Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site pucc-i Path: utzoo!watmath!clyde!burl!ulysses!mhuxl!houxm!ihnp4!inuxc!pur-ee!CS-Mordred!Pucc-H:Pucc-I:ags From: ags@pucc-i (Seaman) Newsgroups: net.puzzle,net.math Subject: Re: Where to cut the chain? Message-ID: <216@pucc-i> Date: Wed, 22-Feb-84 12:20:17 EST Article-I.D.: pucc-i.216 Posted: Wed Feb 22 12:20:17 1984 Date-Received: Fri, 24-Feb-84 00:25:42 EST References: <383@ihuxa.UUCP> Organization: Purdue University Computing Center Lines: 26 > You get a call to bring two lengths of chain with you to a job. Where to > you cut the chain to maximize your chances of both pieces being long enough? Let the two required lengths be x and y. Without loss of generality, x >= y. Therefore the needed lengths can be represented as the coordinates of a point in the first octant of the plane: the area in the first quadrant below the line y=x. In order to answer the question, a probability distribution is needed. There must be a way to assign the relative probability of different points within the designated region. Since the area of the region is infinite and the probability distribution must have an integral of 1 over the region, it follows that some areas are more probable than others -- there is no such thing as a uniform probability distribution over the designated region. The question as stated cannot be answered. If you knew an upper bound to the lengths, you could assign a uniform probability -- but no upper bound is stated. -- Dave Seaman ..!pur-ee!pucc-i:ags "Against people who give vent to their loquacity by extraneous bombastic circumlocution."