= Solution
Applying the optimal single-copy filter independently gives a random number $M_n$ of exact successful targets with <binomial distribution>
$$
M_n\sim\operatorname{Bin}(n,s/t),\qquad
\mathbb E M_n=n\frac{s}{t}.
$$
The <weak law of large numbers> gives $M_n/n\to s/t$ in probability. In contrast, collective <entanglement concentration> and <entanglement dilution> yield asymptotically $n h_2(s)/h_2(t)$ targets with vanishing output error. The improvement is strict, not just nonnegative. Differentiating <binary entropy> gives
$$
\frac{d}{dx}\left(\frac{h_2(x)}x\right)
=\frac{\log_2(1-x)}{x^2}<0\qquad(0<x<1).
$$
Since $s<t$, it follows that
$$
\boxed{\frac{h_2(s)}{h_2(t)}>\frac{s}{t},\qquad
m(n)\sim n\frac{h_2(s)}{h_2(t)}>\mathbb E M_n.}
$$
This is <collective advantage over independent two-qubit filtering>. Independent filtering discards the entanglement in product-state failure branches; the collective reversible asymptotic procedure avoids that extensive loss. The separately filtered targets are exact conditional on success, whereas the collective statement uses the vanishing-error convention from the previous part. Here the counted outputs are target states $|\phi\rangle$; that is also the notation in the original printed question.
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