Chinees. 



I 







| 

 l > 









CHANCES. 



745 



or 



Hence it appear*, that the number of terrai which 

 must be added together to make \(/j + a ) , it the 

 answer to the quettion. Thii being prtminrd, let 



a:l,::\:p- then y= ; therefore, by ubrtitutioB, 

 the above expressions, will become, 



Ice. 



&c. 



1+*.-+ 



*cc. 



1) 1 .m(in j 

 1.2 /T* + 1.2 . 



Now in the present question, where a=zb, we have 



. but in the de- 



/j=l, therefore l+_ = 



velopement of this quantity, the terms equally distant 

 from cither extremity are equal, therefore (1 + 1)"' 

 will consist of half the terms in (1+1)* ; and as the 

 whole number of terms is m-f 1, therefore the answer 



11 L m 4- 1 

 will be 7~- . 



Zt 



Cor. If r represent the number of times that the 

 proposed event is to happen, then, when there is 

 an equality of chance for its happening or failing, 



r= b 7 the problem. Therefore m, the number 



of trials in which it will be an equal chance, whether 

 the event shall happen or not, will be2r 1. 



And therefore in a lottery, in which the number 

 )f prizes is equal to the number of blanks, if it be re- 

 quired to know how many tickets should be bought, 

 in order to have r or more prizes, the answer will be 

 2r 1 tickets; that is, in order to have an equal 

 chance for 1, 2, 3, 4,5, &c. or more prizes, there 

 should be bought 1, 3, 5, 7, 9, &c. tickets. 



PROBLEM XXIX. 



39. In a pack of 39 cards, consisting of 13 hearts, 

 13 spades, and 13 clubs, if m cards be dealt to any 

 how many may he, on an equality of chance, ex 

 p to bt- hearts ? 



SOLUTION. If a, b, and p, represent the same as 



Jt problem, and r be the number required ; then 



because there are two chances for a black card to be 



d.- and but one for a red card, we have 6=2 and 



fl= 1 , and consequently /?=r2. Therefore 



(1 \ m 

 1 ~*"p) = * (*}"' and in this case 



i 772 



&c. ( 10 r terms) = ^ (|)". 



The whole difficulty of the problem is now redu- 



ced to the finding of r, the nun.hr of terms of the 



that must be added together to make ^(^) m . 



Now in the last problem we "found r by a direct pro- 



TOL. V. TART II. 



2)1 / _1_^ 



&c. 



cess, but in this we cannot find it otherwise than by 

 trials. The manner of proceeding may be as follows. 

 (1.) When 7=1, then $n) w =0.75; 

 and when 7n=2, $(l) m zzl.l'25. 

 Now the first term of the series 



Sec. 



2 



is manifestly greater than the former of thee num- 

 bers, but less than the latter ; hence it appears, that, 

 when m=l, then 1 term is greater than | (i) 1 ", 

 when 7=2, then 1 term is less than i(l)" 1 - 



(2.) Again, if ns=4, then ^ (|)'"=2.53, &c. 

 and if m=5, | m m =3.79, &c. 



(Here the values of i(i) w are on 'V f un d to two pla- 

 ces of decimals, these being sufficient for our pur- 

 pose ; and to facilitate the operations, the calcula- 

 tions may be made by logarithms.) Taking now 

 two terms of the series, we have 



1+4x^=3; and 1+5X J=3.5. 

 Hence it appears, that 



when 7?j=4-, then 2 terms are greater than ^ (I)*. 

 when tn=5, 2 terms are less than | (i) m . 



(3.) If wi=:7, then Hl) m = 8.54, &c. 

 andif7=8, |(4)'"=12.31, &c. 



Now, taking three terms of the series, we have 



&c. 



sl^O, &c. 



Hence it appears, that 



when z=7, then 3 terms are greater than 



when 7n=8, 3 terms are less than ( 



(4.) By a like mode of proceeding we find, by ta- 

 king four terms of the series, that 

 when ?n=10, then 4 terms are greater than 

 when 7/i=ll, 4 terms are less than 



and in general, putting x for any number whatever, 

 when m is 



then ,+ 1 terms f greater! fc 



are [less J 



But when the chances for the event's not happening 

 r times in m trials, (which are expressed by the se- 

 ries,) are less than the power; then the chances 

 for its happening must be greater than that number. 

 Therefore, when w=3x-j- 2, there will be more than an 



