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 $\sum_{i=1}^{N1} U_i \leq 1$. 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 longrun 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)[1] # which partial sum first exceeds 1?
}
n < find_n(draws)
draws[1:n]
## [1] 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) {
# mby20 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 number^{1}.
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 2dimensional unit simplex^{2} with area
$$ P(U_1 + U_2 \leq 1) = \int_0^{1}\int_0^{1u_2}du_1du_2 = \int_0^{1}(1u_2)du_2 = 1/2 $$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 3dimensional unit simplex with volume
$$ P(U_1 + U_2 + U_3 \leq 1) = \int_0^{1}\int_0^{1u_3}\int_0^{1u_3u_2}du_1du_2du_3 = \int_0^{1}\frac{1}{2}\left[1  2u_3 + u_3^{2}\right]du_3 = 1/6. $$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 $n\geq 2$,
$$ P(N = n) = P(N \leq n)  P(N \leq n  1) = \left[1  \frac{1}{n!}\right]  \left[1  \frac{1}{(n1)!}\right] = \frac{n  1}{n!}. $$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!

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!
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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.
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