Probability of Making Triangle with Three Stick Pieces

Date: 02/24/2007 at 03:26:40
From: Kiran
Subject: Find probability that 3 parts of a stick form a triangle.

Take a stick of unit length and break it into two pieces, choosing 
the break point at random.  Now break the longer of the two pieces at
a random point.  What is the probability that the three pieces can be
used to form a triangle?

I'm not sure what to do after taking the initial condition.  Let a be
the point of breakage and its distance from one end is x.  Then 
x > 1/2.  The second probability gives y > 1/4.  How to proceed from
there and get the probability?

Date: 02/24/2007 at 09:37:07
From: Doctor Greenie
Subject: Re: Find probability that 3 parts of a stick form a triangle.

Hello, Kiran -

I made a rather extensive search of the Internet and the Dr. Math 
archives and found no satisfactory discussion of this problem.  There 
are several sites I found which discuss the similar problem where the 
stick is broken randomly in two places--unlike your problem, where the 
first break is made and then the second break is made in the longer 
piece.  Even in that other case, I found disagreement about the 
correct answer.  And all of the sites I found used methods of solving 
the problem which seem more difficult than mine.

So here is my solution to your problem....

To make a triangle out of three pieces of a stick of unit length, the 
longest piece must have a length which is less than half a unit.

Let's make our first break and let x be the length of the shorter 
piece.  x can vary between 0 and 0.5.  Let's pick values in this range 
and consider the probability that we can make a triangle for that 
value of x.

Suppose x = 0.2; then the long stick has length 0.8.  The break on 
this stick must be made so that the longest piece has length less than 
0.5.  This means the break must be made between 0.3 and 0.5.  So the 
probability of being able to make a triangle with the three pieces if 
the first break is at 0.2 is 0.2/0.8.

Now let's pick one more value for where we make the first break.  
Suppose x = 0.4; then the long stick has length 0.6.  Again, the break 
on this stick must be made so that the longest piece has length less 
than 0.5.  This means the break must be made between 0.1 and 0.5.  So 
the probability of being able to make a triangle with the three pieces 
if the first break is at 0.4 is 0.4/0.6.

Using our two specific examples, we can see that, in the general case, 
the probability of being able to make a triangle out of the three 
pieces, as a function of the length x of the short piece we get on our 
first break, is

  x/(1-x)

Our variable x can range from 0 to 0.5, so the overall probability 
that we can make a triangle with the three pieces of the stick, given 
your statement of the problem, is

  integral from 0 to 0.5 of (x/(1-x))

This turns out to be

  -.5 - ln(.5) = -.5 + .69314718.... = .19314718...

I hope this helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/