# Probability and graphs from a math contest [closed]

There are $n$ planets. Luke is on planet $A$. For each planet, except planet $B$ and $C$, there are two unidirectional roads that join other planets. From at most one of these two roads, Luke can find a sequence of roads that bring him back to the planet he had just left. Luke's journey ends when he reaches planet $B$ or $C$. When he leaves a planet, he chooses one of the two roads with equal probability. If the probability that he arrives on planet $B$ is $\frac{1}{2016}$, what is the minimum value of $n$?

How would you attack this problem? I have no idea :(.

Any help would be greatly appreciated.

## closed as off-topic by heropup, user223391, Leucippus, Jack, user91500Dec 11 '16 at 6:06

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We can solve the problem with $n=13$, by starting with $$A\to X_1\to X_2\to X_3\to X_4\to X_5\to X_6\to X_7\to X_8\to X_9\to X_{10}\to B$$ and letting all other arrows point to $C$, except that the second arrow of $X_5$ points back to $A$. Then the probability that we ever take the edge $X_5\to X_6$ is $\frac1{2^6}+\frac1{2^{12}}+\frac1{2^{18}}+\ldots = \frac1{63}$. Once we are at $X_6$, the probability to reach $B$ is $2^{-5}$, hence in total we have a probability of $\frac1{63\cdot 32}=\frac1{2016}$ to reach $B$.
On the other hand, there is no solution with $n\le 12$, for in that case the shortest path from $A$ to $B$ (which must exist) would consist of at most $10$ edges (as it cannot pass through $C$), leading to a lower bound of $2^{-10}>\frac1{2016}$ for the probability.