368 THE ALTERNATE VOTE. 



men, who put A first and B second, and similarly for other cases ; 

 and giving at the first scrutiny, as you remember (21), two votes 

 for the first choice, and taking all the preferences expressed for 

 A, all the first preferences AB and AC, and all the second pre- 

 ferences BA, and CA, we get for all the choices expressed tor 

 A, 2 AB + 2 AC + BA + CA. Now we may arrange this in the 

 form (AB + AC + CA) + (AC + AB + BA). In the first bracket we 

 have put all tlie electors or committee-men who prefer A to B, 

 and those were n + c, while in the second bracket we have all who 

 prefer A to C, namely n — h. Hence on the first scrutiny we see 

 that A's score is the sum of all his votes ... ... 2n—b + c 



In the same simple way it is shown that B's score is... 2n — c-i-a 

 and that C's score is ... ... ... ■■. 2n — a + b 



A, B, and C's scores when added up of course gives us 6n, as 

 there are 2n electors, who vote for three applicants A, B, C 

 giving a first rate, and a second rate or preference to each. The 

 average of these three numbers is 2n, and as we have seen before, 

 in the case of any three numbers {Section 10) the highest candi- 

 date will have more than zn votes, and the lowest, who will be 

 rejected at the first scrutiny, will have less than 2n votes. When 

 the votes are counted, the one rejected as lowest will not have 

 more than the average number of votes. 



By the above ingenious method of expressing the scores, 

 which we haA'e seen is absolutely fair, Nanson proves that if a 

 majority of the electors prefer A to B, and also prefer A to C, 

 A will on the above system of scoring get more than 2n votes, 

 that is more than the average, and so will not be rejected at the 

 first count. 



The very simple proof is as follows : — Those voters who prefer 

 A to B are n + c, and are by our supposition in a majority over 

 those who prefer B to A, that is than n — c; thus c is positive, 

 and n + c more than n. Those who prefer A to C are n — b and 

 by our supposition are more than those who prefer C to A, that 

 is more than n+b, so that h must be negative and n—h greater 

 than n. Hence, since n—h is greater than n and also n + c is 

 greater than n, adding these 2n—h + c must be greater than 

 2n, that is A's votes are greater than the average, and so he 

 cannot be thrown out at the first count. Q.E.D. 



Hence either B or C must be the applicant rejected at the first 

 count, and A must win at the second, because according to our 

 supposition he is better than B and better than C. 



The great point is that with Nanson's method, the best man 

 cannot be thrown out at the first count. 



23. In actual practice Nanson makes use of the two cases we 

 considered at the beginning of this paper, namely {Sections 8 

 and 9) that where, on counting the first votes an applicant has an 

 absolute majority, and he is elected at once — first case ; and that 

 where, on countins^ all the first and second votes, the candidate 

 with the least number of votes gets less than one-fourth of the 

 combined sum — second case, that is, has the support of less than 

 half the voters he is rejected, excluded, eliminated because he has 

 an absolute minority. 



By adding up the numbers used in these two counts — that of 



