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 \(E[N]\) equal to Euler’s number \(e\approx2.718282\). 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 \(E[N]=e\), 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 \(\mathbb{R}^n\). Its vertices comprise the origin, the standard basis vectors \(e_1,e_2,\ldots,e_n\), and the sums of two or more of these basis vectors. The convex hull of \(\{0,e_1,e_2,\ldots,e_n\}\) forms an \(n\)-simplex with volume \(1/n!\). The interior of this simplex is precisely the set $$\{X_1,X_2,\ldots,X_n\in[0,1]:X_1+X_2+\cdots+X_n<1\}.$$ 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 \(n\ge1\). Thus, since \(\Pr(N>0)=1\) (and, by convention, \(0!=1\)), we have $$\begin{align} E[N] &= \sum_{n=1}^\infty n\Pr(N=n) \\ &= \sum_{n=1}^\infty n\left(\Pr(N>n-1)-\Pr(N>n)\right) \\ &= \Pr(N>0)+\sum_{n=1}^\infty\Pr(N>n) \\ &= 1+\sum_{n=1}^\infty\frac{1}{n!} \\ &= \sum_{n=0}^\infty\frac{1}{n!} \\ &= e. \end{align}$$ The final equality comes from evaluating the Maclaurin series expansion of \(e^x\) at \(x=1\).