the zeta function has a probabilistic interpretation:
$$ \zeta(2)^{-1}= \prod_p \left( 1- \frac{1 }{p^2 } \right) $$
can we have probabilistic interpretation of $L(1, \frac{\cdot }{ p})$ which also has an Euler product?
$$ L(1, (\frac{\cdot }{ p}))^{-1}=\prod_q \left( 1- (\frac{q}{ p})\frac{1 }{q} \right)^{-1}$$
except that $1+ \frac{1 }{ q}>1 $ is not a probability.