BIPARENTAL INHERITANCE IN PARAMECIUM 403 



Determining by formula (3) the probability that there should 

 be 38 pairs among the 103 left from the 18G hnes in Aliss Cull's 

 case, above cited, we find this to be but 0.00004556; or the odds 

 against this number are 21,948 to 1. 



The result at this stage howe^'er does not give us what we need 

 to know. If 21,949 tickets were placed in a box and one of these 

 drawn out, the odds would be 21,948 to 1 against any particular 

 number, yet some particular number would be drawn, and there 

 would be no ground for surprise that it should be one particular 

 number rather than another. 'Our present case is not entirely 

 similar to this, since a certain number (18) of pairs is more proba- 

 ble than any other." Yet formula (3) shows that the probability 

 for this most probable number is but 0.16685, so that this 'precise 

 number of pairs would be found in but one case out of every 6, 

 and for any given case the odds against it are 5 to 1. 



^Vliat we require to know is, not only what is the most probable 

 number of .pairs, and the deviation from this most probable num- 

 ber, but (and this is the essential point): How probable is it 

 that there should be so great a deviation from this most probable 

 number as that which we find? 



Thus, in Miss Cull's case alread}^ cited, the most probable 

 number of surviving pairs we know to be 28, while the actual 

 number surviving is 38, so that the observed number deviates 

 by 10 from the most probable number. We require to know how 

 probable it is that there should be so high a deviation as 10. 



The probability of a deviation so great as that observed may 

 be determined directly as follows: Determine the probability 

 of all cases showing a deviation less than the given deviation 

 and compare the sum of these probabilities with the sum of all 

 cases showing a deviation as great as the given deviation. In 

 Miss Cull's case, where the most probable number of surviving 

 pairs is 28 and the deviation is 10, we should have to find the 

 sum of the probabilities for all .numbers deviating less than 10 

 (that is, of all numbers from 19 to 37), and compare this with the 

 sum for all numbers deviating more than 10 (all numbers below 

 19 and above 37). 



