Chapter 3, Problem 3.83P

In a certain contest, the players are of equal skill and the probability is $\frac{1}{2}$ that a specified one of the two contestants will be the victor, in a group of $2^n$ players, the 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 contestant, A and B, and define the events $A_i,i\le n,E$ by $A_i$:

A plays in exactly i contests

E: A and B never play each other

1. Find $P\left(A_i\right),i=1,\mathrm{...},n$.
2. Find $P\left(E\right)$.
3. Let $P_n=P\left(E\right)$.
4. Show that $P_n=\frac{1}{2^n-1}+\frac{2^n+2}{2^n-1}{\left(\frac{1}{2}\right)}^2{P}_{n-1}$ are use this formula to check the answer you obtained in part (b).

Hint: Find $P\left(E\right)$ by conditioning on which of the events $P\left(A_i\right),i=1,\mathrm{...},n$ occur. In simplifying your answer, use the algebraic identity ${\displaystyle \underset{i=1}{\overset{n-1}{\sum }}i{x}^{i-1}=\frac{1-n{x}^{n-1}+\left(n-1\right)x^n}{{\left(1-x\right)}^2}}$

For another approach to solving this problem, note that there are a total of $2^n-1$ games played.

1. Explain why $2^n-1$ games are played.

Number these games, and let $B_i$ denote event that A and B play each other in game $i,i=1,\mathrm{...},2^n-1$.

What is $P\left(B_i\right)$.

Use part (e) to find $P\left(E\right)$.