# The Broken Stick Problem (Please check my approach) [duplicate]

There is this famous probability problem called the broken stick problem. The problem is: If a stick of length x is broken into three pieces, what is the probability that the three pieces can be used to construct a triangle? In order to construct a triangle, the longest side should be less than $$\dfrac{x}{2}$$.

Here's my approach:

Let $$E$$ = Event that the stick can be broken into three pieces that can create a valid triangle

$$A$$ = Event stick 1 is longer than $$\dfrac{x}{2}$$

$$B$$ = Event stick 2 is longer than $$\dfrac{x}{2}$$

$$C$$ = Event stick 3 is longer than $$\dfrac{x}{2}$$

If any of the sticks is longer than $$\dfrac{x}{2}$$, a triangle can't be constructed. Thus:

$$E^c$$ = $$A \cup B \cup C$$

$$A$$, $$B$$, and $$C$$ are disjoint, since only one of the sticks can be longer than $$\dfrac{x}{2}$$

$$P(A) = \dfrac{1}{2}$$ which is derived by simple continuous uniform probability, which is also equal to $$P(B)$$ and $$P(C)$$

$$P(E) = 1 - P(E^c) = 1 - P(A \cup B \cup C)$$

Because $$A$$, $$B$$, $$C$$ disjoint:

$$P(E) = 1 - (P(A) + P(B) + P(C))$$

$$P(E) = 1 - 3 \space \times \space \dfrac{1}{2} = -\dfrac{1}{2}$$, which violates the non-negativity rule of probability.

Therefore, this approach is not correct. Can anyone help point out where I made a mistake? How would you solve the broken stick problem?

Picture proof: $$P(A)\ne\frac{1}{2}$$.
• Take a unit stick. Let $x,y$ be the places where the stick is broken. Event $A$ is $y\ge x\ge 1/2$; event $B$ is $y\ge x+1/2$; event $C$ is $x\le y\le 1/2$. – Chrystomath Jul 4 at 8:13