= Solution
Let $U_m$ be the average over all $d=p+1$ coordinate tuples in part (a), and $V_m$ the distinct-tuple average in part (b). Every product summand lies in $[0,1]$. A proportion $\alpha_m=(N)_d/N^d$, where $N=m+1$, of all tuples have distinct coordinates. Thus
$$
U_m=\alpha_mV_m+(1-\alpha_m)W_m,
$$
where $W_m\in[0,1]$ is the average over the remaining tuples. If there are no remaining tuples, the error is zero. Hence $|U_m-V_m|\leq1-\alpha_m$.
For $d$ independent uniform choices from $N$ indices, any particular pair coincides with <probability> $1/N$. A <union bound> over the $\binom d2$ pairs gives
$$
\boxed{|U_m-V_m|\leq1-\frac{(N)_d}{N^d}\leq\frac{\binom d2}{N}\longrightarrow0.}
$$
This deterministic <sampling with and without replacement comparison for bounded products> is the needed link between the two averages.
Part (a) identifies the almost-sure limit of $U_m$ as $\prod_{k=0}^p\mathbb E[\mathbf1_{A_k}(X_k)\mid\mathcal S]$. By part (b) and the <reverse martingale convergence theorem>, $V_m$ converges almost surely to $\mathbb E[\prod_{k=0}^p\mathbf1_{A_k}(X_k)\mid\mathcal S]$. Their difference vanishes, so
$$
\boxed{\mathbb E\!\left[\prod_{k=0}^p\mathbf1_{A_k}(X_k)\mid\mathcal S\right]
=\prod_{k=0}^p\mathbb E[\mathbf1_{A_k}(X_k)\mid\mathcal S]\quad\text{almost surely}.}
$$
This is precisely <conditional independence>. Exchangeability also makes the single-coordinate conditional laws identical, by applying a coordinate transposition and testing against invariant events. The result is the <De Finetti theorem>, established here by finite symmetrization and the vanishing collision fraction.
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