Cases of More than Three Candidates.

It remains now to state and examine the method proposed for the case in which there are more than three candidates.

A series of scrutinies are held on Borda's system of voting, and all candidates who on any scrutiny have not more than the average number of votes polled on that scrutiny are excluded. As many scrutinies are held as may be necessary to exclude all but one of the candidates, and the candidate who remains uneliminated is elected.

The method proposed cannot lead to the rejection of any candidate who is in the opinion of a majority of the electors better than each of the other candidates, nor can it lead to the election of a candidate who is in the opinion of a majority worse than each of the other candidates. These results are an extension of those already proved for the case of three candidates, and they may be proved as follows:—As before, let 2N be the number of electors, and let the candidates be denoted by A, B, C, D, &c. Let the compound symbol ab denote the number of electors who consider A better than B, and let corresponding meanings be given to ac, ad, ba, &c., so that ba will denote the number of electors who prefer B to A, and we shall, therefore, have ab + ba page 28 = 2N. Now suppose that at the commencement of any scrutiny the unexclucled candidates are A, B, C, . . . . P, then the score of A on that scrutiny will be

For suppose that there are n unexcluded candidates, and consider a voting paper on which A now occupies the rth place. For this A gets n — r votes. Now on this paper A stands before n — r other candidates. Thus the n —r votes which A receives may be considered each as due to the fact that A stands before one of the following n — r candidates. Thus we see that on any one voting paper A receives one vote for every candidate placed after him. Summing up for all the voting papers, we see that A receives one vote for each candidate placed after him on each paper. Now ab denotes the number of times B is placed after A on all the papers, and similarly for ac, ad, &c. Thus it is clear that A's score is

This result was stated by Borda,^* but proved only for the case of three candidates.

The whole number of votes polled is

or Nn (n—1). Thus the average polled by all the candidates is N(n—1). Now let us suppose that there is a majority for A as against each of the other candidates, then each of the n—1 numbers ab, ac, ad, . . . . ap is greater than N; thus the sum of these numbers, which is equal to A's score, is necessarily greater than (n—1) N, that is, greater than the average score. Thus A will be above the average on every scrutiny, so that he must win on the proposed method.

Next, let us suppose that there is a majority for each of the other candidates against A. Then each of the numbers ab, ac, . . . ap is less than N, and therefore their sum, which is equal to A's score, is less than (n—1) N, that is, less than the average score. Thus A is below the average, and will, therefore, be excluded at the first scrutiny.

The results which have just been proved are particular cases of a more general theorem, which may be enunciated as follows:—

If the candidates can be divided into two groups, such that each candidate in the first group is, in the opinion of a page 29 majority of the electors, better than each of the candidates in the second group, then the proposed method cannot lead to the election of a candidate of the second group.

The results which have just been proved are obtained from the above by supposing, first, that the first group contains one candidate, and the second group all the rest; and second, that the first group contains all but one of the candidates, and the second group the remaining candidate.

Let the first group consist of the l candidates, A, B, C, &c., and let the second group consist of the m candidates, P, Q, R, etc., and let l + m = n, so that n is the whole number of candidates. Because each of the candidates A, B, C, &c., is better than each of the candidates P, Q, R, &c., each of the numbers ap, aq, ar, &c. . . . bp, bq, &c. . . . &c., is greater than N. Now the scores of A, B, C, D, &c., at the first scrutiny are respectively

If we add together all these numbers, we shall get the sum of the scores of A, B, C, D, &c. Now the numbers in the first l columns can be arranged in pairs, such as ab, ba, and ab + ba = 2N, and then are ½ l (l — 1), of these pairs; thus, the sum of the first l columns is Nl(l — 1). Again, the numbers in the last m columns are each greater than N, and there are lm of these numbers; thus, the sum of the last m columns is greater than Nlm. Thus, the sum of all the numbers is greater than Nl(l—1) + Nlm; that is, than Nl(l + m—1); that is, greater than Nl(n—1). Thus the sum of the scores of the l candidates of the first group is greater than Nl(n — 1). Hence the average score of the candidates of the first group is greater than N(n — 1). Hence the candidates of the first group cannot all be rejected at the first scrutiny. By the same reasoning it follows that those of the first group who survive cannot all be rejected at the second scrutiny; and so on. Thus some candidate of the first group must win on the proposed method; or, in other words, no candidate of the second group can be elected.

If the candidates can be divided into two groups in the manner just indicated, it is quite clear that no candidate of the second group ought to win. At the same time, page 30 whichever of the candidates of the first group wins, the result cannot be shown to be erroneous. If the division into groups can be made in more than one way it is clear that the last statement applies only to the smallest group of the first kind. Now in the proposed method the successful candidate must belong to the smallest group of the first kind. Hence then it is clear that the result of the proposed method cannot be shown to be erroneous in any case.

It is clear that no candidate can have more than N (2n — 2) votes on any scrutiny, n being as before the number of unexcluded candidates at the commencement of that scrutiny. For a candidate could only have this number by obtaining the first place on each voting paper.

Again, if any candidate obtain N (2n — 3) votes on any scrutiny, there is an absolute majority in his favour, so that we can at once elect him. For if a candidate were not put first on half the papers, he could not have so many as (n — 1) N + (n — 2) N votes, this being the number he would have if he were put first on one half of the papers and second on the other half. It is clear, too, that if any candidate has less than N votes there is an absolute majority against him; for if a candidate has less than N votes, he must be last on at least half of the papers. These results are not of much use except in the case of three candidates; for if there be more than three candidates, it is only in cases of remarkable unanimity that a candidate can have so many as N (2n — 3), or so few as N votes. If, however, there be three candidates only, the above results may be stated as follows:—The average is 2N; the largest number of votes any one candidate can have is 4N; if any candidate has 3N votes, or more, there is an absolute majority for him, and we can elect him at once, no matter whether the second candidate is above the average or not; if any candidate has less than N votes, there is an absolute majority against him, so that the result of the proposed method is demonstrably correct.

In the case of any number of candidates it will sometimes save a great deal of trouble if we first examine if there be an absolute majority for or against any candidate. This is easily done, and the results arrived at in the inquiry will be of use in carrying out the proposed method, if such be found necessary. For let A₁, A₂ . . . A., denote the numbers of papers on which A occupies the first, the second ... the last or nth place, and let similar meanings page 31 be given to B₁, B₂, &c., C₁, &c. If A₁ be greater than N, there is an absolute majority for A, and we may at once elect him. If A., be greater than N, there is an absolute majority against A, and we may at once exclude him. If neither of these results hold good for any candidate, we must use the proposed method in its general form. Now As score on that method is

Thus to find As score we must find A₂, A₃ . . . A_n_₁. Now to find these it is not necessary to count all the votes for A. For we have

and A₁, A₃ having been already found, we see that it is sufficient to calculate any n — 3 of the n — 2 quantities, A₂, A₃ . . . A_{n_1}, and the remaining one can then be found from the above equation.

It would, however, in practice be better to calculate each of the n quantities, A₁, A₂ . . . A., and then to use the above equation as a test of the accuracy of the counting of the votes. Similar remarks apply to the numbers B₁, B₂. .. B., c₁, c₂... c., &c.

We have also n equations of the former

where r may have any one of the values 1, 2, 3 . . . n. This gives us n independent tests of the accuracy of the enumeration of the votes. In fact, if we arrange the n² quantities, A₁, A₂ . . . A., B₁ &c., in the form of a square array

the sum of every row and of every column ought to be 2N, so that we have altogether 2n — I independent tests of the accuracy of the enumeration of the votes.

The proposed method is not so laborious as might appear at first sight. The number of scrutinies will not usually be large; for we may reasonably expect to halve the number of candidates at each scrutiny. At each scrutiny we reject all who are not above the average. Now in the long run we may expect to find as many below as above the average on a poll. Thus, if there be eight candidates we should page 32 not, on the average, require more than three scrutinies. There can be no doubt, however, that the method would be tedious if the number of electors were very large, unless the number of candidates was very small indeed. In cases where the number of electors is large Ware's method has great practical advantages; for in that method we only require to count one vote for each paper examined at each scrutiny, and at every scrutiny except the first the number of papers to be examined is but a small fraction of the whole number of papers.