Example.—Take an election where 6000 is the necessary minimum, and suppose A has 8650 votes, composed as follows:
A 600
AB 2,700
AC 4,500
AX 50
AY 200
AZ 600
——-
8,650
Using first the 600 A votes, we are left with 5400 to make up out of the remaining heaps.
1. Suppose B and C have received the quota. The 5400 can be taken from their heaps exclusively, for in their two heaps are 7200 votes; the proportion to be taken from each heap is therefore 5400 out of 7200, which is three quarters. Thus we make up A's number thus:—
A votes 600
Three-quarters of 2,700 AB " 2,025
Three-quarters of 4,500 AC " 3,375
——-
6,000
And transfer the remainder (the AB and AC votes transferred being those stamped with the lowest numbers).
2. Suppose B and X have received the quota. Their two heaps amount to 2750 votes. Using these up, there remain 2650 votes to be made up out of the AC, AY, and AZ heaps. These three heaps together contain 5300 votes; and the proportion to be taken from each heap is 2650 out of 5300, or half. Thus A's number is made up as follows:—
A votes 600
AB " 2,700
AX " 50
Half of 4,500 AC " 2,250
Half of 200 AY " 100
Half of 600 AZ " 300
——-
6,000
And the remaining votes of each of the three last classes—being those stamped with the lowest numbers—will be transferred.
It will be observed that the element of chance is not wholly excluded, since the question, which papers out of the AC heap are transferred, is left to depend upon the order of drawing. To exclude chance wholly, these would have to be sorted into heaps according to the third name upon them, and an equal proportion taken from each heap. The figures in the first half of this paper are sufficient to show that such trouble would be wholly superfluous.