Let $U$ be a uniform random variable on $[0,1]$.
Once $U$ is known, we have additional r.v.s $X_1,\ldots$ also from the Uniform distribution on $[0,1]$.
Let $N$ be the number such that $N$ is the smallest $n$ for $X_n \geq U$.
Find $\mathbb{P}(N=k)$.
This is what I tried:
$$N = \min\{n:X_n \geq U\}.
$$
So
$$\mathbb{P}(N=k) = \mathbb{P}(X_k \geq U) \\ = 1- \mathbb{P}(X_k < U) \\ = 1 - \int_{\text{all} \ x_k} I(0<x_k<1)\ dx_k \\ = 1 - \int_0^u 1 \ d x_k \\ = 1-u
$$
however this seems to be in terms of $u$ and not even $k$....
One can certainly condition on $U=u$ and then integrate over $u$, but here is an approach without integration:
$P(N=k) = P(X_k \ge U > X_i \ \ \forall i \in \{1, 2, ..., k-1\}) = \frac{1}{k+1} \frac{1}{k}$, because this is simply the probably that out of $k+1$ independent and identically distributed (iid) r.v.s, a specific r.v. ($X_k$) comes first and another specific r.v. ($U$) comes second. It doesnt even matter that each of them is uniform. (All we need is to have zero probability of any two r.v.s being exactly equal.)
If you really want integration, let $F(x) = P(X\le x)$ be the CDF, and to avoid exact equality among r.v.s, assume $F$ is differentiable and the PDF is $f(x) = F'(x)$. Then we have $P(N=k | U=u) = F(u)^{k-1} (1-F(u))$. Then integrate over $u$:
$$P(N=k) = \int_{range\ of\ u} F(u)^{k-1} (1 - F(u)) f(u) du \\ = \int^1_0 F(u)^{k-1} (1 - F(u)) dF(u) \\ = \int^1_0 y^{k-1} (1-y) dy = \bigg[\frac{y^k}{k} - \frac{y^{k+1}}{k+1} \bigg]^1_0 = \frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)}$$
and it is clear that the exact distribution doesn't matter as long as it is differentiable / has zero probability of two r.v.s with exactly equal values.