= Solution
The $m$ red positions form a uniformly chosen <subset> of size $m$, so there are $\binom{m+n}{m}$ equally likely colour patterns. A contiguous red block can start at positions $1,2,\ldots,n+1$, giving exactly $n+1$ favourable patterns. Therefore
$$
\boxed{\mathbb P(\text{all reds together})=\frac{n+1}{\binom{m+n}{m}}.}
$$
If the cards are regarded as individually labelled, each colour pattern corresponds to $m!n!$ <permutations>, so the same counting argument still applies. In the edge case $m=1$, the formula correctly gives <probability> one.
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