Another probability question from my textbook Suppose that we have a tennis tournament with 32 players. Players
are matched in a completely random fashion, and we assume that each player
always has probability 1/2 to win a match. What is the probability that two given
players meet each other during the tournament.
 A: Hint 1: Consider the number of other players a particular player meets with what probability: one other with probability $1/2$; two others with probability $1/4$; three others with probability $1/8$; etc.
Hint 2: What is the expected number of other players a particular player meets?
Hint 3: How does Hint 2 relate to the original question?   
Answer: $$ \dfrac{1 \times \dfrac{1}{2} + 2 \times \dfrac{1}{2^2} + 3 \times \dfrac{1}{2^3} + 4 \times \dfrac{1}{2^4} + 5 \times \dfrac{1}{2^4}}{31}  = \dfrac{1}{16}$$
A: Easy general answer for $n$ players in a knockout tournament (and here $n=32$):
There are $\dfrac{n(n-1)}{2}$ potential pairs for matches.
To have one winner, $n-1$ players must be knocked out, so there are $n-1$ actual matches.
So the probability that a particular pair actually have a match is $\dfrac{n-1}{{n(n-1)/2}} = \dfrac{2}{n}$.
A: I get
$$
\frac{1}{31}
+\frac{30}{31}\frac{1}{4}\left(\frac{1}{15}
+\frac{14}{15}\frac{1}{4}\left(\frac{1}{7}
+\frac{6}{7}\frac{1}{4}\left(\frac{1}{3}
+\frac{2}{3}\frac{1}{4}\right)\right)\right)
$$
which equals $\dfrac{1}{16}=6.25\%$. Not sure if this is the most elegant solution, though.
