This solution was sent in by Paul from
Berkhamsted Collegiate School.
To solve this problem we can, without loss of generality, think of
the first player as being at the top of the draw. We MUST also
assume that the chances of winning any given match is $1/2$. Then
the chance of the players meeting in the first round (with $2^n$
participants) is $p=1/(2^{n} - 1)$.
The chances of the two players not meeting in the first round and
winning their first round matches is: $$(1 - {1\over 2^{n} -
1})\times {1\over 4} = {2^{n-1}-1\over 2(2^n - 1)}.$$ It follows
that the chance of the two players being drawn to meet in the
second round is: $$ {2^{n-1}-1\over 2(2^n - 1)}\times
{1\over(2^{n-1} - 1)} = {1\over 2(2^n-1)}= {p\over 2}.$$ So the
odds of them meeting in the second round is half those of meeting
in the first round. By the same reasoning the probability of
meeting in each subsequent round is half the probability of meeting
in the previous round.
For $2^n$ paricipants there will be $n$ rounds, so we can sum ALL
the probabilities of meeting in a certain round, and we get $$
\eqalign{ {1\over 2^n - 1} \left({1\over 1} + {1\over 2} + {1\over
4} + \cdots {1\over 2^{n-1}}\right) &= {1\over 2^n - 1}
\left(2- {1\over 2^{n-1}}\right) \\ &= {1\over 2^n - 1}\left(
{2^n - 1\over 2^{n-1}}\right) \\ &= {1\over 2^{n-1}}.}$$ So the
probability of the two players meeting if there are $2^n$ players
competing is ${1\over 2^{n-1}}$.