How many draws of a random number [0,1] are needed to sum to 1

To review why these would be useful, let χ[T](s) = ⟨ exp(i ω T) ⟩, then χ[A + B] = χ[A] χ[B] for any A and B which are independent, so the characteristic function for a uniform random variable is going to be χ = (exp(i ω) − 1)/(i ω) and thus the sum of n independent uniform random numbers is just χⁿ. Meanwhile the PDF of a random variable is just the inverse Fourier transform of the characteristic function, so we have

fₙ(t) = ∫ dω/(2π) exp(-i ω t) [χ(ω)]ⁿ

Pr(n samples > 1) = ∫{1 → ∞} dt fₙ(t).

You could then interchange those two integrals, solving the convergence problem of exp(-i ω ∞) by giving ω an infinitesimal negative-imaginary component. This translates to moving the ω contour a little below the real line, and then the normal tricks of complex analysis should give you definite values for all of the integrals as residues about the simple pole at 0.

Let's say we draw n times. What's the probability that we get a number at least 1? Well, we want to exclude the cases where we get a number less than 1, so we want the volume of the unit cube (n-tuples of numbers from 0 to 1) minus the volume of the region below the standard simplex (standard simplex is stuff that sums to 1, we're excluding stuff that sums to less than 1.)

Now the volume of the region below the standard simplex in n dimensions is 1/n!. (Brief proof, if you're not familiar: True in 1 dimension. Assume true in n dimensions. Then for n+1 dimensions, have integral over volume in n dimensions, as "sum of first n variables" parameter varies from 0 to 1. But volume varies as length^n in n dimensions, so this is integral from 0 to 1 of x^n/n!, i.e. 1/(n+1)!.)

So after n draws, the probability that we've exceeded 1 is 1-1/n!. So the probability that it requires exactly n draws is 1/(n-1)! - 1/n! (assuming n>1; for n=0 the probability is 0).

So the expected number of draws, then, is the sum over n>=1 of n/(n-1)! - n/n!, but this latter is (assuming n>=2) equal to n/(n-1)! - 1/(n-1)! = (n-1)/(n-1)! = 1/(n-2)!. (For n=1, we get 0.)

I.e. it's the sum over n>=2 of 1/(n-2)!, i.e. the sum over n>=0 of 1/n!, or in other words it's e.