1985 AIME Problems/Problem 12

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Problem

Let $A$, $B$, $C$ and $D$ be the vertices of a regular tetrahedron each of whose edges measures 1 meter. A bug, starting from vertex $A$, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to the vertex at its opposite end. Let $p = \frac n{729}$ be the probability that the bug is at vertex $A$ when it has crawled exactly 7 meters. Find the value of $n$.

Solution

Let $P(n)$ denote the probability that the bug is at $A$ after it has crawled $n$ meters. Since the bug can only be at vertex $A$ if it just left a vertex which is not $A$, we have $P(n + 1) = \frac13 (1 - P(n))$. We also know $P(0) = 1$, so we can quickly compute $P(1)=0$, $P(2) = \frac 13$, $P(3) = \frac29$, $P(4) = \frac7{27}$, $P(5) = \frac{20}{81}$, $P(6) = \frac{61}{243}$ and $P(7) = \frac{182}{729}$, so the answer is $182$. One can solve this recursion fairly easily to determine a closed-form expression for $P(n)$.

There also exists a simple heuristic method to arrive at the answer to this question, due to Simon Rubinstein-Salzedo, as follows: after a couple of moves, the randomness of movement of the bug and smallness of the system ensures that we should expect its probability distribution to be very close to uniform. In particular, we would expect $P(n)$ to be very close to $\frac 14$ for decently-sized $n$, for example $n = 7$. (In fact, from looking at the previous solution we can see that it is already close when $n = 3$, and in fact the earlier values are also the best possible approximations given the restraints on where the bug can be.) Since we know the answer is of the form $\frac n{729}$, we realize that $n$ must be very close to $\frac{729}{4} = 182.25$, as indeed it is.

See also

1985 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AIME Problems and Solutions
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