Harvard-MIT Mathematics Tournament

Solution:

At any point during this competition, we shall say that the situation is living if both players 1 and $2^{n}$ are still in the running. A living situation is far if those two players are diametrically opposite each other, and near otherwise, in which case (as one can check inductively) they must be just one person shy of that maximal separation. At the start of the tournament, the situation is living and near. In each of rounds 1 to $n$, a far situation can never become near, and a near situation can stay near or become far with equal likelihood.

In each of rounds 1 to $n-1$, a living situation has a $1/4$ probability of staying living. Therefore, at the end of round $k$, where $1 \leq k \leq n-1$, the situation is near with probability $1/8^{k}$, and far with probability $1/4^{k} - 1/8^{k}$. In round $n$, a far situation has a $1/4$ probability of staying living, whereas a near situation has only a $1/8$ probability of staying living. But if the situation is living at the beginning of the last round, it can only be far, so we can say with complete generality that, at the end of round $k$, where $1 \leq k \leq n$, the situation is living and far with probability $1/4^{k} - 1/8^{k}$.

We are interested in finding the probability that the situation is living at the end of round $n$ (and hence far); that probability is thus
$$\frac{1}{4^{n}} - \frac{1}{8^{n}} = \frac{2^{n} - 1}{8^{n}}.$$