The Puzzle

You take a stick of unit length and break it at two points, chosen uniformly at random and independently along the stick. This produces three segments.

Question: What is the probability that these three segments can form a triangle?

Triangle Inequality Condition

Let the lengths of the three resulting segments be:

  • \( a \), \( b \), and \( c \), with \( a + b + c = 1 \)

To form a triangle, the sum of the lengths of any two sides must be greater than the third:

\[ a + b > c,\quad b + c > a,\quad c + a > b \]

Using \( a + b + c = 1 \), these inequalities become:

\[ a < \frac{1}{2},\quad b < \frac{1}{2},\quad c < \frac{1}{2} \]

So each piece must be less than 1/2 in length.

Modeling the Cuts

Let the two cut points \( X, Y \) be chosen uniformly and independently from \( [0,1] \). We can assume \( 0 < X < Y < 1 \) by ordering them.

Then the lengths of the segments are:

  • \( L_1 = X \)
  • \( L_2 = Y - X \)
  • \( L_3 = 1 - Y \)

We seek the probability that all of these are less than \( \frac{1}{2} \).

Geometric Probability

This corresponds to the area of the region in the unit square \( [0,1]^2 \) where:

\[ 0 < X < Y < 1,\quad \text{and } X < \frac{1}{2},\quad Y - X < \frac{1}{2},\quad 1 - Y < \frac{1}{2} \]

This region forms a triangle within the unit square with area:

\[ \boxed{\frac{1}{4}} \]

Final Answer

\[ \boxed{\frac{1}{4}} \]

So, there is a 25% chance the three segments can form a triangle.

Reference