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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."