# aliquote

## < a quantity that can be divided into another a whole number of time />

Chris Wellons one asked the following question before providing some code to estimate the result, which I call brute force estimate but I do it a lot too:

How many times should a random number from [0, 1] be drawn to have it sum over 1?

Here is my brute force approach using Scheme and recursion only:

(import (chicken random))

(set-pseudo-random-seed! 101)

(let loop ((n 1)
(sum 0))
(cond ((> sum 1) sum)
(else (loop (add1 n) (+ sum (pseudo-random-real))))))

;; => 1.77838275494088


A more involved solution using (typed) Racket and for/fold is discussed on Stack Overflow.

How about the math? There are several solutions available on math.SO, more or less complicated. See also this application. In essence, the probability that the sum of $n$ variables is $> 1$ while the other $n - 1$ are $< 1$ is:

\begin{aligned} P(S > 1) & = \int_1^n P_{X_1 + \dots + X_n}(u)du - \int_1^n P_{X_1 + \dots + X_{n-1}}(u)du\cr & = \left( 1 - \frac{1}{n!} \right) - \left( 1 - \frac{1}{(n-1)!} \right)\cr & = \frac{1}{n(n-2)!} \end{aligned}

Then it follows that $\sum_{n=1}^{\infty}nP(S > 1)$ is equal to $\sum_{n=1}^{\infty}\frac{1}{(n-2)!}$, or after simplification $e$.

♪ Bauhaus • She’s in Parties