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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?

Agus Bendoro
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  • No, I was asking why is my approach to the problem wrong? Can you point out the mistake? – Agus Bendoro Jul 04 '20 at 02:39
  • I apologize, I meant to link to this question (the above link was to a more general question). This answers your question "How would you solve the broken stick problem?" – Xander Henderson Jul 04 '20 at 02:43
  • Or this question: https://math.stackexchange.com/questions/26424/a-samurai-cuts-a-piece-of-bamboo – Xander Henderson Jul 04 '20 at 02:44
  • Or this question: https://math.stackexchange.com/questions/2862590/ – Xander Henderson Jul 04 '20 at 02:45
  • I'll note that all of the questions to which I have linked were found in the sidebar to the right. – Xander Henderson Jul 04 '20 at 02:46
  • Yeah, I think they solved the "how" question, but I still can't figure out where I made a mistake in my approach. – Agus Bendoro Jul 04 '20 at 02:50
  • @AgusBendoro I am so bad at probability, but I think your problem is your assertion that $P(A) = P(B) = P(C) = 1/2$ is problematic. If you're explicitly assuming that stick 1 is formed first by breaking the stick once (with uniform probability as to the position), then the remaining stick is broken to form 2 and 3, then $P(A) = 1/2$, but the probability of a given pair of length for sticks 2 and 3 depend heavily on the length of stick 1. For instance, if $A$ is true, then $B$ and $C$ cannot be true! – user804886 Jul 04 '20 at 02:56
  • If there was only one cut, making two sticks, then it would be true that $P(A)=P(B)=\frac12$. How do you get $P(A)=\frac12$ when there are two cuts? – David K Jul 04 '20 at 03:23

1 Answers1

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

enter image description here

Chrystomath
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