# The Principles of Parliamentary Representation/Chapter II

Chapter II.
§ 1. Number of Members in House.
There seems to be no sufficient reason, a priori, for any change in this particular. It would probably be best to take 660 as the number to be generally aimed at, though holding ourselves free to modify this as circumstances might require.
§ 2. Number of electoral Districts; whether to be equal or unequal; &c.
The two extreme cases are (1) to have as many Districts as Members, each to return one Member, in which case the Districts should of course be equal; (2) to form the whole Kingdom into one District.
​In the first case (a method that has been much advocated) it is only a bare majority in each District who are represented. For it must not be supposed that all who vote for a Member are duly represented by him. If a District contains 20,001 Electors, so that 10,001 are enough to return a Member, all additional votes are absolutely wasted: hence only 10,001 Electors in that District are represented in Parliament; the other 10,000, whether they vote for the successful Candidate, or for a rival, or even if there be no contest at all, are unrepresented. This method, then, leaves nearly half the whole body of Electors unrepresented.
The injustice of this method may be illustrated from two points of view. Suppose a bare majority of the Electors to be of one party, and the rest of the opposite party; e.g. let 6–11ths be 'red' and 5–11ths 'blue.' Then, as a matter of abstract justice, about 6–11ths of the House ought to be 'red', and 5–11ths 'blue.' But practically this would have no chance of occurring: if the 'reds' and 'blues' were evenly distributed through the Kingdom, a 'red' would be returned in every District, and the whole House would ​be of one party! Yet this distribution is, by the Laws of Probability, more likely than any other one distribution, and, the nearer the distribution to the most probable one, the nearer we come to this monstrous injustice.
The other way of looking at it is almost as telling. Suppose the House to have been elected, and that 6-11ths of the Members are 'red,' and 5-11ths 'blue': all we could learn from this, as to the views of the Electors, would be that 6-22ths (about 28 p. c.) are 'red', and 5-22ths (about 23 p. c.) 'blue': as to the other 49 p. c., we should know absolutely nothing—if they were all 'red' (i.e. if 3-4ths of the Electors were 'red'), or all 'blue' (i.e. 7-10ths of the Electors 'blue'), it would make no difference in the House.
Taking this first extreme, then, as yielding the maximum of injustice which can be effected by arrangement of Districts, and observing that, if each District returned 2 Members, only 1-3rd of the Electors (on the assumption that each Elector has only one vote—an arrangement whose justice we shall hereafter prove) would be unrepresented, if 3 ​Members, only 1-4th, and so on, we see that the fewer and larger the Districts, i.e. the greater the number of Members which, on an average, each District returns, the fairer the result: till we come to the other extreme, where the whole Kingdom is formed into one District returning 660 Members, in which case only 1-661th of the whole body of Electors would be left unrepresented. A general Election, with so gigantic a District, would of course be impracticable: and probably Districts, returning 6 Members each, would be about as large as could be conveniently dealt with: but very small Districts should be, as far as possible, avoided.
I find, in the Standard for October 10, 1884, a very good instance of the injustice done by sub-dividing large electoral Districts. "The Birmingham Conservatives are, a Correspondent telegraphs, keenly discussing the Government Redistribution Scheme. The clause which apportions 6 Members to Birmingham gives much dissatisfaction in Conservative circles. It is contended that, if the borough is to be divided into three electoral Districts, each District to have 2 Members; the Liberals could so manipulate the voters as to be ​certain of returning the whole of the 6 Members." Now, assuming that each Elector is to have one vote only, the Liberals could only do this by mustering more than two-thirds of the votes in each District; i.e. they must be 67 p. c, or more, of the whole body of Electors in Birmingham. But, if the three Districts were made one, it would need about one-seventh of the whole (i.e. 14 and 2-7ths p. c.) to return one Member. Hence 67 p. c. could only return 4 of the 6 Members: it would require 71 p. c. to return as many as 5; and they could not return all 6, unless they were 86 p. c. of the whole body.
Taking it as proved, then, that single-Member Districts should be in all cases avoided, and that all such should be grouped together, so as to form Districts returning at least 2 Members each, and, wherever it is possible, 4 or 5 or even more, we need only add, as a general remark, that, the more we equalise the Districts, the more we equalise the chance that each Elector has of being one of those represented in the House. Thus, in a District, returning 2 Members, the chance is 2-3ds; with 3 Members, it is 3-4ths; and so on.
§3. Formula for determining, for each District, how many Members it shall return.
A preliminary question must here be asked, viz. are we to count population, or Electors only? I do not think it matters much which, as they probably vary nearly together, i.e. a District having twice the population of another would probably have twice as many Electors. The Formula can best be determined for the number of Electors: but if, in using it the number of population be substituted, it will make no important difference in the result.
The formula will of course have to be modified for each case, if it be agreed to give political weight to differences in rateable property, or to the distinction between town and country voters: and for this purpose rules would have to be laid down.
Now, taking 'e' to represent, for any one District, the number of Electors, and 'm' the number of Members to be assigned to that District, and assuming that each Elector has only one vote, we require a formula giving m in terms of e. This formula must evidently be such as will secure that every Member ​in the House shall, as far as possible, represent the same number of Electors.
Now, whatever be the quota of recorded votes, which is necessary and sufficient, before the poll is closed, to make it certain that 'A' will be returned, that is the number of Electors whom A will represent in the House. He cannot represent less, for this number is necessary; and he cannot represent more, for it is sufficient, so that all additional votes are superfluous. Let us call this necessary and sufficient quota 'Q'.
Now, in order that Q may be sufficient, it must not be possible for m other Candidates to obtain Q votes each; i.e. (m + 1). Q must be greater than e; i.e. Q must be greater than e/m + 1. Also, in order that Q may be necessary, it must be the whole number next greater than this fraction. Hence, approximately, Q = e/m + 1; i.e. m =e/Q − 1.
This, then, is the formula required. An example will make it clear. Suppose the universal quota to be 6,000: then a District containing 50,000 Electors would have 7 Members assigned to it.
​We have yet to find a formula for determining Q. Let 'e$_1$' be the number of Electors in District No. 1, 'e$_2$' the number in No. 2, and so on; let 'm$_1$' be the number of Members assigned to District No. 1, 'm$_2$' the number assigned to No. 2, and so on; also let 'E' be the total number of Electors in the Kingdom, 'M' the number of Members in the House, and 'D' the number of Districts. Then we have
(m$_1$ + 1) . Q = e$_1$, (m$_2$ + 1) . Q = e$_2$, &c. ∴ (M + D) . Q = E; i.e. Q = E/M + D; ∴ m = e . M + D/E − 1.
§ 4. Tables calculated by the preceding Formulæ.
Let us suppose the 2,000,000 new Electors to be already enfranchised, thus making the total Electorate about 5,000,000. Let us further assume the number of electoral Districts to be 180, so that each will return, on an average, 3 and 2-3ds of a Member.
Let M = No. of Members in House = 660. D = No. of Districts = 180. e = No. of Electors in a District. E = total No. of Electors = 5,000,000. p = population in a District. P = total population = 36,000,000. Q = universal quota, to be aimed at. m = No. of Members assigned to a District.
Then E/M + D = 5,000,000/840 = about 6,000;
∴ m = e/6,000 − 1 (a)
It will be worth while to contrast with this the 'rough and ready' method of assigning Members in proportion to the number of Electors, so that m:e::M:E. This gives us
m = e.M/E = e.660/5,000,000 = e/7,600 (b)
In the following Table, the second column gives the number of Members to be returned by a District, the first the number of Electors by Formula (a), and the third the same by Formula (b).
TABLE I.
| e$_2$ by (a) | m | e$_2$ by (b) |
| 9,000 | 1 | 4,000 |
| 15,000 | 2 | 11,000 |
| 21,000 | 3 | 19,000 |
| 27,000 | 4 | 27,000 |
| 33,000 | 5 | 34,000 |
| 39,000 | 6 | 42,000 |
| 45,000 | 7 | 49,000 |
| 51,000 | 8 | 57,000 |
| 57,000 | 9 | 65,000 |
| 63,000 | 10 | 72,000 |
| 69,000 | | 80,000 |
The numbers, in the first and third columns, have been calculated by giving to m, in the preceding Formulæ, the successive values one-half, 3-halves, 5-halves, &c. Hence we see that, by Formula (a), a District containing between 9,000 and 15,000 Electors must have between one-half and 3-halves of a Member (i.e. must ​have one Member) assigned to it; and so on. If a District contained almost exactly 15,000, it could not fairly be determined, by this Table, whether it ought to return one Member, or two. In such a case, it would be best to change the boundaries of the District, so as to increase or diminish the number of Electors by 2,000 or so.
Comparing the results of the two Formulæ, we see that, for Districts whose population is about 27,000, it matters very little which Formula we use: but, for small Districts, Formula (b) assigns too many Members, and, for large Districts, too few; e.g. 13,000 Electors ought to return only one Member—Formula (b) gives them two; 60,000 ought to return 9—Formula (b) gives them 8.
We will now examine the effect of counting the population of a District, and not the Electors only.
Here, for E/M + D, we must substitute P/M + D;
    i.e. 36,000,000/840, i.e. about 43,000.
Hence Formula (a) becomes
m = e/43,000 − 1 (c)
​Also Formula (b) becomes
m = e.660/36,000,000 = e/54,500 (d)
TABLE II.
| e$_2$ by (c) | m | e$_2$ by (d) |
| 64,000 | 1 | 27,500 |
| 107,000 | 2 | 82,000 |
| 150,000 | 3 | 136,500 |
| 193,000 | 4 | 191,000 |
| 236,000 | 5 | 245,500 |
| 279,000 | 6 | 300,000 |
| 322,000 | 7 | 354,500 |
| 365,000 | 8 | 409,000 |
| 408,000 | 9 | 463,500 |
| 451,000 | 10 | 518,000 |
| 494,000 | | 572,500 |
Comparing this with Table I, we see that, provided only it be true that the number of Electors in a District is always about 5-36ths ​of the population, the substitution of number of population for number of Electors will suffice for all practical purposes; and, seeing that there is evidently a tendency to go by population, and that it is much more easy to take the population of a District than to estimate what will be the number of its Electors when the Franchise-Bill is passed, the first column of Table II. is probably the best to employ.