



' 





746 



CHANCES. 



CKan.cc*. equal chance for the eveat'sliappening (x + 1 ) times. 



> "V^"' Let us now put r=x-f-l, then ar=r 1 ; and since 



m=3x + 2, .therefore w=3(r l)-f-2=3r 1, and 









That is, if m cards be dealt, it is more 



*M 



than an equal chance that there should be ^~ 



hearts. 



Cor. Because wz=3r 1, therefore, if in a lotte- 

 ry containing two blanks to a prize, it be required to 

 know how many tickets should be bought in order 

 to have an equal chance for r prizes, the answer will 

 be 3r 1 tickets ; that is, in order to have an equal 

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

 must be bought 2, 5, 8, 11, 14, &c. tickets. 



PROBLEM XXX. 



40. In a pack of 52 cards, consisting of 13 of 

 each suit, if m cards be dealt to me, how many may 

 I, on an equality of chance, expect to be trumps ? 



SOLUTION. The same symbols being retained as 

 in the two last problems, then because there are three 

 suits of blanks to one suit of prizes, (or trumps,) 



6=3 and a=l. Therefore ^= 3 } and 



And in this case, 



(1 \" 1 

 +F) 



l+-j- . -f -j ^ ^.|-f-,&c. (torterms)=|(|) w . 



Here, as in the last problem, we must determine r 

 ,by repeated trials. Accordingly, by proceeding as 

 in that problem, we shall fiud that when 



greater 



less 



greater 



less 



greater 



less 



greater 



less 



than 



than 



than 



than 



, a 

 t! * 



= 21 then one term 

 = 3 \ of the series is 

 = 6 1 then 2 terms 

 = 7 J are 

 = 101 then 3 terms 

 = 11 J are 

 = 147 then 4 terms 

 t = 15J are t ._ 



Therefore, in general, when 



f 4x4-27 thenx-J-1 f greater 7 

 ^4x+3j terms arc- {lesser J 

 Therefore, when Wi=4.r4 r 3, there will be some- 

 thing more than an equal chance of the event's hap- 

 pening w + 1 times. Now, since w=:4x-J-3, there- 

 on 3 

 fore x = . And since r = x -f- J , therefore 



m 3 ?4-l _, 



xr 1 ; hence r 1= , and r= - . That 

 4 4 



is, if m cards are dealt, it will be more than an equal 

 : trumps. Therefore, 



T? 



chance that there will be 



in the game of whist, where 13 cards are dealt, there 

 is more than an equal chance for any particular person 



, . 134-1 14 

 having ~ = trumps. 



And since this is more 



than an equal chance, if any player have but three 

 trumps, or less, he may justly conclude that his part- 

 ner has four trumps, or more. 



Cor. Since r- 1 = - , therefore z = 4r I ; 

 4 



Hence it follows, that in a lottery where there are 

 three blanks to a prize, if it be required to kuowjhow 

 many tickets should be bought iji order to have r 



prizes, the answer will be 4r 1 tickets ; thus, in or- Chances. 

 der to have 1, 2, 3, 4., 5, &c. prizes, there must be 

 bought 3, 7, 11, 15, 19, &c. tickets. 



PROBLEM XXXI. 



41. In a lottery which has four blanks to a prize, 

 if a person purchase m tickets, how many prizes may 

 he expect, on an equality of chance ? 



SOLUTION. The same symbols being retained as 

 before, we have, in this case, 6=4 and a=l: and 



(1 V" 

 1 H J = \(i) m . Therefore 

 pi 



in this case we must find r the number of terms of 

 the series 



the sum of which is ^(|.)w. 



By proceeding as in Prob. XXIX. it will be found 

 by trials, that when 

 = 3 i 

 = 4 

 : = 8 ' 

 = 9 

 : = 13 : 

 = 14 

 : = 18' 



m 



greater 

 less 



less 



greater 



less 



than 



> than 



than 



- than 



then one term 



of the series is 



then 2 terms 



are 

 then 3 terms 



are 



then 4 terms '"greater" 

 ^ are "i^less 



'Therefore, in general, when 



then #4-1 ("greater? i i/<>-, 

 terms aVe jfess j than *(*)* 

 Therefore, when m=5.r4 ( -4, there will be some- 

 thing more than an equal chance of the required ef- 

 fect's happening x-\- I times. 



ffl A. 



And since w=5o:4-4, therefore x=z ; also 



5 



since /=.-{- 1, therefore xzzrl, hence r 1 = , 



and r = 



That is, ^ prizes may, on an 

 5 



f'St 



equality of chance, be expected in m tickets. 



Cor. Since r 1 = : , therefore w=5r 1. 

 5 



Hence it appears, that in such a lottery, in order to 

 have an equal chance for r prizes, there must be pur- 

 chased 5r 1 tickets ; that is, 1, 2, 3, 4, 5, &c. pri- 

 zes require 4, 9, 14, 19, 24, &c. tickets. 



PROBLEM XXXII. 



42. If a person playing with a single die determines 

 to cast it m times, how many times out of that num- 

 ber may he, on an equality of chance, undertake to 

 cast an ace. Or, (which is the same thing,) if he cast 

 m dice at once, how many of them may he, on an 

 equality of chance, expect to be aces? 



SOLUTION. Retaining the same symbols, in this 



(1 \ m 

 1+) =()"'. 



We must now find r the number of terms of the se- 

 ries 



r 1 T 1.2 

 which are equivalent to 



-j \ / C)\ 



1 . if. 3 " 



"t 



