Consider a line segment of length L. This segment is cut in three segments at arbitrary points. What is the probability that these three segments could be rearranged as sides of a triangle?Example


2 Answers 2


It is possible to do so with any division where none of the three sections have a length that's over half the total.

This means:

a) first and second cut cannot be different by over half the maximum length

enter image description here

On the graph I colored the 'bad' areas: where the first is over half the maximum bigger then the second, and where the second is over half the maximum smaller then the first.

Simple math would show that's 1/4 that doesn't qualify

b) the first two cuts cuts cannot be both in the first half, nor can they be both in the second half.

This eliminates options colored in blue

enter image description here

Combining those, I think I end up with 1/4 odds for decent solutions. Putting this up now, I'll rethink it after.

  • $\begingroup$ Thanks to Hubble for editing in the link to the related post. I was uncertain what answer I had to end up with, so I'm glad they match. Perhaps the graphical explanation clarifies it a bit further? $\endgroup$ Aug 18, 2014 at 15:31
  • $\begingroup$ The diagram is very helpful yes! $\endgroup$
    – Cruncher
    Aug 19, 2014 at 15:31

The algebraic way of looking at this:

After the first cut the probability (p) that the longest stick will be shorter than length $x$ is $p=2x-1$. The expected length is $(1+0.5)/2=.75$

The second cut has probability $1-x$ of being in the small half (loss) and $2(x-.5)$ of cutting the second half but leaving one stick at least $.5$ long. Adding these together means there is a probability $x$ that the second cut will leave a stick $.5$ or longer. As the expected value for $x$ is $.75$, there is only a $.25$ chance that the two sticks can form a triangle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.