Probability brain teaser with infinite loop
Let $T$ be the time spent in the mine. Conditioning on the first door the miner chooses, we get $$ \mathbb{E}[T]=\frac{1}{3}\cdot3+\frac{1}{3}(5+\mathbb{E}[T])+\frac{1}{3}(7+\mathbb{E}[T])$$ so $$ \mathbb{E}[T]=5+\frac{2}{3}\mathbb{E}[T].$$ If $\mathbb{E}[T]$ is finite, then we can conclude that $\mathbb{E}[T]=15$.
To see that $\mathbb{E}[T]$ is finite, let $X$ be the number of times the miner chooses a door. Then $ \mathbb{P}(X\geq n)=(\frac{2}{3})^{n-1}$ for $n=1,2,3,\dots$, hence $$ \mathbb{E}[X]=\sum_{n=1}^{\infty}\mathbb{P}(X\geq n)=\sum_{n=1}^{\infty}\Big(\frac{2}{3}\Big)^{n-1}<\infty$$ And since $T\leq 7(X-1)+3$, we see that $\mathbb{E}[T]<\infty$ as well.
Let $t$ be the expected time to get out. If he takes the second or third door he returns to the same position as the start, so the expected time after he returns is $t$. Therefore we have $$t=\frac 13(3) + \frac 13(t+5)+\frac 13(t+7)\\\frac 13t=5 \\t=15$$
With probability $1$ he will eventually take the correct door so, $$3*\frac11$$
The wrong doors can be combined, $$ +6*\frac23 $$
and repeated $$ +6*(\frac23)^2 $$
and repeated $$ +6*(\frac23)^3 $$
...
This is can be written as a geometric series
$$3 +\sum (6 * (\frac23)^n)$$
Which is known to simplify to $$ 3+ 4/ (1-\frac23)$$
$$= 3+4*3 $$
$$ = 15$$
This result seems to generalize. For $K$ doors with paths length $A,B,C,...N$ if only $A$ leads out and all other doors lead back to the choice you get $$A + \sum(B*(\frac{1}{K})^n) + \sum(C*(\frac{1}{K})^n) + ...\sum(N*(\frac{1}{K})^n)$$
Which becomes $$A+(\frac{\frac BK}{1-\frac1K})+(\frac{\frac CK}{1-\frac1K})+...(\frac{\frac NK}{1-\frac1K}))$$
$$=A +\frac BK * K + \frac CK * K + ...\frac NK * K$$
$$=A+B+C+...N$$
Going randomly is equivalent of taking each door once. Weird.