Probability distribution of fragment lengths I would like to compute the probability distribution for the length of the fragments which I would obtain by fragmenting a linear rod of length $L$ in the following way:


*

*I choose at random (uniformly) $n$ breakpoints 

*I cut the rod at those breakpoints, creating $(n+1)$ fragments.


Now, while it is easy to see that the probability that a stretch of length $x$ does not contain any breakpoint goes like a negative exponential, I don't know how to throw in the information about the length of the rod.
 A: Let the rod have length $L$ and fix a segment of length $x$.  The chance that any single breakpoint misses the segment equals the proportion of the rod not occupied by the segment, $1−x/L$.  Because the breakpoints are independent, the chance that all of them miss it is the product of $n$ such chances, $(1 - x/L)^n$. 
From comments following the question, it appears that $x$ is intended to be small compared to the rod's length: $x/L \ll 1$.  Let $\xi = L/x$ (assumed to be large) and rewrite $n = \xi(n/\xi)$, leading (purely via substitutions) to
$$\Pr(\text{all miss}) =  (1 - x/L)^n = (1 - 1/\xi)^{\xi(n/\xi)} = \left((1-1/\xi)^\xi\right)^{n/\xi}\text{.}$$
Asymptotically $\xi \to \infty$.  If we assume that $n$ varies in a way that makes $n/\xi$ converge to a constant, this probability approaches a computable limit.  Let this constant be some value $\lambda$ times $x$.  It is the limiting value of $n/\xi/x = n/L$: notice how the length of the rod is involved here and effectively is incorporated in $\lambda$.  Because $\exp(-1) = 1/e$ is the limiting value of $(1-1/\xi)^\xi$ and raising to (positive) powers is a continuous function, it follows readily that the limit is
$$\Pr(\text{all miss}) \to e^{-\lambda x}.$$
One application is when $n$ is a constant, entailing $\lambda = n/L$, and $x \ll L$.  We obtain $$e^{-nx/L}$$ as a good approximation for the probability that all breaks miss the segment.  This analysis shows that the approximation fails as $x$ grows large: the approximation is only as good as the approximation $1/e \sim (1-1/\xi)^\xi$.  Finally, if you set $x = L$, the approximation is clearly wrong because it gives $e^{-n}$ instead of the correct answer, $0$.
A: Let $\{X_i\}$ be the locations of the cuts.
I'd approach this problem by finding the order statistics $\{Y_i\}$ so that $Y_1$ would be the location of the leftmost cut. Then I'd calculate the probability distributions of the differences between the variables $Y_i-Y_{i-1}$. Don't forget to also calculate $Y_1-0$ and $L-Y_n$.
Can anyone think of a better way?
