Players are of equal skill, and in a contest the probability is $\frac{1}{2}$ that a specified one of the two contestants will be the victor. A group of $2^{n}$ players are paired off against each other at random. The $2^{n-1}$ winners are again paired off randomly, and so on, until a single winner remains. Consider two specified contestants, $A$ and $B$, and define the events $A_{i}, i \leq n, E$ by
$A_{i}: A$ plays in exactly $i$ contests;
$E: A$ and $B$ ever play each other.
(a) Find $P\left(A_{i}\right), i=1, \ldots, n$.
(b) Find $P(E)$.
(c) Let $P_{n}=P(E)$. Show that
$$
P_{n}=\frac{1}{2^{n}-1}+\frac{2^{n}-2}{2^{n}-1}\left(\frac{1}{2}\right)^{2} P_{n-1}
$$
and use this to check your answer obtained in part (b).
For another approach for solving this problem, note that there are a total of $2^{n}-1$ games played.
(d) Explain why a total of $2^{n}-1$ games are played.
Number these games and let $B_{i}$ denote the event that $A$ and $B$ play each other in game $i, i=1, \ldots, 2^{n}-1$.
(e) What is $P\left(B_{i}\right)$ ?
(f) Use nart (e) to find $P(E)$.