Uniforms Until the Sum Exceeds One

$N > n$ means the first $n$ draws add up to at most $1$. Find $P(U_1 + \dots + U_n \le 1)$ for $n = 1, 2$ by drawing a picture.

For a count $N$, $E[N] = P(N > 0) + P(N > 1) + P(N > 2) + \dots$

Call the draws $U_1, U_2, \dots$

Step 1: When do we need more than $n$ draws? We are still going after $n$ draws exactly when the total so far has not passed $1$:

So we need $p_n = P(U_1 + \dots + U_n \le 1)$.

Step 2: Why $p_n = \dfrac{1}{n!}$.

Small cases first. $p_1 = P(U_1 \le 1) = 1$. For $n = 2$, the point $(U_1, U_2)$ is uniform in the unit square, and $U_1 + U_2 \le 1$ is the triangle below the diagonal: half the square, so $p_2 = \tfrac12$.

The general case, by sorting. Look at the running totals

The event $S_n \le 1$ says these totals are $n$ points in $[0, 1]$ that come out in increasing order: $S_1 < S_2 < \dots < S_n \le 1$.

Now compare with $n$ independent uniform points $V_1, \dots, V_n$ in $[0, 1]$. Two facts:

1. Turning the draws $(U_1, \dots, U_n)$ into the totals $(S_1, \dots, S_n)$ is a one-to-one change of variables that keeps volume the same (each $S_i$ just adds the earlier draws, a shear). So "the totals land in increasing order inside $[0,1]$" has the same probability as "$V_1 < V_2 < \dots < V_n$".
2. The $n$ points $V_1, \dots, V_n$ can come out in $n!$ different orders, all equally likely, so one particular order has probability $\frac{1}{n!}$.

Therefore

Check: $n = 2$ gives $\tfrac12$, matching the triangle; $n = 3$ gives $\tfrac16$, the corner pyramid of the cube.

Step 3: From these probabilities to $E[N]$.

Why we can add the $P(N > n)$ instead of multiplying by $n$. The definition is $E[N] = \sum_n n \, P(N = n)$. The shortcut rewrites $n$ as a count:

because $N > k$ is true for exactly $N$ values of $k$, namely $k = 0, 1, \dots, N - 1$. Taking expectations of both sides,

Each draw count $n$ still gets weighted by $n$; it's just collected one layer at a time.

Shortcut:

Same answer the long way. $P(N = n) = P(N > n - 1) - P(N > n) = \frac{1}{(n-1)!} - \frac{1}{n!} = \frac{n - 1}{n!}$. Then

Answer: $E[N] = e \approx 2.71828$.

Sanity check: each draw averages $\tfrac12$, so you might guess $2$ draws. But you always need at least $2$, and sometimes $3$ or more, so the average has to be above $2$.