Methods of Election. 217 



candidates, and that A is in reality inferior to each of them, 

 but that the voting is as follows, BA = 5, CA = 4, 

 AB = 1, AC = 1 ; so that B's supporters, in their anxiety 

 to defeat C, put A second, and C's supporters, in their anxiety 

 to defeat B, put A second. The result at the first scrutiny 

 is A 13 votes, B 11 votes, C 9 votes. Thus C is rejected 

 and A. wins in the final scrutiny. A wins because the whole 

 of C's supporters put him second. Had one of C's sup- 

 porters voted according to his real views, and put B second, 

 the result would have been different. 



If the preferential mode of voting were not employed, this 

 objection would be of great force ; for then the supporters of 

 each candidate would put his most formidable opponent at 

 the bottom of their list at the first scrutiny, knowing that 

 they would have at the second scrutiny an opportunity of 

 reviewing their vote. 



A Modification of Proposed Method. 



It may be mentioned that there is another, but in general 

 a more tedious, method of getting at a result, which cannot 

 be shown to be erroneous in any case. This method has 

 been adopted by the Trinity College Dialectic Society. It is 

 as follows : — In the method proposed above, instead of 

 rejecting all the candidates who are not above the average, 

 reject the lowest only. It is obvious from what has been 

 said above that this cannot lead to error. But a second 

 scrutiny will always be required, whereas in the proposed 

 method one scrutiny only may be necessary. There is 

 another disadvantage : the result will not in all cases agree 

 with that of the proposed method. For, let us suppose that 

 a, b, c are all positive, and that a is the least of the 

 three, and at the same time that 2c is less than a + b. On 

 the method proposed, as we have already seen, C would be 

 elected, but on the method now under discussion B would be 

 elected. For the scores of A and B at the first scrutiny are 

 2N — b + c, 2N — c + a, respectively, and the first of them is 

 the smallest, because 2c is less than a + b, and therefore c — b 

 is less than a — c. Thus A would be thrown out at the first 

 scrutiny, and a second scrutiny would be held to decide 

 between B and C, and B would win because a is positive. 

 Thus the result is that which would follow from abandoning 

 the proposition " A is better than B," which is affirmed by a 

 majority of 2c, whereas the result of the proposed method is 

 that which would follow from abandoning the proposition 



