lOS MONTMORT. 



gets the seven. There are 50 cards left and Peter takes one of 

 these instead of his seven ; 39 cards out of the 50 are favour- 

 able to Paul, namely 3 sevens, 4< Kings, 4 nines, 4i eights, 4 sixes, 

 ... 4 aces. 



Proceeding in this way we find for Paul's chance 



4 47 + 43 + 39 + 35 + 31 ,, , . 780 



that IS 



51" 50 ' 51.50' 



In this case Paul's chance can be estimated without speculating 

 upon the conduct of Peter, because there can be no doubt as to 

 what that conduct will be. 



II. Paul has a seven; required his chance if he retains the 

 seven. 



The chance in this case depends upon the conduct of Peter. 

 Now it appears to be tacitly allowed by the disputants that if 

 Peter has a nine or a higher card he will retain it, and if he has a 

 seve7i or a lower card he will take another instead. The dispute 

 turns on what he will do if he has an eight. 



(1) Suppose that Peter's rule is to retain an eight 



Paul's chance arises from the hypotheses that Peter has a seven, 



six, five, four, three, two, or ace, for which he proceeds to take 



another card. 



We shall find now, by the same method as before, that Paul's 



chance is 



3^ 24 ^ 27 j^ 27 _£ 27 4 27 ^ 27 ^ 27 

 51 * 50 "^ 51 ' 50 "^ 51 * 50 "^ 51 • 50 "^ 51 • 50 "^ 51 ' 50 "^ 51 • 50' 



that is 



51.50* 



(2) Suppose that Peter's rule is to change an eight 



4 24 

 We have then to add -pr • ^t: to the preceding result ; and thus 



51 oU 



we obtain for Paul's chance - 



51.50' 



780 

 Thus we find that in Case I. Paul's chance is , and that 



51 . 50 



in Case II. it is either -,- ^ or .^ —.7: . If it be an even chance 



51 .50 51 . oU 



