Uniform sums and Euler’s number

Suppose I sample values uniformly at random from the unit interval. How many samples should I expect to take before the sum of my sampled values exceeds unity?¹

Let $N$ be the (random) number of samples taken when the sum first exceeds unity. Then $N$ has expected value $E\left[N\right]$ equal to Euler’s number $e\approx 2.718282$. This can be verified approximately via simulation:

simulate <- function(run) {
  tot <- 0
  N <- 0
  while (tot < 1) {
    tot <- tot + runif(1)
    N <- N + 1
  }
  N
}

set.seed(0)
mean(sapply(1:1e5, simulate))

## [1] 2.7183

To see why $E\left[N\right]=e$, let $(X_i{)}_{i=1}^{\infty }$ be an infinite sequence of random variables with uniform distributions over the unit interval. Then the probability that $N$ exceeds any non-negative integer $n$ is $$Pr(N>n)=Pr(X_1+X_2+\cdots +X_n<1).$$ Consider the unit (hyper)cube in $R^n$. Its vertices comprise the origin, the standard basis vectors $e_1,e_2,\dots ,e_n$, and the sums of two or more of these basis vectors. The convex hull of $\{0,e_1,e_2,\dots ,e_n\}$ forms an $n$-simplex with volume $1/n!$. The interior of this simplex is precisely the set $$\{X_1,X_2,\dots ,X_n\in [0,1]:X_1+X_2+\cdots +X_n<1\}.$$ It follows that $Pr(X_1+X_2+\cdots +X_n<1)=1/n!$ and therefore $Pr(N>n)=1/n!$ from above. Now $$Pr(N=n)=Pr(N>n-1)-Pr(N>n)$$ for each $n\ge 1$. Thus, since $Pr(N>0)=1$ (and, by convention, $0!=1$), we have $$\begin{array}{cc}E\left[N\right]& =\sum _{n=1}^{\infty }nPr(N=n)\\ & =\sum _{n=1}^{\infty }n(Pr(N>n-1)-Pr(N>n\left)\right)\\ & =Pr(N>0)+\sum _{n=1}^{\infty }Pr(N>n)\\ & =1+\sum _{n=1}^{\infty }\frac{1}{n!}\\ & =\sum _{n=0}^{\infty }\frac{1}{n!}\\ & =e.\end{array}$$ The final equality comes from evaluating the Maclaurin series expansion of $e^x$ at $x=1$.