9588. (Charles L. Dodgson, M.A.)—A random point being taken on a given line, find the chance of its dividing the line into two parts (1) commensurable, (2) incommensurable.
Comment on Solutions
(1) In reply to the Rev. T. C. Simmons, if, instead of dividing his line by 2, 3, &c., he will divide it by , , &c., and if, where he has written “commensurable,” he will write “incommensurable,” he will find his argument quite as sound as before, and, instead of proving the two chances to be “zero” and “unity,” he will prove them to be “unity” and “zero.” An argument that proves with equal ease either of two contradictories, needs very cautious handling.
(2) In reply to Professor Tanner, I must respectfully decline to explain how a thing can happen which I say cannot happen at all! No “aggregate of points,” as I believe, can ever “make up the whole of a line,” or any portion of it: so I must refer him, for the explanation he desires, to the “opposition,” who are so ready to explain how “an aggregate of absolute zeros may be unity.” While their hand is in, they may as well do the other little job. In the latter part of his letter he asserts, unless I misunderstand him, that, if the chance of a random point coinciding with one assigned point be δ, then its chance of coinciding with one or other of such points is unity. I suppose he would say, taking 10 bags, each containing 1 white counter and 9 black, that, since the chance of drawing a white from one bag is , the chance of drawing a white from one or other of the 10 bags is unity. Does he accept this as a fair instance of the theorem?