes at 1/3 transfer value
each, or 4/3 votes.
Perm 6 transfers 5 x 1/3 = 5/3 votes from C to A. Perm 7 transfers 3 x 1/3 =
1 vote from C to B. This transfer of C's surplus is shown in table 1, stage
2. Adding the surplus transfer, to the remaining candidates' previous tally,
is enough for B to reach a quota and be declared elected ( in stage 2
continued ).
Candidates.  Stage 1.  Stage 2.  Stage 2 ctd.  Keep values. 

1st preferences: C elected 
Transfer of C's surplus at 4/12 = 1/3  B elected  Divide quota by votes candidates achieve.  
D  2  4/3  3 1/3  8/3.33 = 2.4 
A  5  5/3  6 2/3  8/6.67 = 1.2 
E  6  6  8/6 = 1.33  
C  12  8  8/12 = .67  
B  7  3/3  8  8/8 = 1 
total vote:  32  32 
Table 2 is derived from table 0 in just the same way as table 1 was. The only difference is that table 0 is read from last to first preferences ( right to left ), instead of from first to last.
Table 2: Unpreference count. Quota = 8.
Candidates.  Stage 1.  Stage 2.  Stage 2 Ctd.  Keep values. 

Last preferences  B surpasses unpreference quota. B's last prefs transfered @ 4/12 = 1/3 surplus value to next least prefered candidates.  Add B's surplus to B's voters' next least prefered candidates. No further candidate's unpreference elected.  Divide quota by candidates' achieved votes.  
D  3  5/3  4.67  8/4.67 = 1.71 
A  7  7  8/7 = 1.14  
E  3  5/3  4.67  8/4.67 = 1.71 
C  7  2/3  7.67  8/7.67 = 1.04 
B  12  8  8/12 = .67  
Total votes:  32  32 
Table 1, being of preference, is an election count. Table 2, of unpreference, is an exclusion count. But inverting the keep values of exclusion effectively makes them election keep values. Table 3 then multiplies the election keep values of the candidates by their inverse exclusion keep values to derive overall keep values. These are a more informative index of the candidates' relative performance.
(0) Candidates  (1) Election keep values  (2) Inverted exclusion keep values  (3) Election times inverse exclusion keep values  (4) Overall (geometric mean) keep values: Square root of col. (3). 
D  8/3.33 = 2.4  .58  1.39  1.18 
A  8/6.67 = 1.2  .88  1.06  1.03 
E  8/6 = 1.33  .58  0.77  .88 
C  8/12 = .67  .96  0.64  .8 
B  8/8 = 1  1.49  1.49  1.22 
Thus, the first order binomial STV gives about the same result, in the same order: C, E, A, as, in the section below, a second order binomial expansion of the count, using four controled counts, instead of just two, here. If there is any advantage in the more complicated second order expansion of counts over this first order count, it is that candidate A is confirmed as elected with a keep value of less than one. Here, A has a keep value of just over one and is therefore technicly not quite elected to the third seat.
I should explain why this is the simpler first order expansion, tho it is easier to appreciate in the context of the second order expansion. The bi in binomial refers to two terms, which may be labeled p for preference and u for unpreference. The second order expansion would be the result of squaring the sum of p and u. That is ( p + u )². The ( noncommutative ) expansion of that formula determines the counts. All that needs be said here is that the first order expansion is ( p + u ), which is implicitly to the power of one. And this signifies one preference count and one unpreference count, which is what the above example of a first order binomial STV count consists of..
Footnote ( 2 december 2005 ): Strictly speaking, the square roots of the overall keep values should be taken, because the elective and effectively elective ( inverse exclusion ) keep values may be considered as the endpoints of a range of values. Taking the square root of the multiple of two end points gives an average called the geometric mean. First order Binomial STV's two counts thus combine for a more typical or representative keep value, for each candidate than either on their own.
( 8 dec. '05 ) Not only did the cause of simplicity stay me from completing, with the square root, the geometric mean of the keep values. But also I overlooked, as long as I could, the fact these keep values are only relative. By this I mean that a candidate might get a poor preference keep value well over unity ( say 6 ) but so long as the candidate had a relatively good inverted exclusion keep value ( say 1/6 or less ) this would serve to give the candidate an overall keep value of one or less and so finally "elect" that candidate.
It is not entirely out of the question that such a "relativistic" system
might be logicly justified. One could argue that it doesnt matter that a
candidate is "elected" with very little support given that there is even less
objection amongst the voters to his exclusion. There is tacit or passive
consent for want of active opposition.
However, this state of affairs would seem to encourage the kind of corruption
that comes from power that does not need popular participation. And it is
easy to see that such a system would not be acceptable.
Therefore, we move to "Plan B" that there must be limits to a candidate's elective keep value, beyond which he cannot be elected with a poor elective keep value, nor unelected with a good elective keep value, no matter how great a counterweight from the inverse exclusion keep value.
The simplest amendment, I can think of ( but not I think the statisticly correct one ) is that a candidate with an elective keep value of 3/4 or less, in a three member constituency, should be deemed elected. And not be unelectable by the inverse exclusion keep value.
Supposing that to be the case, we can also estimate the upper bound of election keep values, beyond which a compensating inverse exclusion keep value cannot be allowed to help elect a candidate. The geometric mean is the square root of the multiple of upper and lower bounds to a possible series of values. Or, GM = Sq. rt. ( upper bound times lower bound ). That is one squared equals upper bound times threequarters. Or, upper bound equals fourthirds.
In general, such lower and upper bounds of qualification of the election keep value are R/(R+1) and (R+1)/R respectively, where R is the number of representatives or seats in the multimember constituency.
Therefore, any candidate getting an election keep value of over 4/3 = 1.333... is deemed too much over unity to be within the bounds of qualifying for a possibly compensating inverse exclusion keep value. It so happens that in table three above, that candidate E's election keep value is exactly this value of 1.333... That is E is on the outer limits of qualifying for inverse exclusion as an effective election at .77. This is low enough to bring E's overall keep value below unity and qualify for election to the third seat.
Had E's election keep value been just over 1.333... then E would not have
qualified. Candidate A, with an election keep value of 1.2, would qualify for
compensation from its effective election keep value. But this value of .88 is
not quite enough to bring below unity A's overall keep value, which is 1.03.
However, this value is considerably better than B's overall value at 1.22. B
just qualifies for election with an elective keep value of 1. But any range
of unelectability, about one, subjects B to possible defeat. Defeat that
occurs because of the large unpreference vote for B.
By first order Binomial STV and, as it happens, also with its qualified
version, C and E are elected with A qualifying as the runnerup.
I should conclude with the tedious but necessary detail that the above simple bounds, for allowing candidates to be elected or unelected with reference to effectively elective ( from inverted exclusion ) keep values, are probably statisticly too generous. This is discussed on my page of a Further example of first and second order Binomial STV.
Tho the four counts of the Retransferable Vote have a certain logic, I eventually realised that it is not the logic of the second order binomial theorem expressed as a transferable voting count. It turned out that only two of the counts in RV were the same as required for the second order binomial expansion. The other two counts in RV turned out to be the two counts in the first order binomial expansion.
RV's four counts can be renamed in terms suitable for a binomial count.
The first count was akin to STV, that is an election of the most prefered. (
This was the "preference" count, a first order binomial count, which may be
given the symbol p. )
The second count turned round the voters' preferences, starting from the most
unprefered, to see how many candidates could be excluded on reaching quotas
of unpreference. But that unpreference count was qualified by any candidate,
in the first tabled count, who had already received an election quota. That
candidate could not be excluded and so all the voters, who least prefered
that candidate, had their votes redistributed to count against their next
least prefered candidates. ( This second count is renamed the "preference
qualified unpreference" count. Being a count qualified by a count, it amounts
to a second order binomial count, which may be symbolised: pu. )
The retransferable vote's third count was similar to the second count,
except that the count of unpreference was not qualified by the first count of
preference. ( In binomial terms, this was just an "unpreference" count,
unqualified by another count, and therefore a first order count.
)
The fourth count turned the voters' preferences round again, back to starting
with highest orders of choice. But instead of being a straight preference
count, like table 1, table 4 was qualified by excluding any first candidate
to reach a quota of unpreference, in table 3. ( RV's fourth count is
binomially renamed the "unpreference qualified preference" count, another
second order count, symbolised: up. )
The question now is what are the four counts we need to modify the retransferable vote to a second order binomial count? Well, we already have the two counts for a first order binomial count. The first order binomial expansion of the algebraic terms p and u, standing for preference count and unpreference count, is simply p and u, because a term or a combination of terms, to the power of one, is simply itself.
Using keep values, including the novelty of deficit candidates' keep values, as a measure of the result, amounts to a ratio of the quota divided by the maximum vote a candidate achieves, whether or not that is a vote in surplus of a quota or in deficit of it. As each voter has one vote, the keep value ( the part of a vote that a candidate gets to keep ) is less than one for a candidate who has a surplus vote, over the elective quota, transferable to a next prefered candidate. The larger the candidate's surplus vote, the larger the "transfer value" as a part of a given unit vote that the elected candidate does not get to keep.
The novelty I introduced was to have keep values for candidates in deficit of a quota, which means for candidates with negative surpluses. As keep value plus transfer value equals one, a deficit candidate has a keep value of more than one.
The point is that a candidate who is in deficit of a quota in one count may be in surplus of a quota in a system using more than one count run. Indeed, it is an advantage to be in deficit of a quota when the quota is of least preference votes. The overall result is what finally counts, from count runs in conformity with the logic of a binomial distribution of combinations of preference counts and unpreference counts.
The second order binomial expansion is the well known quadratic
equation:
( p + u )² = p² + 2pu + u²
However, this is the expansion that conforms to the rules of arithmetic. The
rules of logic differ slightly in that "pu" is not the same as "up". The
order of p and u is of import, so p and u do not "commute" or exchange, here:
Rather, the second order expansion is: pp + pu + up + uu.
Also the plus sign does not mean what it does in arithmetic. The plus here is used simply in the commonsense way we mean "and". Logic has its own symbol for "and" ( ^ ) with its own specific meaning in logical operations.
It is legitimate to use the binomial theorem as a logical, rather than an arithmetic, expansion. It applies to "truth tables" found in symbolic logic or finite mathematics. Indeed, p and u could be considered as equivalent to the terms true and false ( that is true and false preference ) which are used to construct tables of all the logical possibilities of truth and falsehood to a statement that determine whether its logic is true or false.
Having logicly expanded the second order binomial theorem, we see that the retransferable vote differed from it only by possessing the terms p and u, rather than pp and uu. If we convert p to pp and u to uu, then we have the four counts needed for second order binomial STV.
The count pp means the preference count qualified by the preference count.
In other words, a preference count, p, like the original STV count, is
repeated with the qualification that the candidate, with the largest surplus
of votes, has already been elected and so cannot be elected again. That
candidates' votes are redistributed to help elect their next prefered
candidates.
An analgous process takes place to derive the "unpreference qualified
unpreference" count, uu.
In the qualified count, only the one candidate with the biggest surplus ( of preference or unpreference ) can be excluded from that count. This exclusion is only a suspension for the purposes of the counting procedure, not a final decision on the candidate's prefered or unprefered status. ( "Excluded" here is rather in the sense of "controlled" for, to use the jargon of experimental procedure ). That way, as the relevant count tables show below, there are just enough  but not too many  votes to go round for all the other candidates, in turn, to reach a quota, if all the voters have stated complete sets of preference for every candidate.
In the election example, I keep using, there are 32 voters electing 3 candidates on a quota of 32/(3+1) = 8 votes. In a qualified count, the foremost candidate is excluded as if already elected ( whether to preference or unpreference ) so that there are still 32 votes to go round, but only for 4 remaining candidates. This means that if all voters express a preference for every candidate, those remaining candidates will all eventually in turn reach the quota  the last candidate only just reaching the quota, in those most extreme circumstances. That is the limiting case of 4 candidates times the 8 votes quota equals all 32 voters. Therefore, no more than one candidate can be excluded, or there would be more votes than needed for each candidate to achieve a quota each and the quota condition of equal election would be broken.
Practise makes perfect, and only trial elections will really assure that Binomial STV works well. My guess is that for most practical purposes, the second order binomial count should clarify the collective order of choice for the candidates. When voters do not express many preferences for all the candidates, then there may be a struggle for enough candidates to be prefered to all ( of a large number of ) vacant seats.
If the preference shortfall is serious enough, given a lot of candidates unfamiliar to the voters, then a third order binomial count may come into its own. When voters cease to register further preferences ( or further unpreferences given a "Resumed preference" vote ) then the total vote may be effectively reduced, so that it does not exceed the quota times the number of remaining candidates, who have not been pulled out of the binomial STV count, to pass on their voters' next preferences ( or unpreferences ) to those remaining candidates.
To begin, the votes are added according to how many voters share a given order of choice for the candidates. This is set out in a table of the permutations of preference, that the voters happen to have chosen. I call this table 0 ( that is table zero ) of voters' permutations.
Votes per permutation.  Permutations.  Permutation row number  

2  D  E  A  C  B  1 
5  A  C  E  D  B  2 
6  E  B  C  D  A  3 
1  C  D  E  B  A  4 
3  C  D  A  B  E  5 
5  C  A  D  E  B  6 
3  C  B  E  A  D  7 
7  B  E  A  D  C  8 
32 votes total. 
The returning officer starts table 1 for preference election, by adding up how many first preferences go to each candidate. Candidate C has more votes than the quota and is declared elected in stage 1. The value of C's surplus is proportionly shared between all C's voters' next preferences, which each have a transfer value, calculated as (12  8)/12 = 4/12 = 1/3 of a vote.
Table 0 shows 12 first preference voters for C fall into four different
permutations. 1 voter, for perm 4, and three voters, for perm 5, both make D
their second choice. So D gets 4 transferable votes at 1/3 transfer value
each, or 4/3 votes.
Perm 6 transfers 5 x 1/3 = 5/3 votes from C to A. Perm 7 transfers 3 x 1/3 =
1 vote from C to B. This transfer of C's surplus is shown in table 1, stage
2. Adding the surplus transfer, to the remaining candidates' previous tally,
is enough for B to reach a quota and be declared elected ( in stage 2
continued ).
Candidates.  Stage 1.  Stage 2.  Stage 2 ctd.  Keep values. 

1st preferences: C elected 
Transfer of C's surplus at 4/12 = 1/3  B elected  Divide quota by votes candidates achieve.  
D  2  4/3  3 1/3  8/3.33 = 2.4 
A  5  5/3  6 2/3  8/6.67 = 1.2 
E  6  6  8/6 = 1.33  
C  12  8  8/12 = .67  
B  7  3/3  8  8/8 = 1 
total vote:  32  32 
There is one more candidate to elect to the third available seat but no more surplus votes to transfer.
Table 1 of the preference count is a first order binomial count not required for the second order binomial count. But it is required as the qualifying count of a repeated preference count, that redistributes the votes of the leading elected candidate, if any, to derive the "preference qualified preference count", which is the first of the four second order counts required for the second order binomial count. See table 2.
Candidates.  Stage 1.  Stage 2.  Stage 2 ctd.  Stage 3.  Keep values. 

1st preferences: C already elected in table 1.  Exclude C, redist. 12 votes.  A elected. B already elected in table 1.  2 surplus votes each from A and B eventually go to D and E.  Divide quota by votes candidates achieve.  
D  2  4  6  8  8/8 = 1 
A  5  5  10  8  8/10 = .8 
E  6  6  8  8/8 = 1  
C  12  0  0  8/12 = .67  
B  7  3  10  8  8/10 = .8 
total vote:  32  32  32 
Table 3 is another count qualified by table 1. This time the preference count qualifies the unpreference count. The unpreference count carries out the quota count on the candidates in order of their least preference with the voters. For this purpose, preferences are read from right to left in table zero. But this is done with the qualification that candidate C is already elected in table 1 and so cannot be excluded in table 3. Therefore the 7 voters who least like C have their unpreference votes redistributed to their next least prefered candidates.
These 7 votes from C all go to D ( See table 0, perm row 8. )
D is excluded with 10 votes. ( See table 3, stage 2. ) D's two surplus votes
have a 2/10 transfer value. All 10 of D's votes go to A, who therefore gets
10 x 2/10 = 2 votes. This brings A from 7 to 9 votes, just over the exclusion
quota. ( See table 3, stage 2 ctd. )
Notice that in the last column of this table 3 and the other four binomial
count tables that the higher surplus a candidate achieves, this goes into the
denominator of that candidate's keep value, to lower the ratio with respect
to the quota, in this example, 8 votes.
Thus, with respect to this second order binomial STV election, the lowest
keep values for preference ( as in tables 2 and 4 ) help to determine the
most prefered candidates, and the lowest keep values for unpreference ( as in
tables 3 and 6 ) help determine the least prefered candidates.
Candids.  Stage 1.  Stage 2.  Stage 2 ctd.  Stage 3.  Stage 3 ctd.  Keep values. 

Last prefs. 
C already elected, redist. C's 7 votes.  Excluded D's surplus transfer @ 2/10.  Excluded B's surplus transfer @ 4/12.  Add B's surplus.  Divide quota by candidates' achieved votes.  
D  3  10  8  8  8/8 = 1  
A  7  7  9  2/3  9.667  8/9.667 = .83 
E  3  3  3  10/3  6.333  8/8 = 1 
C  7  0  0  0  8/7 = 1.14  
B  12  12  12  8  8/12 = .67  
Total votes: 
32  32  32  32.000 
As with table 3, table 4 starts with the last preferences. But now the object is simply to "elect" or determine the most unprefered candidates. Like table 1, table 4 is a count in its own right, albeit of unpreferences. Hence, the first stage is to transfer any surpluses of last preferences, as is done for candidate B with 4 votes above the quota of unpreference or unpopularity.
Just as table 1 served to qualify table 2, the preference qualified preference count, so table 4 here serves to qualify table 5, which will be the unpreference qualified unpreference count.
Candidates.  Stage 1.  Stage 2.  Stage 2 Ctd.  Keep values. 

Last preferences  B surpasses unpreference quota. B's last prefs transfered @ 4/12 = 1/3 surplus value to next least prefered candidates.  Add B's surplus to B's voters' next least prefered candidates. No further candidate reaches unpreference quota.  Divide quota by candidates' achieved votes.  
D  3  5/3  4.67  8/4.67 = 1.71 
A  7  7  8/7 = 1.14  
E  3  5/3  4.67  8/4.67 = 1.71 
C  7  2/3  7.67  8/7.67 = 1.04 
B  12  8  8/12 = .67  
Total votes:  32  32 
Note that table 5's count run gives candidate A a default victory as the last to reach a quota of unpreference.
Candidates.  Stage 1.  Stage 2.  Stage 2 Ctd.  Keep values. 

Last preferences  B quotaunprefered in table 4. B's last prefs. redistribute to next least prefered.  Add B's vote of last prefs. to B's voters' next least prefered candidates. D, E and C reach unpreference quota.  Divide quota by candidates' achieved votes. ( C's surplus vote transfers to A.) 

D  3  5  8  8/8 = 1 
A  7  7  8/8 = 1  
E  3  5  8  8/8 = 1 
C  7  2  9  8/9 = .89 
B  12  0  8/12 = .67  
Total votes:  32  32 
The second order binomial expansion, of preference and unpreference counts, was explained above to consist of four combinations. So far, pp ( table 2 ) and pu ( table 3 ) and uu ( table 5 ) have been calculated. Table 6, below is for "up" unpreference qualified preference. That "pu" and "up" do not commute, in the binomial expansion count, means that table 3 and table 6 are not the same counts.
"up" table 6 is qualified by "u" table 4, as "pu" table 3 is qualified by "p" table 1.
Cands.  Stage 1.  Stage 1 ctd.  Stage 2  Stage 2 ctd.  Stage 3  Stage 3 ctd. 
Keep values. 

1st prefs.  Thru table 4's unpref. quota, redist. B's 7 votes.  Transfer of E's surplus @ 5/13.  Add E's surplus transfer 
Transfer C's surplus @ 6.31/14.31 = .44  Add C's surplus. 
Divide quota by candidates' maximum votes. (C's surplus passes to D.)  
D  2  2  2  6.31x.44 = 2.78  4.78  8/8 = 1  
A  5  5  7x5/13 = 2.69  7.69  8x.44 = 3.53  11.22  8/11.22 = .71 
E  6  13  8  8  8/13 = .62  
C  12  12  6x5/13 = 2.31  14.31  8  8/14.31 = .56  
B  7  0  0  0  8/7 = 1.14  
total vote:  32  32  32.00  32.00 
Table 7 tabulates the keep values for the four kinds of count given by the
second order binomial expansion. The two qualified preference keep values and
the two qualified unpreference keep values are respectively multiplied to
form the combined preference keep value and the combined unpreference keep
value. These two combinations then form a ratio of preference to
unpreference, by which the most prefered candidates have the lowest keep
values. A keep value ratio of one or less implies an elective quota
achieved.
Candidates.  (1) From table 2: pref. qual. pref.: pp  (2) From table 3: pref. qual. unpref.: pu  (3) From table 6: unpref. qual. pref.: up  (4) From table 5: unpref. qual. unpref.: uu  (5) pp x up  (6) pu x uu  (7) Pref. to unpref. ratio: (pp x up)/(pu x uu) 

D  1  1  1  1  1  1  1 
A  1  .83  .71  1  .71  .83  .86 
E  1  1  .62  1  .62  1  .62 
C  .89  1.14  .56  .89  .498  1.01  .49 
B  .67  .67  1.14  .67  .764  .44  1.74 
This revised version of the example, of the Retransferable Vote, to comply with a second order binomial distribution of counts, shows a similar result. The candidates' keep value ratios are all in the same order. The same three candidates are elected: C, A and E. This is pleasing because it suggests the result has some stability for two moderately varying ways of looking at the logic of a retransferable vote.
Remember, also, from the above section that first order Binomial STV gives C and E elected with A qualifying as the runnerup.
Just a year after writing this page, I decided to amend table 7 with table 8. Table 8 is yet more ( dare I say it ) complicated than table 7 but I am not as comfortable with the short cut as I was a year ago. Tho, in this example there is no change in the relative positions of the candidates, comparing the overall keep values in the last columns of tables 7 and 8.
The table 8 amendation means that each time you multiply two keep values, as in table 7 columns (5) and (6), you have to take their square roots to restore them as an average keep value of the two multiplied keep values. And finally, when you multiply the average elective keep value by the inverse of the average exclusive keep value, you have to take that multiple's square root to reach an average of the two averages. That is to say each candidate's overall or final keep value is an average of two averages. This averaging of averaging is a process of approaching the most typical or representative possible keep value for each candidate.
The average involved is known as the geometric mean. It consists, in this
case, of treating two keep values, such as pp and up, as the endpoints of a
range of values, whose square root is taken.
Candidates  (1) Square root of (pp x up)  (2) Square root of (pu x uu)  (3) Inverse of col. (2)  (4) Cols. (1) x (3)  Overall keep values: Square root of col. (4) 
D  1  1  1  1  1 
A  .843  .911  1.098  .926  .962 
E  .787  1  1  .787  .887 
C  .706  1.005  .995  .702  .838 
B  .874  .663  1.508  1.318  1.148 
The above example is one where voters equally unprefer as prefer candidates, stating all lesser preferences as well as greater preferences. ( By the way, there is the logical possibility that voters might omit a first preference, as well as a last preference. But that is an arcane point that can be left for many more pressing considerations. ) Equal voting for preference and unpreference implies that the normal or symmetrical version of the binomial distribution is applied to the logic of the count, as has been done in the above example.
If there is a significantly greater show of preference than unpreference for the candidates, or vice versa, in the preference voting pattern, then the STV binomial distribution count can be skewed to represent that. Say there were twice as much preference as unpreference in the votes. Then, the binomial theorem could do an expansion of ( 2p + u )² if the degree of approximation required was a second order count.
The degree of preference to unpreference, to arrive at this weighting of the binomial theorem, could itself be calculated, perhaps, by comparing weighted averages of the ranges of preference voting and unpreference voting. For instance, say, a couple of per cent of the voters might ballot the last 30 as well as the first 30 of, say, 60 candidates. But most of the voters might only register ( with a "Resumed Preference" vote ), say 60 per cent and 30 per cent, respectively, the last 5 or 10 candidates. A weighted average can take such differences into account to arrive at a representative average. Say, the voters on average prefered the first 16 candidates and on average unprefered the last 8 candidates. Then there is the above mentioned twotoone skew on the binomial expansion of the count.
All this is statistical approximation, which has not been contemplated before. But other statistical approximation has been used for STV, even where strictly speaking, it is not necessary, as when representative random samples are taken to transfer surplus votes.
When the binomial distribution is extremely skewed, it becomes the Poisson
distribution, based on the exponential series instead of the binomial series.
Indeed, the whole exponential family of distributions, and other series,
might be matched to different possible preferential voting patterns.
Mathematical distributions become choice distributions.
The high information content of the single transferable vote allows the above kind of intensive analysis. Statisticians have a whole array of lesser measures or tests for collected data with poorer information content. Nevertheless, binomial STV has brought home to me that any rule makers, for a returning officer's count, are putting an interpretation on the structure of the voting data.
The great virtue of STV, compared to other voting methods, has been its liberation of the vote from some presumption of what it is, imposed in the count. For example, the simple majority system assumes that single member constituencies are most representative, when they are only most local. A local monopoly on choice has usurped freedom of choice. Likewise, party list systems monopolise representation, because the preference vote is monopolised by the party boss making out his party list. These list systems are miscalled proportional representation when they are only proportional partisanship.
I dont think there is a practical problem with closer approximation of the STV count, by higher orders of binomial count. The binomial distribution ( with allowance for skewness of the distribution if the preference voting distribution warrants it ) is very common to patterns of natural and social behavior.
It's also worth mentioning, that in natural science, approximation calculations are very common. The rapidly increasing efficiency of computers has made them the norm. Typicly, calculations were only done to the first or second order of approximation. That may have been as much because no greater level of accuracy was required. I believe that some such first or second order Binomial STV count would be enough with the run of political STV elections.
But it doesnt really matter if third or higher order approximations were needed, because Binomial STV counts are strictly for a computer program. You only have to look at the eight tables, above, it took me to workup a piffling number of preference permutations into a result. And then it was only a traditional handcount. Meek's method, that follows every possible surplus transfer, without any traditional shortcuts, would have to be included in the Binomial STV program.
Our commonplace political assumptions are based on just such a perceived pattern that most votes are to be found in the centre of politics and that the right and left wing parties consign themselves to oblivion if they move too far to one or other extreme.
Much of the dynamic for electoral reform in Britain in the last quarter of
the twentieth century was based on the perception that a single
nontransferable vote polarises support between two dogmaticly
opposed "wing" parties, who alternately grabbed center stage from each other,
because the first past the post system exaggerates small swings in voter
support with large swings in the number of single member constituencies
changing hands between the parties.
That prophet of science, H G Wells said this early in the same century, and
he derived his knowledge of voting method from fifty years research since
Thomas Hare and John Stuart Mill.
To appreciate the future of voting research, I think we have to go back even further to the early nineteenth century and the origins of the transferable vote. Thomas Wright Hill asked his school boys to line up behind the most popular boys to be elected on a committee. The queues they formed were of uneven length, till some of the boys realised they could transfer themselves from the longest queues to help elect their next prefered candidate. And other boys, lined up behind the shortest queues of nohope candidates, moved to support boys with a better chance.
This was not only a first occasion for transferable voting. It was also a first for interactive voting. That is to say you have a realtime election where voters can respond to what each other is doing. This is, potentially anyway, much more subtle than a rigid set of mathematical rules under which large scale elections must be held to be managable. In fact, in Hill's interactive STV election, the participants dont even have to be literate or numerate. They see how long are the candidates' queues of support and react accordingly in this election game. There might be a chalk line that marks the quota a queue must reach but need not surpass for its candidate to be elected.
The rise of interactive computer games and digital realities means that informal elections like Hill's can be much more conveniently studied without its restrictions on scale and without the mass elections' restrictions by mathematical rules that have to be rigidly observed. Voting and reactive votings could be visually observed, as they happen, perhaps on screen by a pictorial representation of shifting queues of support. The mass interactive election would only be limited by the current limits of computer technology.
There is nothing original about the idea of computer interactive elections. The point being made is that they are required as a complementary research tool to match mathematicly structured counts which may be theoreticly supposed to satisfy a great variety of possible voting patterns.
This brings me to a joke to tell against myself. One of the members of the
STVvoting email group, Brian White responded, to my idea of a
retransferable vote, with another suggestion of letting the voters decide the
best voting system. He called this electoral reform as an experimental
science. As he had sidetracked my email, I returned the favor by
sidetracking his sidetrack. As he is a citizen of British Columbia, I
innocently suggested an experiment, that had just been performed in his
province, namely that a Citizens Assembly be set up and its members be given
the better part of a year to study voting methods...
I asked if this rang any bells.
Brian White made a verbal submission and a vocal presentation to the Citizens Assembly. Having thirty years experience of STV in Ireland, as one of the Protestant minority, he was able to assure assembly members that the system does protect minorities. Thousands of people took part in the provincewide debate and the Citizens Assembly did eventually recommend BCSTV, their own excellently suited version of STV for British Columbia.
Mr White's participation in Irish STV elections also made him believe that there was a need to be able to exclude candidates you don't want elected. This gave me some evidence that the power to exclude candidates, on a quota of least preferences, perhaps does meet a need.
No sooner had I got off that email to Brian than I realised he was right
about the "experimental science" for electoral reform. And I sent off another
message admitting "the joke is on me". That was because, I had just managed
to work out how the binomial theorem could be applied as successive
approximations to an STV count. This was accompanied by the realisation that
a count could be considered as an interpretation of various voting patterns
by various possibly matching mathematical structures.
In short, a count is to theory as a vote is to experiment. Theory and
experiment are the two basic and complementary ingredients of a science.
Therefore, one side, to a future electoral reform or research, is to hold
experimental votes, cutting back on restricting mathematical assumptions
about the nature of the count these involved, to achieve a wider
understanding.
Richard Lung
21 November 2004;
amended 2; 8 december 2005; reorganised January 2006.