212 Methods of Election. 



the first scrutiny he cannot fail to win at the second scrutiny. 

 Let therefore the whole number of electors be 2N, and let 

 the number who prefer B to C be N + a, and consequently 

 the number who prefer C to B be N — a ; similarly, let the 

 number who prefer C to A be N + b, and therefore the 

 number who prefer A to C be N — b, and let the number who 

 prefer A to B be N + c, and therefore the number who prefer 

 B to A be N — c. Then it is easy to see that the numbers of 

 votes polled by A, B, C at the first scrutiny will be 



2N — b 4- c, 2N — c + a, 2N — a + b 



respectively. For if the compound symbol A B be used to 

 denote the number of electors who put A first and B second, 

 and similarly for other cases, it is clear that A's score at the 

 first scrutiny will be 



2AB + 2AC + BA + CA. 



Now this expression can be written in the form 



(AB + AC + CA) + (AC + AB + BA), 



and it is clear that the three terms in the first pair of 

 brackets represent precisely the number of electors who 

 prefer A to B, which number has already been denoted 

 by N + c. In the same way the remaining three terms 

 represent the number of electors who prefer A to C, which 

 number has been denoted by N — b. Hence the score of 

 A on the first scrutiny is 2N — b + c. In exactly the 

 same way it may be shown that the scores of B, C are 2N — 

 c + a and 2N — a + b respectively. The sum of these 

 three numbers is 6N, as it ought to be. Thus 2N is the 

 mean or average of these three numbers, and consequently 

 the highest of the three candidates must have more than 

 2N votes, and the lowest must have less than 2N votes. 

 Now, let us suppose that a majority of the electors prefer A 

 to B, and likewise that a majority prefer A to C ; then c 

 must be positive, and b must be negative. Hence the score 

 of A, which has been shown to be 2N — b + c, is neces- 

 sarily greater than 2N, for it exceeds 2N by the sum of the 

 two positive quantities — b and c. Thus A has more than 

 2N votes, that is, more than one-third, or the average of 

 the votes polled. He cannot, therefore, be rejected at the 

 first scrutiny, so that B or C or both must be rejected at the 

 first scrutiny. If either of the two, B and C, be not rejected, 

 A must win at the second scrutiny, for there is a majority 

 for A against B, and also against C. Hence, then, it has 



