Well, that was unexpected

I am grateful to Plinio Sd for pointing out a gap in the previous version of this proof.

Let $X_n$ be the value of the $n$ th random variable, and let $S_n = \sum_{j=1}^n X_j$ be the sum of the first $n$ of them. Let $N$ be the number of RVs that are needed to make the sum greater than $1$ , so that the random variable we want is $S=S_N$ . So long as $0<x\le1$ , $\mathbb{P}[S_n<x]=\int_0^x \mathbb{P}[S_{n-1}<x-y]\,dy$ so a simple induction gives $\mathbb{P}[S_n<x]=\frac{x^n}{n!} \hspace{1cm}0<x\le1$ Thus $\mathbb{P}[N=n]=\mathbb[S_{n-1}<1\le S_n]=\frac{1}{(n-1)!}-\frac{1}{n!}=\frac{n-1}{n!} \hspace{1cm}n \ge2$ Although we do not have the full probability density function $f_n$ for $S_n$ it is clear that $f_n(x) \; = \; \frac{x^{n-1}}{(n-1)!} \hspace{1cm} 0 \le x \le 1$ Moreover $\begin{aligned} \mathbb{P}[S \le 1+x \; \mathrm{and}\; N = n] & = \; \int_0^1 f_{n-1}(y)\mathbb{P}[1-y \le X_n \le 1+x-y]\,dy \\ & = \; \int_0^x f_{n-1}(y)\mathbb{P}[1-y \le X_n \le 1]\,dy + \int_x^1 f_{n-1}(y) \mathbb{P}[1-y \le X_n \le 1+x-y]\,dy \\ & = \; \int_0^x f_{n-1}(y)y\,dy + x\int_x^1 f_{n-1}(y)\,dy \; = \; \frac{x}{(n-1)!} - \frac{x^n}{n!} \end{aligned}$ and hence $\begin{aligned} \mathbb{E}[S|N=n]\,\mathbb{P}[N=n] & = \; \int_0^1 (1+x)\,\frac{d}{dx}\left[\frac{x}{(n-1)!} - \frac{x^n}{n!}\right]\,dx \; = \; \int_0^1 (1+x)\left[\frac{1}{(n-1)!} - \frac{x^{n-1}}{(n-1)!}\right]\,dx \\ & = \; \frac{3n^2 - n - 2}{2(n+1)!} \end{aligned}$ for any $n \ge 2$ , and hence $\begin{aligned} \mathbb{E}[S]& =\; \mathbb{E}[\mathbb{E}[S|N]]\;=\;\sum_{n \ge 2} \mathbb{E}[S|N=n]\,\mathbb{P}[N=n] \\ & = \; \sum_{n \ge 2} \frac{3n^2 - n - 2}{2(n+1)!} \; = \; \frac12\sum_{n \ge 2} \left[ \frac{3}{(n-1)!} - \frac{4}{n!} + \frac{2}{(n+1)!}\right] \\ & = \; \tfrac12e \end{aligned}$ making the answer $\boxed{1.359140}$ .

Let the process stop after the $n$ th time .

The condition for this is we pick $x_{1},x_{2}..........x_{n}$ and $x_{n+1}$ such that $\displaystyle x_{1}+x_{2}+.....x_{n}<1$ and $\displaystyle x_{1}+x_{2}+x_{3}+.....x_{n}+x_{n+1}>1$ .

Let $A$ denote the event that $\displaystyle x_{1}+x_{2}+........x_{n}<1$ and $B$ denote the event that $\displaystyle x_{1}+x_{2}+x_{3}+.....+x_{n}+x_{n+1}>1$ .

We need $P(A\cap B)$ in order to calculate expectation.

$P(A\cap B) = P(A)\cdot P(B\vert A)$ . [Here $P(B\vert A)$ denotes the probability of event $B$ given that event $A$ has happened] .

So let us calculate $P(A)$ .

The probability of this event is given by $\displaystyle \frac{\text{Volume of a standard n-simplex}}{\text{Volume of unit Hypercube in n-dimension}}$ .

The volume of the standard n-simplex can be found using calculus.

$\displaystyle \int_{0}^{1}\int_{0}^{1-x_{n}}\int_{0}^{1-x_{n-1}-x_{n}}......\int_{0}^{1-x_{3}-......-x_{n-1}-x_{n}}\int_{0}^{1-x_{2}-x_{3}-......-x_{n-1}-x_{n}} \,dx_{1}\,dx_{2}\,dx_{3}......dx_{n-1}\,dx_{n} = \frac{1^{n}}{n!}=\frac{1}{n!}$ .

Hence $\displaystyle P(A) = \frac{1}{n!}$

Now to find $P(B\vert A)$ let us call $\left(x_{1}+x_{2}+........x_{n}\right) = z$ .

So the $P(B\vert A)$ is identical to finding the probability of $P(z+x_{n+1}) <1$ where $z$ and $x_{n+1}$ are uniformly and randomly selected from $(0,1)$ which is an easy two dimensional problem. We can even apply basic geometry to solve it . It would be the area of the triangle formed by the line $x+y=1$ and the coordinate axis . But let us do it by the above simplex concept as a $2\text{-simplex}$ is just a line.

$\displaystyle P(B\vert A) =\int_{0}^{1}\int_{0}^{1-x_{n+1}}\,dz\,dx_{n+1} = \frac{1}{2}$ .

Hence $\displaystyle P(A\cap B) = \frac{1}{2\cdot n!}$

Hence the expectation $\text{E}(S)$ is given by $\displaystyle \text{E}(S)=\sum_{n=1}^{\infty} n\cdot P(A\cap B) = \sum_{n=1}^{\infty}\frac{n}{2\cdot n!} = \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{(n-1)!} = \frac{\text{e}}{2}$ .

Useful wiki :- Simplex .