TITLE: Choose a random number between $0$ and $1$ and record its value. Keep doing it until the sum of the numbers exceeds $1$. How many tries do we need?
QUESTION [62 upvotes]: Choose a random number between $0$ and $1$ and record its value. Do this again and add the second number to the first number. Keep doing this until the sum of the numbers exceeds $1$. What's the expected value of the number of random numbers needed to accomplish this?

REPLY [8 votes]: The derivations involving the multidimensional integral for the probability of finding a sum smaller than 1, can be simplified by restoring permutation symmetry in the integration before evaluating it. The probability that the sum of $n$ uniformly distributed random numbers between 0 and 1 does not exceed 1 is given by:
$$P(n) = \int_{0\leq s_1\leq s_2\leq\cdots \leq s_n\leq 1} ds_1 ds_2\cdots ds_n $$
The integration variables here are the partial sums of the first $n$ random variables, which are therefore ordered as indicated. But because these are dummy variables, we can permute these variables in any way we like. This means that integrating over the variables without imposing the ordering will yield $n!$ times the value of the original integral. We thus have:
$$P_n = \frac{1}{n!} \int_0^1\int_0^1\cdots\int_0^1 ds_1 ds_2\cdots ds_n = \frac{1}{n!}$$
The expectation value then follows e.g. using the derivation given in this answer by Dinesh.