Consider a game with 2 stages. In the first stage, the player tosss a coin until a head is appears

Question

Consider a game with 2 stages. In the first stage, the player tosss a coin until a head is appears. Say that it takes j flips to get a head. Then, in the second stage, he random draws an integer between 1 and $ 2^j $ to get his winnings (so for example if he draws 2,he gets 2 dollars). a) What is the average amount of dollars he gets? b)Given that he wins 4 dollars, what is the probability that the coin was flipped exactly 6 times,i.e,that j=6?
My work for question a is I have the probability $\frac{1}{2^j}$ until getting a head,and each time I have a head,i calculate the expected value of money in range from 1 to j,so:
\begin{equation} E(X) = \sum_{j=1}^\infty \frac{1}{2^j}\sum_{i=1}^{j}i\frac{1}{2^j} \end{equation} b) I call A is the event he gets 4 dollars and B is the event he flipped the coin exactly 6 times,so
\begin{equation} P(B|A) = \frac{P(B \cap A)}{P(A)} \end{equation} with
\begin{equation} P(B \cap A) = \frac{1}{2^6} \ and \ P(A)=\sum_{i=2}^\infty \frac{1}{2^i} \end{equation} is it a right way to do? if it's right, how can i calculate the sums?

Answer 1

(a) Your inner summation should go from $1$ to $2^j$. You can also simplify the summations:

$$ \begin{equation} E(X) = \sum_{j=1}^\infty \frac{1}{2^j}\sum_{i=1}^{2^j}i\frac{1}{2^j} = \sum_{j=1}^\infty \left( \frac{1}{2^j}\cdot \frac{1+2^j}{2} \right) =\frac{1}{2}\sum_{j=1}^\infty \left( \frac{1}{2^j} + 1 \right) = +\infty \end{equation} $$

(b) How did you get $P(B \cap A) = \frac{1}{2^6}$? Isn't that $P(B)$? Since $P(A)$ and $P(A \cap B)$ are hard to computer I suggest computing the conditional this way:

$$ \begin{equation} P(flip=6|won=$4) = \frac{P(won=$4|flip=6)P(flip=6)}{\sum\limits_{i=1}^\infty P(won=$4|flip=i)P(flip=i)} \end{equation} $$