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: $P(A)\ne\frac{1}{2}$.

enter image description here

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  • $\begingroup$ Do you mind explaining in words as well? $\endgroup$ – Agus Bendoro Jul 4 at 8:08
  • $\begingroup$ 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 $x\le y\le 1/2$. $\endgroup$ – Chrystomath Jul 4 at 8:13

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