A finite sample can miss a predicted outcome with positive probability: an elementary product-model counterexample

Let $p\in (0,1)$ and let ${\mathrm{\Omega }}_0:=\{0,1\}$ with weights $w_0(1):=p$ and $w_0(0):=1-p$; these weights are nonnegative and sum to $1$, so $({\mathrm{\Omega }}_0,w_0)$ is a finite probability space in the sense of [L1], and the event $A_0:=\{1\}$ has $\mathbb{P}(A_0)=p$ by [L3], with $A_0\mathrm{\ne }\mathrm{\varnothing }$ and $\mathbb{P}(A_0)\in (0,1)$.

For $n\ge 1$ form the finite product of $n$ copies of $({\mathrm{\Omega }}_0,w_0)$: the outcome set is $\mathrm{\Omega }:=\{0,1{\}}^n$ with weights $w(\omega )={\prod }_{k=1}^n{w}_0({\omega }_k)$, a finite probability space by [L2]; the single-observation event of the refuted statement corresponds to the coordinate event $C_k:=\{{\omega }_k=1\}$ in the $k$-th coordinate, with coordinate probability $p$ by mutual independence of the coordinates in [L2], and setting ${\epsilon }_k(\omega ):={\omega }_k$ in the operational model of [L4] gives $Y_k=\mathrm{\Phi }+{\epsilon }_k={\omega }_k$ as the $k$-th observation. The family $(Y_k{)}_{k\le n}$ identically distributed by construction and mutually independent by [L2], so it satisfies the IID sampling hypothesis of Repeated-observation protocols of the force branch: adopted IID and second-moment sampling hypotheses and the sample mean; each $Y_k$ is bounded, so the second-moment hypothesis holds as well.

The event "no outcome in $A$ occurs in the finite sample" is $E:=\{\omega \in \mathrm{\Omega }:{\omega }_k=0\text{ for every }k\le n\}={\bigcap }_{k=1}^n\{{\omega }_k=0\}$; by the product formula of [L2] applied to the coordinate events $\{{\omega }_k=0\}$, whose coordinate probabilities are $1-p$, one has $\mathbb{P}(E)={\prod }_{k=1}^n(1-p)=(1-p{)}^n$.

Since $0<p<1$ one has $0<1-p<1$ and therefore $0<(1-p{)}^n<1$ for every $n\ge 1$; the event $E$ of a finite sample with no outcome in $A$ has strictly positive probability.

Under the product model of steps 1.2 and 2.1 the single observation lies in the designated event with probability $p\in (0,1)$, the coordinates are mutually independent, and yet the finite sample avoids the event with positive probability $(1-p{)}^n$ by step 3.1; therefore the implication asserted in the refuted statement is false: a finite sample with no outcome in $A$ can occur with positive probability under the declared sampling model, so its occurrence does not refute that model.

The probability that at least one outcome in $A$ is seen in the finite sample is $1-(1-p{)}^n$, which tends to $1$ as $n$ grows but equals $1$ for no finite $n$ when $0<p<1$; the failure to exhibit the feature in a finite sample is a positive-probability event of the model, not a refutation of it.