TREMBLE Y. 431 



803. As a subsidiary part of his solution Trembley gives 

 a tedious numerical investigation which might be easily spared. 

 He wishes to shew that supposing James to have a higher card 

 than both Peter and Paul, it is an even chance whether Peter 

 or Paul is excluded. He might have proceeded thus, which will 

 be easily intelligible to a person who reads the description of the 

 game in Montmort, pages 278, 279 : 



Let n denote the number of James's card. 



I. Suppose n — r and n — s the other two cards ; where r and 

 s are positive integers and different. Then either Paul or Peter 

 may have the lower of the two n — r and n — s\ that is, there are 

 as many cases favourable to one as the other. 



II. Peter's card may also be n\ then Paul's must be 1, or 

 2, or 3, ... or ?i — 1. Here are n — 1 cases favourable to Peter. 



III. Peter and Paul may both have a card with the same 

 mark n — r\ this will give n — 1 cases favourable to Paul. 



Thus II. and III. balance. 



