Sum of Uniforms Below One

#A volume inside the cube

The vector $(X_1, \dots, X_n)$ is uniform on the unit cube $[0,1]^n$, so every probability is just a volume. The event $\{X_1 + \cdots + X_n < 1\}$ is the corner simplex $\{x_i \ge 0,\ \sum_i x_i < 1\}$, which sits entirely inside the cube, so the probability equals that simplex's volume.

#Partial sums straighten the simplex

Send each point to its running totals, $y_k = x_1 + x_2 + \cdots + x_k$. This map is linear with a triangular matrix whose diagonal is all ones, so its determinant is $1$ and it preserves volume. It carries the simplex $\{x_i > 0,\ \sum_i x_i < 1\}$ exactly onto the sorted region $\{0 < y_1 < y_2 < \cdots < y_n < 1\}$.

#Count the orderings

The cube breaks into $n!$ congruent pieces, one for each ordering of the coordinates, all of equal volume by symmetry. The sorted region is a single one of them, so it has volume $\tfrac{1}{n!}$, and therefore

For $n = 2$ the chance is $\tfrac{1}{2}$ and for $n = 3$ it is $\tfrac{1}{6}$, the factorial shrinking the corner simplex away to nothing as the dimension grows.

#The general statement by induction

The same answer drops out of a sharper claim, that $\PP(S_n \le t) = \tfrac{t^{n}}{n!}$ for every $0 \le t \le 1$, where $S_n = X_1 + \cdots + X_n$. The base case $n = 1$ is just $\PP(X_1 \le t) = t$. Assuming the claim at $n$, condition on the final draw $X_{n+1}$, whose density is $1$ on $[0,1]$,

where the integrand drops to zero once $x > t$ because then $S_n$ cannot be negative. Setting $t = 1$ recovers $\tfrac{1}{n!}$, now as the value of a polynomial that also gives the law of the partial sum at every cutoff below $1$.