180 Prof. Edgeworth's Problems in Probabilities. 



versa, if each of the marks in each subject had been deter- 

 mined by a jury ; the number of candidates, say m, and the 

 number of subjects, S, being large. The number of candi- 

 dates and prizes being large, we may assume that the mark 

 of the nth candidate in a descending order of merit, that is 

 the Honour line, will be constant for any particular set of 

 examiners ; for much the same reason that, if two large 

 batches of similar objects, e. g. statures of the same nation, 

 be taken at random, the quartiles, octiles, deciles, &c. remain 

 constant. Thus the problem is reduced to Axyz which — as 

 one case of Axz — has been solved. 



3xyz. This problem is related to the preceding Bxyz, as 

 Axz to Kxz. 



~Bxyz. This case differs from ~Bxyz in that the number of 

 the competitors (and prizes) is small. First, let there be only 

 two competitors, and one prize. The problem is : What is 

 the probability that the verdict of any particular set of S ex- 

 aminers would be reversed, if the two papers in each of the S 

 subjects had been marked by a jury. Let the (compound) 

 marks of the two candidates differ by I. The probability of 

 reversal is identical with the probability that the difference 

 between the candidates in a narticular direction, or with its 

 sign, according to the actual set of examiners, should differ by 

 as much as I from the difference in the same direction under 

 the jury-system. If, as before, C is the Mean Deviation in 

 each subject, then, upon principles to which allusion has been 

 made, the required probability is 



T 00 1_ 



_ *2_ 



e 2SC2 dx. 



Next, let there be three candidates ; and, to fix the ideas, 

 let there be two prizes, and let the question be, What is the 

 probability that the second prize-man would fail to obtain a 

 prize, if the work were marked by other examiners ? In order 

 that the original second should become third, it is evident 

 that he must come out below the original third. Thus, I being 

 the distance between the original second and third, the solu- 

 tion would be the same as for the simple case, if the original 

 first were not liable to move relative to the original second. 

 But, in virtue of this liability, a certain proportion of cases in 

 which the original second comes out below the original third 

 are not failures for the original second. Thus /, the distance 

 from the first of the unsuccessful, or Honour-line as it may 

 be called, being the same, the probability of failure decreases 

 with the increased number of candidates. The limiting case 



