5

My friend told me that the answer to this question was $e$, which intrigued me, but he refused to tell me why.

My initial intuition was completely wrong. I thought that since the expected value of any one of these numbers (uniformly distributed on $[0,1]$) was $1/2$, the expected number of them you'd need to get a sum of $1$ would be two. I realized this was rather stupid.

This is my attempt at calculating the expected value term-by term. Let $N$ be the number of numbers between $0$ and $1$ necessary for their sum to exceed $1$, and let $P(N=k)$ be the probability that it takes $k$ such numbers. Then the expected value of $N$ is $$E(N) = P(N=2) \cdot 2 + P(N=3) \cdot 3 + P(N=4) \cdot 4 + \cdots$$

I then calculated the first term:

$$P(N=2) \cdot 2 = \frac{1}{2} \cdot 2 = 1$$

Unfortunately, the second and higher terms aren't as easy ...

Zubin Mukerjee
  • 18,311
  • 1
    First of all, it can't be $2$, because the probability that you reach $1$ in one choice is $0$, so $2$ is minimum value for the number, so the average value of the number would have to be more than $2$ unless it was almost always $2$... – Thomas Andrews Jul 02 '14 at 21:10
  • 3
    http://math.stackexchange.com/questions/111314/choose-a-random-number-between-0-and-1-and-record-its-value-and-keep-doing-it-u –  Jul 02 '14 at 21:12

1 Answers1

3

It might be easier to write this as $P(N>0)+P(N>1) + P(N>2)\dots$.

Then you need to see that $P(N>k)=\frac{1}{k!}$, which is essentially a geometric argument - the ($k$-dimensional) volume of $\{(x_1,\dots,x_k)\in[0,1]^k:x_1+\dots+x_k<1\}$ is $\frac{1}{k!}$.

For the first line:

$$\begin{align}P(N>0) &= P(N=1)+&P(N=2)+&P(N=3)+\dots\\ P(N>1)&=&P(N=2)+&P(N=3)+\dots\\ P(N>2)&=&&P(N=3)+\dots \end{align}$$

Add them up and you get $\sum_{k=0}^\infty P(N>k) = \sum_{k=1}^\infty kP(N=k) = E(N)$.

Thomas Andrews
  • 186,215