between Stars forming Binary or Multiple Groups. 425 



The case may be illustrated by a reference to one of those 

 games of chance which serve best to fix the ideas as to every 

 kind of hazard, adopting of course Mitchell's opinions on the 

 mode of conceiving the fundamental question. Let there be 

 as many dice as stars, and let n be that number. Let each 

 die have/? sides, numbered 1, 2, 3, &c. I say that the chance 

 of doublets occurring amongst the n dice thrown at once is 

 the same as that of two stars being found at a less distance 

 than the radius of a small circle of the sphere which includes 



an area — th of the entire surface of the sphere. For take any 



star, and describe round it as a centre the small circle which 



includes -th of the spherical surface, a second star falling 



within this circle will constitute it a double star; and the 

 chance against its doing so is evidently p to 1, or the chance 

 of not throwing doublets with two dice havings sides. But 

 there are n stars altogether, and each may duplicate the other 

 under like conditions, so that the probability of the occurrence 

 of a doublet amongst n dice is the same as the chance that 

 two of n stars shall duplicate one another, or shall be included 

 in a circle 1° in diameter ; the occurrence of a triplet repre- 

 sents three stars being within distance of one another, and so 

 forth. 



Llaving myself no aptitude or practice in calculating chances, 

 and knowing the oversights sometimes made in such cases 

 even by persons who have been considered as authorities on 

 the subject, I requested a mathematical friend, whose skill 

 in these matters gives the utmost attainable assurance of his 

 accuracy, to solve this and some other questions to which I 

 shall have occasion to refer. He kindly complied with my 

 request, and has given me the following solution. 



The number of arrangements which the n dice can form 

 'without duplication is the same as the number of permutations 

 of/? things taken n together, or 



p(p-l)(p~2)....{p-n+l); 



also the total number of falls is p n . Consequently the chance 

 of a fall without duplication is 



p{p-l)..*(p-n+l) 



And the chance that two or more of the dice show the same 

 face is 



i P(p-l)»»(p-n+l) 



1 p n 



