  We consider the problem of communicating over a channel that randomly "tears"
the message block into small pieces of different sizes and shuffles them. For
the binary torn-paper channel with block length $n$ and pieces of length ${\rm
Geometric}(p_n)$, we characterize the capacity as $C = e^{-\alpha}$, where
$\alpha = \lim_{n\to\infty} p_n \log n$. Our results show that the case of
${\rm Geometric}(p_n)$-length fragments and the case of deterministic
length-$(1/p_n)$ fragments are qualitatively different and, surprisingly, the
capacity of the former is larger. Intuitively, this is due to the fact that, in
the random fragments case, large fragments are sometimes observed, which boosts
the capacity.
