BIPARENTAL INHERITANCE IN PARAMECIUM 399 



equals 2/19, ( or :, ). We had found from the previous source 



\ m — 1/ 



a chance of 1/19 for a single pair, so the total chance for one pair 



when 3 are drawn is 1/19 + 2/19, or 3/19 f in general, r + 



\ m — 1 



2 3 \ / 



7 = ). The remainder of the chances ( 16/19, or 



m — 1 m — 1 / \ 



TH — ' 4\ 



I are for pairs when 3 are drawn. Thus, when we draw 



7n — 1/ 



3 from a given number ???, and repeat the drawing many times, 



3 



the average number of pairs obtained will be . 



m — 1 



B}^ a continuation of this line of reasoning, which becomes 



somewhat complicated as the number drawn becomes larger, 



we discover that the average number of pairs that will be obtained 



when n units are drawn is 



1 + 2 + 3 + 4 ■ ■ ■ ■ to (n-l) _ i n («- 1.) 

 171 — 1 m — 1 



Thus, when 9 are drawn from 20, the average number of pairs to 

 be obtained, if the drawing is repeated an indefinitely great num- 

 ber of times, is 



1(9X8) 36 17 



■^— ^ , or — = 1 — 



19 ' 19 19 



This average, like any other average, would be obtained in a 

 concrete case, by multiplying each number of pairs by the num- 

 ber of times it is drawn, and dividing by the number of drawings. 

 Thus, in the case gi\'en above, if there are a great number of 

 drawings of 9 from 20, the various number of pairs will be ob- 

 tained in the following proportions: 



pairs in 384 drawings (= pairs) 



1 pairs in 3,456 drawings (= 3,456 pairs) 



2 pairs in 6,048 drawings (= 12,096 pairs) 



3 pairs in 2,520 drawings (= 7,560 pairs) 



4 pairs in 189 drawings (= 756 pairs) 



12,597 drawings 23,868 pairs 



