If a $1$ meter rope is cut at two uniformly randomly chosen points (to give three pieces), what is the average length of the smallest piece?

I got this question as a mathematical puzzle from a friend. It looks similar to MathOverflow question If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4.

However, in this case, I have to find the expected length of the smallest segment. The two points where the rope is cut are selected uniformly at random.

I tried simulating it and I got an average value of $0.1114$. I suspect the answer is $1/9$ but I don’t have any rigorous math to back it up.

How do I solve this problem?

**Answer**

My approach is maybe more naive than the others posted.

Break the unit interval at $x$ and $y$ where $x < y$. Our lengths are then $x$, $y – x$, and $1 – y$. It’s not hard to show that they all have probability $1/3$ of being the shortest. In any case, our joint PDF is given by $f(x,y) = 6$ (since $x$ and $y$ remain uniform random variables on $1/6$th of the square $[0,1] \times [0,1]$). Each triangle in the diagram below corresponds to the domain of the PDF for one of the three cases.

I’ll take care of the case when $x$ is shortest, that is, $x \leq y – x$ and $x \leq 1 – y$. This is the leftmost triangle. Since we’re assuming $x$ is least, we are looking for

$$E[x] = \int_0^{1/3} \int_{2x}^{1 – x} 6x \;dy \;dx = 1/9$$

The cases when $y – x$ and $1 – y$ are shortest are similar.

**Attribution***Source : Link , Question Author : svenkatr , Answer Author : Hans Parshall*