```Date: Jul 12, 2004 7:49 PM
Author: Van L. Nguyen
Subject: Re: probability of a triangle (SPOILER)

I believe that the answer to part (b) should still be 1/4.1.  Heuristically or intuitively, although the 2 sequences of eventsas "described" in part (a) and part (b) may "appear" to be different,in the end they really amount both to the same event of breaking aunit length into 3 pieces, whether by making 2 "simultaneous" cuts(part a) or making 1 cut followed by a second one (part b).2.  Mathematically, suppose the first cut in (b) results in 2 pieces,and the piece picked to be further broken has length X, with the otherpiece of length c = (1-X).  Then supposed X is further divided in 2pieces of length b = y and a = (X-y).      Clearly,                  0 < y < X    In the cartesian plane (X, y), the domain of all possible points(X, y) is the right triangle intersection of the 3 areas y>0, y<X andX<1.    The subdomain of all points (X, y) resulting in a, b and c forminga triangle must satisfy the 3 conditions a < 1/2, b < 1/2 and c < 1/2,or:     y > X - 1/2,  y < 1/2  and X > 1/2      The area of this triangular subdomain is 1/4 the area of the entiredomain, thus the probability remains 1/4, conforming with whatintuition indicated.On Jun  7 12:48:12 1996, Daniel A. Asimov wrote:>In article <9604298333.AA833373730@ccmail.odedodea.edu>Pat_Ballew@ccmail.odedodea.edu writes:>>     b)  If the length is broken at a random point, and then one ofthe two >>     pieces is randomly selected and broken at a random point on itslength >>     what is the probability that the three pieces will form atriangle>>     >>     Pat Ballew>>     Misawa, Japan>>     Pat_Ballew@ccmail.odedodea.edu>----------------------------------------------------------------------->>Here's my answer to question b):>>After the first random break, we can (without loss of generality)skip the step >of randomly selecting one of the pieces, since the distribution ofthe left>and right pieces will be identical anyway.  This simplifies theproblem a bit.>>Say we always choose the leftmost piece; call its length x.  >>At this stage we have 2 pieces: [0,x] and [x,1].>>Now we want to break [0,x] "at a random point", which amounts tochoosing >(independent of x) a number y at random in the interval [0,1], wherey >represents the fraction of x where the break will occur.>>So, we end up with 3 pieces:  [0,xy], [xy, x], and [x,1].  Theirlengths are>clearly xy, x-xy = x(1-y), and 1-x.>>The distribution on x and y is uniform on the unit square[0,1]x[0,1].>>And the region where the 3 pieces form a triangle corresponds (asMark Burkey, >and via e-mail, Pat Ballew have pointed out) to precisely thecondition that >all 3 sides are of length < 1/2.>>The subset of the square, then, where xy < 1/2, x(1-y) < 1/2, and 1-x< 1/2 >(i.e. x > 1/2) is the thorn-shaped region bounded on the left by theline >x = 1/2 and on the right by the 2 hyperbolas xy = 1/2 and x(1-y) =1/2.>>Calculus then gives the area of this region -- which must be theprobability we>are seeking -- as Prob(triangle) = ln(2) - 1/2 =.1931471805599453094172321....>>--Dan Asimov
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