Expected number of (0,1) distributed continuous random variables required to sum upto 1 I define $X_i$ as a random variable that is uniformly distributed between (0,1). What is the expected number of such variables I require to make the sum go just higher than 1. 
Thanks
 A: I asked a related question in MathOverflow a while back. Here is the link. Let $N$ be the number of $U(0,1)$ random variables needed for the sum to cross $1$. It is in fact possible to derive the pmf of $N$ and compute $E(N)$ as David Bar Moshe has indicated. But there is a nifty derivation if we are only interested in $E(N)$ that involves conditioning on the very first random variable $U_1$ that we pick. I have sketched an outline below. Let me know if any part of the derivation isn't clear.
Let $N(x)$ be expected number of $U(0,1)$ random variables needed for the sum to cross $x$ where $0 \leq x \leq 1$. Then, a recursion can be derived for $f(x) = E(N(x))$ as
$f(x) = \int_{0}^{1} E(N(x) \mid U_1 = y) dy$.
This integral naturally splits into two - that between $0$ and $x$ and that between $x$ and $1$. If $U_1 < x$, $E(N(x))$ is simply $1 + f(x-U_1)$. If $U_1 \> x$, $E(N(x))$ is just $1$. This will give us an integral equation for $f(x)$ that can be solved to get the solution (with suitable initial condition) as $f(x) = e^x$. 
A: This problem is well-known. For an answer, see, for example, the first part of Section 2 in http://myweb.facstaff.wwu.edu/curgus/Papers/27Unexpected.pdf, or Equations (7)-(10) in http://mathworld.wolfram.com/UniformSumDistribution.html
A: I don't know the exact value, but by Wald's equation it is between 2 and 4.
Edit: Let $N$ be the number of terms required.  $N$ is a stopping time for the sequence $X_1, X_2, \dots$, so Wald's equation gives us $$E[X_1 + \dots + X_N] = E[N] E[X_1].$$
We have $E[X_1] = 1/2$, and $1 \le X_1 + \dots + X_N \le 2$, so rearranging gives us $2 \le E[N] \le 4$.
