Problem Description

A stick is broken randomly into 3 pieces.  What is the probability that the pieces can be assembled into a triangle? 

Background & Techniques

I ran across this problem in one of the early chapters of "Mathematical Recreations and Essays", Ball and Coxeter, Dover Publications, presented as a paradox.  They describe it this way: " It is possible to put the stick together into the shape of a triangle provided the length of the longest piece is less than the sum of the other two (Euclid - Book 1, Proposition 20), that is,  provided the length of the longest piece is less than 1/2 of the total length.  But the probability that a fragment of the stick shall be less than half of the stick  equals 1/2.  Thus the probability that a triangle can be constructed of  out the the three pieces into which the stick is broken would appear to be 1/2."

 Or, try my rationale:   Common sense says that the probability is 1/2 since when we make the first random cut, one piece is longer than 1/2 the total length and  the other piece less than or equal to 1/2 the total length.  When we randomly make the second cut, there is a 50-50 chance that we'll cut the piece that is more than half the length of the stick. In that case, a triangle can be formed no matter where we cut it . If we happen to cut the shorter  piece, then no triangle can be formed no matter where it is.  So the probability of forming a triangle is 1/2.

But common sense doesn't always guarantee a correct decision! 

This program runs batches of 100,000 trials and accumulates the number that would successfully create a triangle.   As you might suspect by now, the average number of successes is not 50%.   More a case of false reasoning than a real paradox, but an interesting exercise in any case.

I added an option to run single trials and display the resulting triangle if one is possible. 

Run a few trials and try to explain the result.  A separate "Explanation"  tab gives the correct reasoning. 

 I'll call this a Beginner's level program even though it exceeds our 50 lines of code limit - it has about 70 user written lines.   But there are two distinct parts  - The first part of the  code  generates random cuts and tests whether they would form a triangle about 30 lines. This is incorporated in the MakeTriangle function.   The Make 100,000 Trials button calls MakeTriangle 100,000 times and counts how many created valid triangles.

The graphics portion (40 lines of code) calls MakeTriangle to make the cuts, but then must actually determine the coordinates of the vertices in  order to draw the triangles.  A Triangle from 3 Lines  page over in the Math section derives the equations for calculating  the coordinates. 

August 28, 2011:  A viewer recently browsed through the source code for this program and asked about another strategy for cutting the stick which I had coded but decided to comment out.  Just for fun, here's an update which discusses three other ways to "randomly" choose where to make the two cuts.  They all make the first cut randomly at any point on the the stick.  The differences are how to make the 2nd cut.

For the best of these strategies, Strategy 2, we can analytically verify the experimental results. In what follows, we assume that the original length of the stick is 1 unit.  We'll call the initial cut  P1, and  we can assume that the P1 is less than or equal to 0.5 (since "without loss of generality" as the math guys say) if P1 is greater than 0.5, we could simply turn the stick around or walk around to the other side of the table to continue the analysis.  We will reference the location of both cut points (P1 and P2) to the left end of the stick, i.e. point Px is Px units from the left end.  

I claim (or perhaps discovered) that the probability of forming a triangle with P2, the second cut, made randomly on the longer right hand stick piece,  is P1/(1-P1).   In fact, a triangle is formed only if P2 lies between 0.5 and 0.5+P1. The probability of a random point lying with any band of length L on a line of length N is L/N. In this case, that is P1/(1-P1).

Let's prove it by contradiction. First, assume that P2 is between P1 and 0.5 and that we can still form a triangle. . This implies that the rightmost piece is of length 1-P2 is <0.5, otherwise we could not form a triangle. Since P2 is < 0.5  (by assumption), the length 1-P2 > 0.5, a contradiction to the assumption that it is <= 0.5 in order to form a triangle.  On the other hand, assume that P2 is greater than 0.5+P1, then the left hand piece produced by this cut is longer than the distance from P1 to 0.5+P1, i.e. is greater than 0.5, again a contradiction.  I guess we also need to prove that for 0.5 <P2<.5+P1 that the segments produced by the P2 cut are both less the 0.5 units long. The left hand segment length is P2-P1 and  P2<0.5+P1 so P2-P1<0.5 +P1 -P1 = 0.5.  The right hand side length is 1-P2 and  P2>0.5  so 1-P2 must be < 0.5.    

Running/Exploring the Program 

Suggestions for Further Explorations

August 28, 2011: Done! There is another variation of the problem that asks for the probability of forming a triangle if we make the cuts sequentially, i.e. make the first cut, then randomly select one of those pieces to make a second random cut.   Surprisingly, at least to me, the probabilities are not the same as the original problem.