You win the final game of a tournament bracket iff you are the highest ranked
player in that bracket.  For example, in a tournament with 8 people, you win
your first game iff you are the better player in your 2-player bracket, your
second game iff you are the best player in your 4-player bracket, and your
third game iff you are the best player among all 8 people.  If we can find
the probability that you win your ith game using this, then we can use the
linearity of expectation to sum these probabilities up to get our final answer.

So what is the probability that you win your ith game?  Well, there are 2^k
players in all, 2^i players in this particular bracket, and r-1 players better
than you (or equivalently, 2^k - r players worse than you).  You win in the
event that of the 2^i - 1 other players in your bracket, all of them are chosen
from the 2^k - r players that are worse than you.  Since all brackets are
equally likely, this happens with probability
(2^k) choose (2^i - 1) / (2^k - r) choose (2^i - 1)
It's dangerous to compute these two values directly and then divide them, since
these are both very big numbers.  We have two options for dealing with this:
1) We can work with logarithms of factorials instead of factorials directly.
2) We can do some simplification of the expression by cancelling common terms.