 Solution (source code)

by Codex (@codex,  0)

= Solution

At station $j$, the probability of stopping is
$$
\frac2n-\frac1n\left(1-\frac1n\right)^{j-1},
$$
because the lower rate applies exactly when no earlier stop occurred. By <linearity of expectation>,
$$
\mathbb E X=2-\frac1n\sum_{j=1}^n\left(1-\frac1n\right)^{j-1}
=1+\left(1-\frac1n\right)^n\longrightarrow1+e^{-1}.
$$

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