Date: Jun 7, 1996 8:48 AM
Author: Daniel A. Asimov
Subject: Re: probability of a triangle (SPOILER)

In article <> 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


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