A Surprising Place $e$ Appears In

Occurance

It takes an average of $e$ randomly generated numbers in $[0,1]$ for the sum to be more than $1$

Meta

I was first told of this problem from Emilia on discord, and I was determined to find a solution with almost no calculus.

Proof

Let $s_n$ be the probability distribution of the sum of $n$ numbers in $[0,1]$. Let $N$ be the probability distribution of the number of numbers you need to sum to get more than $1$. Let $P(x)$ be the probability of the predicate $x$ holding true.

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Lemma: $P(s_n < 1) = 1/n!$

Proof: Subdivide the interval $[0,1]$ into $m$ parts, and say we can only pick the numbers $X = {x_1,x_2\cdots x_n}$ in multiples of $1/m$. There are $m^n$ ways to choose $X$. However, there's only $\text{Binomial}(m,n)$ ways to pick $X$ such that $\text{sum}(X) < 1$, which can be shown via the stars and bars method. Now the probability that $\text{sum}(X)<1$ is $P = \text{Binomial}(m,n)/m^n$, and as $m\rightarrow \infty$, $P\rightarrow 1/n!$. Hence proved.

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Now,

\[\begin{aligned} P(s_n < 1) &= P(N > n) = \frac{1}{n!} \\ P(N = n) &= P(N > n-1) - P(N > n) = \frac{1}{(n-1)!} - \frac{1}{n!} \end{aligned}\]

You can see how $1 = \sum_{n=2}^{\infty} P(N = n)$ because it telescopes into 1.

Therefore, the expected value of $N$ can be found as such:

\[\begin{aligned} E(n) &= \sum_{n=2}^{\infty} P(N = n) \times n \\ &= \sum_{n=2}^{\infty} \left[ \frac{1}{(n-1)!} - \frac{1}{n!} \right] \times n \\ &= \sum_{n=2}^{\infty} \frac{1}{(n-2)!} \\ &= \sum_{n=0}^{\infty} \frac{1}{n!} \\ &= e &\blacksquare \end{aligned}\]