62 JAMES BERNOULLI. 



Suppose A to have m counters, and B to have n counters ; let their 

 chances of winning in a single game be as a to 6 ; the loser in each 

 game is to give a counter to his adversary : required the chance of 

 each player for winning all the counters of his adversary. In the 

 case taken by Huygens m and n were equal. 



It will be convenient to give the modern form of solution of 

 the problem. 



Let u^ denote J.'s chance of winning all his adversary's count- 

 ers when he has himself w counters. In the next game A must 

 either win or lose a counter; his chances for these two contin- 

 gencies are r and t- respectively: and then his chances 



of winning all his adversary's counters are u^_^_^ and u^_^ respectively. 



Hence 



_ a h 



This equation is thus obtained in the manner exemplified by 

 Huygens in his fourteenth proposition; see Art. 34. 



The equation in Finite Differences may be solved in the or- 

 dinary way; thus we shall obtain 



where C^ and C^ are arbitrary constants. To determine these 

 constants we observe " that ^'s chance is zero when he has no 

 counters, and that it is unity when he has all the counters. Thus 

 u^ is equal to when x is 0, and is equal to 1 when x is m + n. 

 Hence we have 



0=0.+ a„ 1 = 0.+ c,g) 



«x 



«!+n 



therefore ^i — ~ ^2~ 





m+n 1 m+n ' 



Hence u^ = 



^m+n _ ^m+n-:c J^, 



X rn-^n J vi+n 



To determine ^'s chance at the beginning of the game we 

 must put x = m; thus we obtain 



7/ = 



