24iS EULER. 



Suppose tlie tickets of the first class to have been drawn, and 

 that the prizes have fallen on certain n tickets A, B, G ... 



Let the tickets of the second class be now drawn. Required 

 the chance that the prizes will fall on the same n tickets as 

 before. The chance is 



1.2 71 



(m + 1) (m+ 2) {m + n) ' 



And in like manner the chance that the prizes in all the 

 classes will fall on the same tickets as in the first class, is obtained 

 by raising the fraction just given to the power k — 1. 



Let {(m + 1) (m + 2) {m + n)Y~'=M, 



and {1.2 nY'' = a. 



Then -^ is the chance that all the prizes will fall on the same 



n tickets. In this case there are m persons who obtain no prize, 

 and so the managers of the lottery have to pay m ducats. 



445. Now consider the case in which there are m — 1 persons 

 who obtain no prize at all. Here besides the n tickets A, B, G, ... 

 which gained in the first class, one of the other tickets, of which 

 the number is m, gains in some one or more of the remaining 

 classes. Denote the number of ways in which this can happen by 

 ^m. Now If denotes the whole number of cases which can 

 happen after the first class has been drawn. Moreover /3 is in- 

 dependent of m. This statement involves the essence of Euler's 

 solution. The reason of the statement is, that all the cases 

 which can occur will be produced by distributing in various 

 ways the fresh ticket among A, B, G, ... excluding one of these 

 to make way for it. 



In like manner, in the case in which there are m — 2 persons 

 who obtain no prize at all, there are two tickets out of the m 

 which failed at first that gain prizes once or oftener in the remain- 

 ing classes. The number of ways in which this can occur may 

 be denoted by <ym {in — 1), where 7 is indejjendent of m. 



Proceeding in this way we have from the consideration that 

 the sum of all possible cases is M 



M= a + jSin + ym (in - 1) + 3m [m — 1) {in - 2) -f . . .. 



