USE OF TABLES. 79 



a + 6 : extract the square root of the quotient., and 

 multiply it by 3 1 : subtract 65, and divide by 1 30 : the 

 nearest whole number is the answer required. Thus in 

 the preceding instance, where the square root is 81*65, 

 multiply this by 31, which gives 2531 '15 ; and 65 less 

 is. 2466-15, which divided by 130 gives 18'97- Hence 

 it is very little less than an even chance that the aces 

 in 6000 throws shall be between 1000 +19 and 

 1000 19, or 1019 and 981. 



Having found the limits of departure for which there 

 is an even chance, we can now use Table II. as follows. 

 The values of t in Table II. are the proportions of 

 various departures (each increased by -5) to that depar- 

 ture which has an even chance, as just ascertained, and 

 also increased by *5 : the values of K are the proba- 

 bilities of the departures answering to those of t. Having 

 then ascertained 18 '97 to be the departure for which 

 there is an even chance, suppose I ask what is that limit 

 of departure within which it is two to one that the 

 aces shall be contained. Two to one gives J for the 

 chance, or -66667 : I look into Table II., and find that 

 when K is -66667, t is 1-43433, found as follows:- 



K= -66667 t= 1-43 

 Next below -66521 



Tab. Biff. 337) 146000(433 t= 1-43433 

 1348 



1120 

 1011 



1090 

 1011 



79 



This is the proportion which the departure in ques- 

 tion, increased by *5, bears to 18*97 increased by -5 or 

 19-47. Multiply 1-43433 fcy 19'47, giving 27*93; 

 from which subtract -5, giving 27*43 for the limit of 

 departure, the same as in page 78. 



