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 events

as "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 a

unit 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 other

piece of length c = (1-X). Then supposed X is further divided in 2

pieces 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 and

X<1.

The subdomain of all points (X, y) resulting in a, b and c forming

a 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 entire

domain, thus the probability remains 1/4, conforming with what

intuition 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 of

the two

>> pieces is randomly selected and broken at a random point on its

length

>> what is the probability that the three pieces will form a

triangle

>>

>> 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 of

the left

>and right pieces will be identical anyway. This simplifies the

problem 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 to

choosing

>(independent of x) a number y at random in the interval [0,1], where

y

>represents the fraction of x where the break will occur.

>

>So, we end up with 3 pieces: [0,xy], [xy, x], and [x,1]. Their

lengths 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 (as

Mark Burkey,

>and via e-mail, Pat Ballew have pointed out) to precisely the

condition 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 the

line

>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 the

probability we

>are seeking -- as Prob(triangle) = ln(2) - 1/2 =

.1931471805599453094172321....

>

>--Dan Asimov