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 $$x=1$$.

1. Grant Sanderson mentions this problem in this Numberphile video. ↩︎