I was thinking about this question after a related problem: what's the expected number of draws for the sum to exceed 1? For that problem, the answer is known and is a surprising result <http://mathworld.wolfram.com/UniformSumDistribution.html>

My intuition is the final draw will be larger than the average draw. This is because larger draws are more likely to make the sum exceed 1, skewing the expectation to be above average.

I did a simulation in a spreadsheet with 1 million draws and the estimate I found was the final draw was about 0.655.

I have tried many techniques to try and solve this analytically. But none of my answers are close to 0.655, so either my simulation is wrong or my math is wrong (I'm thinking it's my math).

Anyone able to help how to write out this expectation? 

My current method is this: the expectation of the final draw will be the sum from k=2 to infinity of the probability the kth draw makes the sum exceed 1 (this is 1/[k(k-2)!] as explained [here](http://mathworld.wolfram.com/UniformSumDistribution.html)) times the expected value of the final draw--which would range from a lower value of 1 - (sum on turn k-1, given this sum is less than 1) to the upper value of 1. It's this term that is giving me trouble.

Or maybe I'm approaching this completely wrong. Any pointers would be great.

Also, I'm planning on using this as a puzzle for my blog. I will definitely credit anyone that helps. Thanks.