Expected stopping time Let $X_1,X_2,\ldots,$ be iid uniform in $[0,1]$ and define $S_n:=\sum_{i=1}^n X_i$. Let $\tau$ be the smallest $n$ for which $S_n>1$. It is known that $E[\tau]=e\approx 2.7183$, and this can be easily shown using the
Irwin-Hall https://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution 
distribution. Indeed,
$$ P(\tau\ge n) = P(S_{n-1}< 1)=1/(n-1)!,$$
whence
$$ E[\tau]=\sum_{n=1}^\infty P(\tau\ge n) = \sum_{n=0}^\infty 1/n! = e.$$
Here, all of the heavy-hitting is done by the
Irwin-Hall distribution, and I'm wondering: is there a clever "soft" martingale argument that gives the answer with less heavy-hitting?

For $x \in [0, 1)$, let $\tau(x)$ denote the smallest $n$ for which $S_n > x$.  Let $g(x) := E[\tau(x)]$.  We're interested in $g(1)$.
Using its definition, $g$ satisfies an integral recurrence
$$ g(x) \ = \  \int_{y = 0}^{x} [1 + g(x - y)] \ dy   \ + \ \int_{y = x}^{1} 1 \ dy   $$
which if I'm not mistaken, is satisfied by the function $e^{x}$.  It also satisfies the "limit boundary condition" $\lim_{x \rightarrow 0} g(x) = 1$, which I believe can be established using crude arguments.  I am not going to attempt to argue that these two conditions force the unique solution $g(x) = e^{x}$, but it looks true and doable.
(Note, this might be isomorphic to some of the work on Irwin-Hall; I haven't checked.)
