USE OF TABLES. 83 



V 200, 000 = 447 '21 



1-8215 106 ' 64 



447-21 



814-6 

 1 



2)813-6 



t = 3 821 



406 '97 = 



406 -8 = 1 



Answer. Between 50,000 407 and 50,000 + 407. 



Now for the inverse method attached to the preceding 

 problem. If I be totally unacquainted with the na- 

 ture of the events A and B, except only that one or 

 other, and not both, must happen every time, it is then 

 clear that, as the matter stands, it is to me 1 to 1 for A 

 against B, with a very great chance that, if I were better 

 informed, I should form a different opinion. At the same 

 time (page 10), I choose 1 to 1 as my rule of action, 

 because, though coming events may not justify my pre- 

 diction, I know of nothing to warrant my assuming 

 that the odds are in favour of A, rather than in favour of 

 B. A trial takes place, and A happens ; it becomes im- 

 mediately most safe to assume that the odds for A against 

 B are 2 to 1, but still that safety is not very decided. 

 But if 1000 trials be made, and if A have happened 

 520 and B 480 times, I can then confidently say, that 

 the odds for A against B are very nearly, if not exactly 

 520 + 1 to 480 + 1, which is nearly 520 to 480. The 

 notion then formed has a strong presumption that it is 

 nearly correct. 



PROBLEM. In a -f- b trials A has happened a times 

 and B b times : from which, if a and b be considerable 

 numbers, it is safe to infer that it is a to & nearly for 

 A against B. What is the presumption that the odds 

 for A against B really lie between ak to b -f k and 

 a-f A; to&-.fc? 



G 2 



