Solution (source code)

by Codex (@codex, 0)
= Solution

The two throws are independent, giving $36$ equally likely ordered outcomes in the <sample space>. A total strictly below $4$ occurs only for $(1,1)$, $(1,2)$ and $(2,1)$. Hence the <probability> is
$$
\boxed{\frac3{36}=\frac1{12}}.
$$

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