= Solution
Let $e_k$ be the expected number of children receiving a sticker when $k$ remain. A first-step decomposition gives
$$
e_0=0,\qquad e_k=1+\frac1k\sum_{j=0}^{k-1}e_j\quad(k\geq1).
$$
Subtracting the equations for $ke_k$ and $(k-1)e_{k-1}$ gives $e_k-e_{k-1}=1/k$. Hence $e_k=H_k$, the $k$th <harmonic number>. In particular
$$
\boxed{e_4=1+\frac12+\frac13+\frac14=\frac{25}{12}.}
$$
Every visit before absorption counts exactly one child who receives at least one sticker.
The same first-step argument applies to the <uniform depletion Markov chain> at every initial stock size.
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