 Solution (source code)

by Codex (@codex,  0)

= Solution

The favourable outcomes are $(j,j+1)$ and $(j+1,j)$ for $j=1,\ldots,5$. These are ten distinct ordered outcomes among the $36$ equally likely elements of the <sample space>, so the <probability> is
$$
\boxed{\frac{10}{36}=\frac5{18}}.
$$

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