Arsthbt

Let $mathcalH$ be a separable Hilbert space. Let $v$ be a random variable taking values in $mathcalH$ such that $P(v perp h) < 1$ for all $h in mathcalH.$ Suppose we sample an infinite sequence $v_1, v_2, ldots.$ Is it the case that, almost surely, the closed span of $v_1, v_2, ldots$ is all of $mathcalH?$

(This may turn out to be a simplified version of J. E. Pascoe's answer).

With probability one, the closure of the random set $V = v_1, v_2, ldots$ is equal to $operatornamesupp v$, the support of the distribution of $v$ ($operatornamesupp v$ is the set of those $h$ such that $P(v in B(h, varepsilon)) > 0$ for every $varepsilon > 0$).

For every $h$, we have $P(h perp v) < 1$, and therefore $h$ is not orthogonal to $operatornamesupp v$. It follows that the closed span of $operatornamesupp v$ is $mathcalH$.

It remains to note that the closed span of $V$ is the same as the closed span of the closure of $V$, equal to $operatornamesupp v$ with probability one.

Another Try

We say a $mathcalH$-valued random variable $h$ is a random vector if $P(h perp g)<1$ for all $gin mathcalH.$

If $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors,
then, almost surely, the closed span of the $h_i$ is equal to $mathcalH.$

First we will need a lemma.

Lemma 1
Let $h$ be a random vector.
There is a countable subset $A$ of $mathcalH$ such that the closed span of the elements of $A$ is equal to $mathcalH$
and for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a.$

Proof
For any subset $A$ such that for every point $ain A,$ $P(hin U)>0$ for any neighborhood $U$ of $a,$ and
the closed span of the elements of $A$ is not equal to $mathcalH,$
we will show that we can grow $A$ by a single element which is not in closed span of the elements of $A.$
We can only do this a countable number of times because the Hilbert space dimension of $mathcalH$ is countable.
(Otherwise, via Gram-Schmidt, we could construct an uncountable orthonormal set by transfinite induction.)

Choose $g$ such that $g perp a$ for all $ain A.$ Now, $P(h perp g)<1.$ So there must be a point $b$ such that
$P(hin U) >0$ for every neighborhood of $b$ and $b$ is not perpendicular to $g,$ therefore, $b$ is not in the span of the elements of $A.$ QED

Suppose $h_1, h_2, ldots$ is a sequence independent identically distributed of random vectors.
Let $A$ be as in Lemma 1. Index $A$ a a sequence $a_n.$
Let $B_m,n$ be a ball of radius $1/m$ centered at $a_n$
Almost surely, the sequence $h_i$ must visit $B_m,n$ infinitely often,
as $P(h_iin B_m,n)>0$. Therefore $A$ is a subset of the closure of the values of the sequence. (We have essentially the fact that a random function $f:mathbbNrightarrow mathbbN^2$ is surjective with infinite multiplicity.)