28 



ON THE MATHEMATICAL PROBABILITY 



We will first examine the simplest case, in which there arc m identical words, and m 

 identical meanings in the two languages. 



Suppose the words arc similarly arranged in each language. Then, among the possible 

 permutations of the meanings, there is only one in which they will all coincide, and none 

 in which m — 1 will coincide. 



If all but two coincide, those two must change places. There arc, therefore, as many 



such arrangements as we can make selections of 2 out of m, or, m ■ \?"" • 



If all but three coincide, those three must change places in such a way that neither will 

 occupy its original place. This can be done in two ways with each group of 3, and there 

 are, therefore, twice as many such arrangements as the number of selections that we can 



make of 3 out of m, or, 2 x ~~ 



Tabulating: and differencing these results, it will be seen that the number of arrange- 



ments in which there are n displacements out of m, is A" 0! 

 The value of this expression can be readily ascertained, for 



> — »+ ') 

 w! 



A" 0!= 



n! 

 "01 



mA"- ] 0!= 



n \ , 

 " IT 27 



n\ n 1 , n\ . ( Vl+1.1 



Ol 1 1 21 



Hence, by subtraction, 



A" ! = nA"- 1 ! ( — )"' 1 



ml 



There will .\ be m displacements, or coincidences, in A m 0!= "7=.36788 ml 

 arrangements, leaving for the probability of one or more coincidences, .63212 ml out of 

 ml possible; arrangements, — which is a probability of about t. There will, therefore, 

 probably be at least one coincidence. 



There will also be m — 1 displacements, or 1 coincidence, in .36788 ml arrangements. 

 Deducting this amount from .63212 ml there remains only a chance of t %%%%% for more 

 than one coincidence. If the words, and also the meanings, were identical in two 

 languages, there would .-. probably be 1 accidental coincidence, and no more, as stated 

 above. 



In order to investigate the other supposed case, it will be necessary to consider the 

 more general expression, A" ml, the value of which can Ik; conveniently ascertained in two 

 ways. For, 



A n m ! = A n (1 + A)'"' ! 



( A n m /\ n+ 1 j_ m (m — 1) /yn+2 ^ A m + 

 j 01 + 1 ! 



m (m— 1) ^n+2 

 2! 



"} 



01 



0) 



or, A m\ = m\ 

 A 1 tn ! =)7i.77i ! 



A 2 m ! ==m A'+ A + A'. 

 (m+1) A' + A. 



A 

 A' 



A" 



