If a 1 meter rope is cut at two uniformly randomly chosen points, what is the average length of the smallest piece?

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.

Each triangle corresponds to which length is shortest

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

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