TITLE: Explanation of numeric experiment that approximates e?
QUESTION [6 upvotes]: Recently I found this post on Reddit. It describes the following algorithm to find e: 

Here is an example of e turning up
  unexpectedly. Select a random number
  between 0 and 1. Now select another
  and add it to the first. Keep doing
  this, piling on random numbers. How
  many random numbers, on average, do
  you need to make the total greater
  than 1? The answer is e.

This means that you need on average ~2.7 random real numbers to make the sum greater than 1.
However, a random number between 0 and 1 would on average be equal to 0.5. So intuitively I would think that, on average, only 2 random numbers would be required to have a sum > 1.
So where did I go wrong in my thinking? 
Update
I just figured it out myself: You need at least two numbers to have a sum > 1, but often you'll need three, sometimes you'll need four, sometimes five, etc... So it only natural that the average required numbers is above 2.
Thanks for the replies!

REPLY [3 votes]: The other answers should have resolved why the number is >2, but for the exact reason why it is e:
For having N random numbers whose sum is <1, the probability is exactly the same as the volume of a unit standard N-simplex, i.e. 1/N!.
Therefore, the probability that it takes exactly N random numbers for a sum of ≥1, the probability is
$$ \frac1{(N-1)!} - \frac1{N!} = \frac1{N(N-2)!} $$
Therefore the expected number is
$$ \sum_{N=1}^\infty \frac N{N(N-2)!} = e $$