Probability :Knock Out Tournament Of Ranked Players 
Thirty-two players ranked 1 to 32 are playing in a knockout
  tournament. Assume that in every match between any two players, the
  better-ranked player wins, the probability that ranked 1 and ranked 2
  players are winner and runner up respectively, is ?

I dont understand. Should'nt the player ranked 1 always be the winner as the better ranked guy always wins?
 A: If you picture the usual tree diagram for a tournament bracket, you'll see that the 1 and 2 seeds will meet in the final if and only if they start on opposite sides of the bracket.  By symmetry, it doesn't matter where you position the 1 seed (so you may as well place her say at the top left).  This leaves $31$ possible starting positions for the 2 seed, $16$ of which are on the opposite side from the 1 seed.  So the probability they'll meet in the final (where the 1 seed will prevail) is
$${16\over31}\approx.516129$$
A: Yes, the player ranked number $1$ will win all of his matches, and will eventually win the tournament.
But you are asked what the probability is that the players ranked $1$ and $2$ will meet in the final (because only in this scenario, the ranked $1$ player will win, and the ranked $2$ player will be the runner-up).
For the two players to meet in the final, you need to avoid that they meet before the final, and this is what you need to find the probability of.
A: The top ranked player will always win, but the second ranked player will only reach the finals if he doesn't play the top ranked player in the first 4 games. In a knockout tournament with a bracket that means that they must be on opposite halves of the bracket, but in a tournament without a set bracket you are looking for the probability that the player ranked second doesn't play the player ranked first in each round.
$R_{1} * R_{2} * R_{3} * R_{4}$, where $R_{n}$ is the probability of surviving each round, or:
$\frac{30\text{ wins}}{31\text{ possible opponents}} * \frac{14}{15} * \frac{6}{7} * \frac{2}{3} = 0.516129$
A: The only way the $2nd$ ranked player is not runner up is if the draws are randomly put.
The $1st$ ranked player can be anywhere, and will win
The $2nd$ ranked player, however, must be in the other half to be runner up,
hence $Pr = \dfrac{16}{31}$ 
