204 ON THE SOLUTION OF EQUATIONS 



bability is^?), this becomes y^:,^} if lie lose it (of which the 

 probability is q), it becomes y^i.,^, and therefore 



This is the equation of the problem. It is clear that 



y* = 0, 2/ a , = ........................ (2), 



as the party ceases as soon as M or N has a counters. Again, 

 y^ = o unless x = 1, and # 1-0 = p ............ (3) ; 



for if JV have more than one counter he cannot lose them all 

 at the next game; and if he have only one, his chance of his 

 being left without any is p. 



Let us assume y M = <v x ........................... (4), 



a being arbitrary. Then 



Of this a solution is 



X 



Vx ~ \q) i n ^ a W ' ' ^ " 



where C and co are arbitrary, and /JL such that 



a ..................... (7): 



this form of solution is therefore real if a 2 is less than 

 In order that (4) may satisfy the conditions (2), we must have 



L 



x~\2 



i/ = C sin a) = 0, v a =Cij sin (pa + <a) = ...... (8). 



It is impossible to satisfy these two conditions without making 

 C 0, which would give a nugatory result, unless sin ^a = 



or fju , r being an integer. Let us therefore assume this 

 value for p ; and then, by (7), 



a. = 2*/(pq) cos ..................... (9). 



Gt 



In order to satisfy (8), we have now only to make o> = 0, and 

 then, substituting the values of v x and a in (4), we get 



in x cos .......... (10). 



