N-length string loop cut N times: piece length as N $\to \infty$? Suppose I take a loop of string of length N units, and mark N points on it at independant and uniformly random (not necessarily integral) points along its length.
(So for example suppose I have a 3 inch string:  I might make marks at 0.425 inches, 0.924 inches and 2.4155 inches clockwise from some arbitrary origin point on the loop.)
Then lets say I cut the string loop at each of the N points.  I will be left with N pieces of string.  The average length of these pieces of string will be 1 unit (as their total length is N units, and there are N of them: $N/N = 1$).
As N approaches infinity, what is the distribution of the length of these pieces of string?
I think the answer is a probability density function with center of mass at 1, and total integral 1.
What is this function?  Does it have a name?
 A: Let us start by considering a fixed finite $N$. By symmetry, we only need to consider the segment clockwise from the first cut. That segment's length is determined by the minimum of the clockwise distances from the first cut to the $k$'th cut for $k>1$. The $k$'th cut will have clockwise distance $<x$ from the first cut with probability $\frac{x}{N}$, so the distribution of the minimum has a CDF of:
$$C_N(x) = 1 - \left(1-\frac{x}{N}\right)^{N-1}$$
Differentiating gives us a PDF of:
$$P_N(x) = \frac{N-1}{N}\left(1-\frac{x}{N}\right)^{N-2}$$
The distribution as $N\rightarrow \infty$ is given by:
$$
\begin{align}
P(x) &= \lim_{N\rightarrow\infty}P_N(x) \\
&= \lim_{N\rightarrow\infty}\frac{N-1}{N}\left(1-\frac{x}{N}\right)^{N-2} \\
&= \left(\lim_{N\rightarrow\infty}\frac{N-1}{N}\left(1-\frac{x}{N}\right)^{-2}\right)\left(\lim_{N\rightarrow\infty}\left(1-\frac{x}{N}\right)^{N}\right) \\
&= \left(1\right)\left(e^{-x}\right) \\
&= e^{-x}
\end{align}
$$
Here, we made use of the fact that the limit of a product is the product of the limits (provided the limits exist), and the fact that $\lim_{n\rightarrow\infty}(1+x/n)^n = e^x$.
It is easy to check that this function $P(x)=e^{-x}$ has center of mass at 1 and total integral 1. Its name is the exponential function.
