Keep-value Averaged STV.
Reply to apology for anarchy of voting methods.

Five voting systems elect five different candidates.

The kind of distribution of preferences in the table 1 below was first raised by the eighteenth century French filosofs Condorcet and Jean-Charles de Borda. Such a pattern of preferences among voters shows how different voting methods, to elect a single representative, could each elect a different candidate, A, B, C, D, or E.

First Past The Post elects A.
The Supplementary Vote elects B ( by a run-off between the two candidates with the most first preferences ).
The Alternative Vote elects C ( by successively re-distributing the votes of the candidate, with the least first preferences, to his next preference ).
Borda's method elects D ( by a system of giving one more point for each higher preference ).
Condorcet's method elects E ( as the winner of all one-to-one contests ).
The controversy, between Condorcet and Borda, is discussed on my web page about Choice Voting America? There it is explained why transferable voting for multi-member contests supercedes Borda's method for single member contests.

Table 1: 110 voters' preferences for 5 candidates.
voters' order of choice: 1st 2nd 3rd 4th 5th
36 voters prefer: A D E C B
24 " " B E D C A
20 " " C B E D A
18 " " D C E B A
8 " " E B D C A
4 " " E C D B A

Applying Binomial STV to the example for one seat.

Up till now, STV has not been considered applicable to single seat elections. That is because a transferable vote implies at least two seats, in that when one candidate has enough votes to be elected, a surplus is transferable to help next prefered candidates contest a second or more seats.
If Binomial STV turns out to be a valid procedure, it appears to over-come this limitation, presumably thru its keep-value weighting. I'm more sure that the the averaging of Condorcet keep-values is valid. But I dont think it is an independent method in its own right, when you know the over-all range of preferences between more than pairs of candidates. I think something like Binomial STV with its results for the full range of preferences has to complement weighted Condorcet pairing to make fuller use of the preference information.

No candidate's first preferences come close to the quota for one vacancy, 110/(1+1) = 55 votes. So, we cannot give them the benefit of the doubt, with help from an unpreference count that showed them "not unpopular. " Accordingly, we pass over the first order binomial STV count.

The above example is similar to Riker's example on a previous page: the, at first, most prefered candidate, A, is also an even more, at last, the unprefered candidate. As a consequence, it turns out we can repeat the trick we used to obtain a second order count. In Riker's example, W was in the preference situation A is in here.

In both examples, second order binomial STV's two preference counts, whether qualified by preference or unpreference happen to be the same. Because, in both instances, candidate A qualifies as the candidate who has most first preferences and most last preferences. However, only the unpreference-qualified preference count is allowed, because A has a quota of unpreferences to be excluded but not a quota of preferences to be elected.

No doubt, if both preference counts could be used the result would be more representative, as a geometric mean of the two qualified preference counts. Anyway, we persist as best we can with table 2 of the unpreference-qualified preference count.
We also add for good measure, the one of the second order count's two qualified unpreference counts allowed, for similar reasons ( as in Riker's example ). That is the unpreference-qualified unpreference count and not the preference-qualified unpreference count. Tho, as with the two qualified preference counts, they both make the same tabulation.

Table 2: 2nd order count of unpreference-qualified preference ( up ) and unpreference-qualified unpreference ( uu ).
(0) Candidates. (1) 1st preferences. Re-distribute excluded A's 1st prefs. (2) up keep values. (3) Re-distrib. excluded A's last preferences. B excluded. Re-dist. B's surplus @ 3/58. (4) uu keep values.
A - 55/36 - 55/74
B 24 55/24 58 - 3 = 55 55/58
C 20 55/20 32 + 3x36/58 = 33.862 55/33.862
D 54 55/54 20 + 3x4/58 = 20.207 55/20.207
E 12 55/12 0 + 3x18/58 = .931 55/.931
Total votes: 110 110

Table 2 shows that candidate D almost has the quota, 54 out of 55 votes needed for an over-all majority. Statisticly, D is well worth giving the benefit of the doubt. And help is forthcoming for D is also the second least unpopular candidate. My normal procedure would be to invert D's unpreference keep value to 20.207/55. Multiply this by the 55/54 preference keep value and take the multiple's square root for their geometric mean keep value. But obviously, a keep value well below the elective threshold of one is achieved for D.
Candidate E is the least unpopular but also has way the least first preferences.

Keep-value averaged Condorcet pairing.

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Should anyone remark on the 5 contrary results this example is supposed to generate, let us show otherwise by using a different method of Condorcet pairing than that traditionly used to show E is the Condorcet winner. In fact that result is only the result of inaccurate method. When Condorcet pairing is properly quantified with keep values, the result differs, as table 3 shows.

Table 3: Geometric mean keep values from Condorcet pairings.
1st prefs. 2nd pref. A 2nd pref. B 2nd pref. C 2nd pref. D 2nd pref. E Condorcet geometric mean keep values: 4th root of multiplied four vote scores, per candidate, divided into 55.
A 36 36 36 36 55/36 = 1.528
B 74 32 52 44 55/48.246 = 1.14
C 74 78 24 38 55/47.9 = 1.148
D 74 58 86 54 55/66.817 = .8231
E 74 66 72 56 55/66.615 = .8256

Traditional Condorcet pairing shows candidate E wins every one-to-one contest and is therefore the Condorcet winner. Averaged Condorcet pairing shows that, in this example, candidate E is over-hauled by candidate D, albeit by a tiny margin. But then the example was designed on narrow margins to make possible seemingly contrary results. Actually, D wins all pairings but with E, who only wins by the narrowest possible margin. Traditional ( winner-takes-all ) Condorcet pairing hides such quantitative margins which can and do distort the result.

In fact, Borda's method and this more accurate, weighted Condorcet method agree that D is the winner. The discrepancies of First past the post, the Supplementary Vote, the Alternative Vote and the less quantitative traditional form of Condorcet winner are all due to their imprecise counting methods which discard too much preference information.

Borda's method is more quantitative than the other four traditional methods and comes up with the right result here. But even Borda's method uses only an estimated weighting of the vote that is no better than a guess that can be out.

Moreover, even keep-valued Condorcet pairing is not the whole picture. Altho it gives a systematic comparison between each candidate, it discards the over-all orders of preferences voters choose for the candidates. Therefore, Condorcet counting can only be a supplement of a count of the over-all orders, as by binomial STV.

Table 2 of the second order count confirms that D is the over-all order count winner.

So, we dont actually need to do another table, table 4, of the averaged Binomial with Condorcet count to find a clear winner. If we did, we would have to weight the respective keep values in proportion to the number of candidates there were to share the votes between. In Condorcet pairing, there are only two contestants. The over-all election is of four candidates, twice as many. So you must allow for the extra difficulty of the candidates to get votes by giving the four-ways contest proportionately more weight.

As explained previously, a geometric mean weighting must be in terms of powers or geometric multiples of two to one ( not arithmetic multiples of two to one ). In this example, if X was a binomial keep value, it would be weighted X to the power of two-thirds. If Y was a Condorcet keep value, it would be weighted Y to the power of one-third. The two keep values would be multiplied for a unitary geometric keep value as the final result for each candidate.

The example for two seats.

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We are going to use this testing example not only on a one-seat election but for two seats, and in the next section for three seats. I must admit the results are awkward.

Table 4: First order binomial STV count for 2 seats. Quota = 110/(2+1) 37 ( rounded up to whole number).
(0) Candidates. (1) 1st preferences. No-one elected. (2) Preference keep values. (2) Last preferences. A excluded. (3) Transfer A's surplus @ 37/74 = 1/2. (4) Transfer B's surplus @ 10/47. (5) Unpreference keep values. (6) Effectively elective keep values: inverse of col. (5). (7) Over-all keep values: cols. (2) x (6).
A 36 37/36 = 1.03 74 37 37 37/74 = .5 2 2.06
B 24 37/24 = 1.54 36 47 37 37/47 = .79 1.27 1.96
C 20 37/20 = 1.85 0 16 23.66 37/23.66 = 1.56 .64 1.184
D 18 37/18 = 2.06 0 10 10.43 37/10.43 = 3.55 .28 .58
E 12 37/12 = 3.08 0 0 1.91 37/1.91 = 19.37 .05 .154
Total votes: 110 110 110 110

From column (7), table 4 sends a clear message that D and E are the two elected candidates. But a voting system is only as good as its assumptions. And this result depends on the assumption that it is equally important not to be unpopular as it is to be popular. This may apply in some circumstances but most people will feel that this is not true for political elections. I'm inclined to agree. The arithmetic allows winners from those who dont have to be popular, provided they are not so much unpopular, as appears to be the case with candidate E, here.

Statistics can show whether any of the candidates are close enough to the elective quota to be worth the benefit of the doubt as representatives. And then, if they are not unpopular, by enough, they might be deemed elected. Unfortunately, no candidate meets both these criteria. Candidate A is certainly close enough to be given the benefit of the doubt as elected but is much too unpopular to gain any extra support from that direction.

Moreover, the other candidates closer to election are further from being not unpopular. Of course, the election has been contrived by critics in this unrealistic way. Nevertheless, we must have rules for every possible scenario, if there is always to be an agreed result between opposing parties.

We must not forget, that provided we have the rules in place for such an eventuality, there is nothing wrong in saying that the voters have not given a clear indication of choice, and perhaps a new set of candidates are needed. Some voting theorists are perhaps too apt to assume that every election must result in a definite sub-set of the candidates being elected. I do not believe this is necessarily the case.

Anyway, we have the option to go onto the second order binomial STV count and hope that is more helpful. Using the same procedure outlined for the one-seat case, table 5 is an unpreference-qualified preference count. Again we are not able to avail ourselves of the preference-qualified preference count. No doubt this reduces the value of the second order count as a statistical count of averaged keep values.

For what it is worth, and some will quite rightly question whether it is legitimate, we now have an elected candidate in D. Yet D was the winner of the one-seat count not only by this method of 2nd order binomial count but also by the keep-value weighted Condorcet count.
This leaves candidate A as the runner-up with almost an elective quota. Despite A's widespread unpopularity, no other candidate comes near the election quota.

Table 5: 2nd order count ( unpreference-qualified preference ) for 2 seats.
(0) Candidates. (1) 1st preferences with A's vote re-distributed. D elected. (2) Transfer D's surplus @ 17/54. (3) new totals (4) up keep values.
A - - - 37/36
B 24 24 37/24
C 20 18x17/54 = 5.67 25.67 37/25.67
D 54 37 37/54
E 12 36x17/54 = 11.33 23.33 37/23.33
Total votes: 110 110

Example for three seats.

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To add to my embarrassment, if possible, let's turn to the example for three seats. The quota is now 110/(3+1) = 28 ( rounded up to the nearest whole number ).
The first order count doesnt improve matters much. A is elected with a surplus of 8 votes, which go to D, now with 26 votes. This might be deemed close enough to the quota to qualify D for election with the help of a "not unpopular" vote. But the taker of the third seat would not be settled.

So, we essentially, take this process further ( some would say too far to be equitable ) and go onto the second order count of unpreference-qualified preference.

Table 6: 2nd order count of unpreference-qualified preference ( up ) for 3 seats. Quota = 28.
(0) Candidates. (1) First preferences with A's vote redistributed. (2) D's surplus transfer'd @ 26/54. (3) up keep values.
A - - 28/36 = .78
B 24 24 28/28 = 1
C 20 20 + (18x26/54) = 28 2/3 28/29.64 = .94
D 18+36= 54 54-26 = 28 28/54 = .52
E 12 12 + (36x26/54) = 29 1/3 28/29.33 = .95
Total votes: 110 110

(21 1/3)/22 = .97 of E's surplus passes to C before B, improving C's keep value. Because all the preferences are marked, even the last candidate B gets an elective quota of 28 votes and keep value of 1 by the end of the count.

Table 7 is the unpreference-qualified unpreference ( uu ) count.

Table 7: unpreference-qualified unpreference ( uu ) count.
(0) Candidates. (1) Last preferences. Redistribute A's last prefs. (2) Transfer surpluses: B's @ 30/58; C's @ 4/32 = 1/8. (3) uu keep values.
A - - 28/74
B 36 + 22 = 58 58-30=28 28/58
C 32 32-4=28 28/50.62
D 20 20 + 2.07 + 4 = 26.07 28/26.07
E 0 27.93 28/27.93
Total votes: 110 110

C, who is excluded by A's 32 vote re-distribution, is further embarrassed by B's surplus transfer of 36x30/58 = 18.62, worsening C's unpreference keep value to 28/50.62. The 18.62 then goes to E, who also gets 18x30/58 = 9.31, for a total of 27.93.
D gets the remaining 4x30/58 = 2.07 surplus from B.

C's own surplus of 4 goes to D.

In effect, table 7 of the uu count disqualifies, with a high unpreference vote, candidate C, who was only marginally elected ( in table 6 ). C's preference keep value is .94. On previous workings, this safety margin of .06 below the elective value of 1, was assessed as follows. An lower bound was taken to be the PR, namely 3/4 for a three member constituency. This is a difference of 1/4 from 1. The rule for measuring deviation is to square the difference, that is 1/16 or .0625.

So a candidate with a keep value of less than .9375 would be more than one deviation away from unity as the mean or average. Candidate C is not even outside this limit. So, on grounds of chance, C's keep value could quite possibly have been on the unelective side of unity, with a keep value of more than one.

Two deviations or 2 x 1/16 = 1/8 or .125 might be agreed beforehand as the limit beyond which an elected candidate could not be unelected. That is on the grounds that such a good keep value was unlikely to be a chance misrepresentation of less than elective support. So, any candidate with a preference keep value lower than .875 could not be unelected by a poor "not unpopular" keep value.

This clears the way for elected candidates D, A, E to take the three seats. A's 36 first preferences are well enough clear of the elective quota, to remove any threat from A's high unpreference vote. E's election is marginal enough to be called into question but E's exclusion vote shows E is not unpopular. D is elected by a safe margin and could not have been unelected by his unpreference keep value, anyway.

This result feels just a bit too good to be true.
One possible line of justification for qualified counts of the second order is that they anticipate the transfer of greater surpluses as one increases the number of seats and lowers the quota. That is as has been done with this example for one, two and three seats. Maybe this example was fortuitous. But one can imagine that qualified counts of preference and unpreference are like surfing the biggest wave in and out, taking one further than one could otherwise travel to an inherent result.

One has to have faith in one's ideas to carry them thru and I have churned out examples of Binomial STV as if nothing was wrong with it. ( I apologise here for any arithmetic mistakes or carelessness that may have crept into all those tabulations. ) Having done that I can allow myself a measure of self-doubt, if only to be frank with the reader. Certainly, Binomial STV will need a lot more trials before it can be judged ( more or less ) sound. And that is to say nothing of all the practical details that a new counting method has to acquire before one can be confident it meets all eventualities. In fact, returning officers' rule books of accepted voting systems are continually modified.

Source for the example:
John Allen Paulos: Beyond Numeracy. ( Entry on voting methods. )

Richard Lung.
5 December 2006.

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