370 CONDORCET. 



and that v, e, i are the probabiHties that a voter will vote for A, B, C 

 respectively. Since there are 6^+1 voters the three candidates 

 cannot be bracketed, but any two of them may be bracketed. We 

 may consider three problems. 



I. Find the probability that neither B nor C stands singly at 

 the head. 



II. Find the probability that neither B nor C is hefore A, 



III. Find the probability that A stands singly at the head. 



These three probabilities are in descending order of magnitude. 

 In III. we have all the cases in which A decisively beats his two 

 opponents. In II. we have, in addition to the cases in III., those 

 in which A is bracketed with one opponent and beats the other. 

 In I. we have, in addition to the cases in II., those in which A is 

 beaten by both his opponents, who are themselves bracketed, so 

 that neither of the two beats the other. 



Suppose for example that q = l. We may expand {v + e + iy 

 and pick out the terms which will constitute the solution of each 

 of our problems. 



For III. we shall have 



v' + 7v' {e + i) + 21v' (e + if + 85y* (e + if + ^ov' ^ii\ 



For II. we shall have in addition to these 



For I. we shaJl have in addition to the terms in II. 



7v 1^eH\ 



These three problems Condorcet briefly considers. He denotes 

 the probabilities respectively by IF ^ TF/, and W"^. It will scarcely 

 be believed that he immediately proceeds to a fourth problem in 

 which he denotes the probability by TF/^, which is nothing hut the 

 second problem over again. Such however is the fact. His enun- 

 ciations appear to be so obscure as even to have misled himself 

 But it will be seen on examination that his second and fourth 

 problems are identical, and the final expressions which he gives 

 for the probabilities agree, after allowing for some misprints. 



