Probability in knockout games. Suppose in a knockout tournament 32 players p1 , p2 .....p32 participate.
In each round players are divided into pairs at random and winner goes to the next round.
If p5 reaches semifinal what is the probability that p1 wins the tournament?
All players are equally skilled.
 A: I'll take the statement "All players are equally skilled." to mean that every player has a $50\%$ chance to win against every other player.
$7$ players have been eliminated by $p_5$; the probability that $p_1$ was not among them is $24/31$. If $p_1$ has not been eliminated, the identity of the winner of $p_5$'s branch of the tournament is irrelevant to her; her chances of winning the tournament are still $1/32$. So the chance of $p_1$ winning the tournament, given that $p_5$ reaches the semifinal, is
$$
\frac{24}{31}\cdot\frac1{32}=\frac3{124}\;.
$$
You can also view this along the lines of Conrado's (now deleted) answer: The chance that $p_1$ reaches one of the remaining three spots in the semifinal is $3/31$, and the chance of winning the tournament once the semifinal is reached is $1/4$, again yielding
$$\frac3{31}\cdot\frac14=\frac3{124}\;.$$
And a third way to calculate: $p_5$ now has a chance of $1/4$ to win. Everyone else's chances must add up to the remaining $3/4$, and there are $31$ of them, which again yields
$$
\frac1{31}\cdot\frac34=\frac3{124}\;.
$$
