400 H. S. JENNINGS AND K. S. LASHLEY 



Thus a total of 23,868 pairs will have been obtained in 12,597 



17 

 drawings, giving an average of 1 — (or 1.8947) pairs for one 



drawing. In an actual case of 100 drawings of 9 from 20, the 

 average was 1.91 pairs. 



This gives us our first formula. If we let k be the average 

 number of pairs, our formula for determining it is 



k = '^^"-l^ (1) 



TO —1 



As a rule this average number of pairs gives also the most 

 probable number of pairs that will be drawn — this being the 

 integral number nearest the average. In the above case, for 

 example, the most probable number of pairs when 9 are drawn 

 from 20 is 2. In Miss Cull's case, already cited, where 83 died 

 out of 186, the average number of pairs included would be 



^ (83 X 82) 



. or 18.39, so that the most probable number is 18 

 185 



(in place of 28, the number actually found). (Five actual draw- 

 ings of 83 tickets from 186 gave respectively 16, 18, 18, 18, 20 pairs, 

 the most frequent being 18, as our formula demands.) For most 

 practical purposes formula (1) is quite adequate for determining 

 the most probable number of pairs. 



But it happens in rare cases that the integer nearest the aver- 

 age is not the most probable number of pairs. It is apparently 

 always within 1 of the probable number, and this is usuall}^ a 

 sufficiently close approximation, since the probability will be 

 nearly the same for the two numbers that differ only by unity. 

 To take an example, when we draw 5 from 20, the average num- 



ber of pairs to be obtained from repeated drawings is 



H5 X 4) 



19 



= — . Now, the nearest integer to — is 1, yet complete analysis 



shows that is slightly more probable than 1 (the relative proba- 

 bilities of 0, 1 and 2 pairs are in this case as 168, 140 and 15). 



