12 ON PROBABILITY. 
and the probability of A's winning the set in not more than m + n games 
will appear to be 
m. (m+ 1) m.(m~1).(m+2)..(m~+n—1) 
mil +m —_— —m Iq 
{1+ gt Tk 1.2.83. .; Nn... 7} 
If A, in order to win the set, must win m games before B wins n games, 
A must win m games out of m —-n — 1, and the probability of this event is 
m.(m + 1) m.(m+1). (m+ 2)..(m+n—2) 
_m 1 i eligi =Z n=11{¢ i 
Lm 5 ¢ ee. LT } ~ 
and the probability of B’s winning n games out of m + n — 1 is — 
n.(n+1) n.(n+1).(n+2)..(n4m—2) 
nl] ee SS J pam = 
"1+ np + 127 2,8 eu? ) 
The same result may be obtained from the following considerations. If 
the play be supposed to continue without end, the probability that A will 
gain a single game 
=p+patpe +. o=pio orl, [: 
and the probability that A will win any finite number of games as m will be | 
4 ne 
fed by bo 
represented by Gi -) 
m.(m-+1) 
mp iltmgt— sro... 0), oy 
This probability is made up of the partial probabilities that A will win m redu 
games in m exactly, in m ~- 1 exactly, &c. The probability that A will win I 
m games in m -- @ exactly must have p™. ¢” for its argument, and, there- take 
fore, since the above-written series contains all these partial probabilities, (a+ 
and no others, and consists solely of terms whose arguments are of the form § 
p™ . ¢%, each of these partial probabilities will be rightly represented by the the | 
corresponding term in that series, for it can be exhibited in no other shape, two 
having the arguments of all its terms of this necessary form. Therefore the can | 
probability that A will win m out of m +n games consists of the first terms to in 
of that series up to that inclusively, whose argument is p™ . ¢", the same as the f 
before obtained. S08 
If A wants m games of being up, and B n games, and they agree to leave bh 
off playing, the stakes should be divided between them in the proportions of 
their probabilities. Co 
This problem is celebrated in the history of the theory of probabilities, 
and was the first of any difficulty which was solved. It was proposed to 
Pascal by the Chev. de Méré, with some others relating to games with dice. 
Ex. 13. A bag contains 2 + 1 balls which are marked by the numbers 
0.1.2.3...n, a ball is drawn and afterwards replaced in the bag. Required 
the probability that after ¢ drawings the sum of the numbers drawn is equal . 
to s. 
A little consideration will show that the probability required will be the : 
: : : tat Fat... Fa) the 
coefficient of &* in the expansion of Er YE because that 
! n-+41 
coefficient will be made up of all the different ways in which the different 
indexes of x can be combined in this developement, so as to equal the 
required index s. 
~~