= Solution
Conditional on a contiguous red block, its $n+1$ possible starting positions remain equally likely because their original colour patterns had equal <probabilities>. The first card is red exactly when that block starts at position $1$. Hence
$$
\boxed{\mathbb P(\text{first red}\mid\text{all reds together})=\frac1{n+1}.}
$$
This also follows by dividing the <probability> of the single favourable block position, $1/\binom{m+n}{m}$, by the <probability> of all $n+1$ block positions.
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