
Re: probability of a triangle (SPOILER)
Posted:
Jun 7, 1996 8:48 AM


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, xxy = x(1y), and 1x.
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 email, 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(1y) < 1/2, and 1x < 1/2 (i.e. x > 1/2) is the thornshaped region bounded on the left by the line x = 1/2 and on the right by the 2 hyperbolas xy = 1/2 and x(1y) = 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

