# The Sum Over 1 Problem

October 30, 2014

I recently came across this fun probability problem which I've dubbed the 'Sum Over 1' problem.

It goes like this:

On average, how many random draws from the interval [0, 1] are necessary to ensure that their sum exceeds 1?

To answer this question we'll use some probability theory (probability rules, distribution functions, expected value) and a dash of calculus (integration).

## Let's Begin!

Let $U_i$ follow a Uniform distribution over the interval $[0, 1]$, where $i = 1, 2, \ldots$ denotes the $i^{\text{th}}$ draw.

Let $N$ denote the number of draws such that $\sum_{i=1}^{N} U_i > 1$ and $\sum_{i=1}^{N-1} U_i \leq 1$. Thus, $N$ is a discrete random variable taking values $n=2,3,\ldots$ ($n=1$ is impossible because no single draw from $[0, 1]$ can exceed 1).

We seek the long-run average, or expected value, of $N$ denoted $E(N)$. A good first step for a numerical problem like this is to utilize computer simulations to point us in the right direction.

## Simulations with R

Let's generate, say, 20 random draws from $[0,1]$, $(u_1, \ldots, u_{20})$.

draws <- runif(20)  # vector of u_1, ..., u_20

Consider the partial sums of these draws $(u_1, u_1 + u_2, \sum_{i=1}^3 u_i, \ldots, \sum_{i=1}^{20} u_i)$.

If we index this partial sum vector from 1 up to 20, then $n$ here is simply the index of the first component of this vector whose value exceeds 1.

The find_n function accomplishes this for any vector of random draws u:

find_n <- function(u) {
which(cumsum(u) > 1)  # which partial sum first exceeds 1?
}

n <- find_n(draws)
draws[1:n]
##  0.2858056 0.1689087 0.6259122

In this instance, it took 3 random draws from $[0, 1]$ to sum over 1. We can sample from the probability distribution of $N$ by repeating this process many more times. The sample_from_N function takes m samples from the distribution of $N$:

sample_from_N <- function(m) {
# m-by-20 matrix of random [0,1] draws
draws <- matrix(runif(m * 20), ncol = 20)
# for each row of 20 draws, find n
n.many <- apply(draws, 1, find_n)
# estimate P(N = n), n = 2, 3, ..., 10
# (truncated b/c probs are miniscule for n > 10)
estimated.probs <- table(factor(n.many, levels = 2:10)) / m
return(estimated.probs)
}

m <- 1000
approx.N <- sample_from_N(m)  # approximate prob. dist'n of N
approx.N
##
##     2     3     4     5     6     7     8     9    10
## 0.514 0.330 0.121 0.026 0.005 0.004 0.000 0.000 0.000

So, we estimate $P(N = 2)$ by $\hat{P}(N = 2) = 0.514$, $P(N = 3)$ by $\hat{P}(N = 3) = 0.33$, and so forth.

These are just point estimates for $P(N = n)$, but we can estimate their standard errors as well by running sample_from_N many more times. The approximate probability distribution of $N$ is depicted below.

The dotted line here marks the estimated expected value of $N$, $\hat{E}(N) = \sum_{n=2}^{\infty} n\hat{P}(N=n) \approx 2.71962$. This value is tantalizingly close to the mathematical constant $e\approx 2.7183$ known as Euler's number1.

Could the answer to our probability puzzle truly be $e$?

## Proof

Let's start small by finding the probability that $N=2$. The probability that it takes exactly 2 random draws from $[0, 1]$ for their sum to exceed 1 is simply the complement of the probability that the sum of our 2 draws fall short of exceeding 1. That is,

$$P(N = 2) = P(U_1 + U_2 > 1) = 1 - P(U_1 + U_2 \leq 1).$$

We can interpret the probability that $U_1 + U_2 \leq 1$ geometrically. Since $U_1$ and $U_2$ are Uniform random variables on $[0, 1]$, the point $(U_1, U_2)$ can arise anywhere in the unit square with equal probability. Thus, $P(U_1 + U_2 \leq 1)$ corresponds exactly to the proportion of the unit square shaded in the following image.

This shaded region is referred to as a 2-dimensional unit simplex2 with area

$$P(U_1 + U_2 \leq 1) = \int_0^{1}\int_0^{1-u_2}du_1du_2 = \int_0^{1}(1-u_2)du_2 = 1/2$$

So $P(N=2) = 1 - 1/2 = 1/2$, which agrees nicely with our simulation estimate $\hat{P}(N = 2) = 0.514$.

Now for the probability that $N=3$, it is tempting to proceed by finding the exact probability

$$P(N = 3) = P(U_1 + U_2 + U_3 > 1, U_1 + U_2 \leq 1),$$

but dealing with both inequality constraints simultaneously is annoying. It is helpful to recognize that $P(N = 3) = P(N \leq 3) - P(N = 2)$ where

$$P(N \leq 3) = P(U_1 + U_2 + U_3 > 1) = 1 - P(U_1 + U_2 + U_3 \leq 1).$$

Similar to the previous case, the probability that the sum of three random draws from [0,1], $U_1, U_2,$ and $U_3$, does not exceed 1 is given by the proportion of the unit cube shaded in the following image.

This shaded region is a 3-dimensional unit simplex with volume

$$P(U_1 + U_2 + U_3 \leq 1) = \int_0^{1}\int_0^{1-u_3}\int_0^{1-u_3-u_2}du_1du_2du_3 = \int_0^{1}\frac{1}{2}\left[1 - 2u_3 + u_3^{2}\right]du_3 = 1/6.$$

So $P(N=3) = P(N\leq 3) - P(N=2) = (1 - 1/6) - 1/2 = 1/3$ which again agrees with our simulation estimate of $\hat{P}(N = 3) = 0.33$.

Foregoing a formal proof, we can generalize our results on the unit simplex: the $n$-dimensional unit simplex has "volume" given by $1 / n!$. For our purposes, $P(N \leq n) = 1 - 1/n!$. Thus, for any $n\geq 2$,

$$P(N = n) = P(N \leq n) - P(N \leq n - 1) = \left[1 - \frac{1}{n!}\right] - \left[1 - \frac{1}{(n-1)!}\right] = \frac{n - 1}{n!}.$$

We've got everything we need to find the expected value of $N$.

$$E(N) = \sum_{n = 2}^\infty n P(N = n) = \sum_{n = 2}^\infty n\left(\frac{n - 1}{n!}\right) = \sum_{n = 2}^\infty \frac{1}{(n - 2)!} = \sum_{n = 0}^\infty \frac{1}{n!}$$

Recall the exponential series is given by $e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}$.

Thus, $E(N)$ is equivalently $e$.

Yes that's right: we expect, on average, $e$ random draws from the interval $[0, 1]$ to ensure their sum exceeds 1!

1. A convenient way to remember $e$: Take 0.3 from 3 (2.7) followed by the election year of former U.S. president (and horrible human being) Andrew Jackson twice in a row (1828). That is, $e \approx 2.718281828$. Okay, maybe it's not that helpful, but you're slightly more prepared for American presidential history trivia!

2. An $n$-dimensional unit simplex is the region of $n$-dimensional space for which $x_1 + x_2 + \cdots + x_n \leq 1$ and $x_i > 0, i = 1, 2, \ldots, n$. A unit simplex in one dimension (1D) is boring; it's just the unit interval $[0, 1]$. In 2D, the unit simplex takes the form of a triangle bounded by $x_1 + x_2 \leq 1, x_1 > 0, x_2 > 0$. In 3D, we get a tetrahedron, in 4D a hypertetrahedron, and in $n$ dimensions — well, an $n$-dimensional simplex.