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 ${1\over 2^{n-1}}$.