Lucky game of a success and failure probability. (Advanced probability exams problem) Problem :

One random experiment has two possible results : Success (S) with probability $p$ and Failure (F) with probability $q=1-p$. We consider the following lucky game : We conduct independent repetitions of the experiment until we come across the first Success. If the success comes at the first repetition, we win $2\$$, if it comes at the second repetition, we win $1\$$, if it comes at the third one then we win $1/2 \$$ and so on.
$(a)$ Find the expected profit for a player in this game.
$(b)$ If $p=1/3$, is it fair that taking part in such a game costs $1\$$ ?

Discussion :
The problem above was an exams question in one of the first "advanced" probability courses on the Mathematics Department of the School of Applied Mathematics and Physics of NTUA.
My thoughts on it, is first to describe the function of the winning amount in terms of the winning repetition step :
$$w_n = \frac{2}{2^n} = 2^{1-n}$$
and we observe that the winning amount gets cut in the half as described in the problem form.
Now, judging from the form of the game, I guess the distribution that it follows is a geometric one, which means that :
$$P\{X=k\} = (1-p)^{k-1}p$$
How would I find the expected profit for a player though ? I wasn't able to solve the problem in the exams, I would really appreciate if someone could provide a hint or elaborate thoroughly so I can grasp what I couldn't understand.
 A: You are correct that this follows a geometric distribution; if $W$ is the winnings of the game, then the expectation can be written by conditioning on the round in which the game is won; let $N \sim \text{Geom}(p)$ be the round in which the game is won, then
$$\mathbf E[W] = \sum_{n=1}^\infty \mathbf E[W \, | \, N = n] \mathbf P[N=n].$$
You have already provided the formula for $\mathbf P[N = n] = (1-p)^{n-1}p$, as well as the formula for the value won if the game ends in round $n$
$$\mathbf E [ W \, | \, N = n] = 2^{2-n},$$
(note that my formula differs from yours by a factor of $2$, since I assume that the first round is $n = 1$, which matches the standard definition of a geometric variable. Hence the expectation is
$$ \mathbf E[W] = 2p \sum_{n=1}^\infty \left( \frac{1-p}{2} \right)^{n-1} 
 = 2p \sum_{n=0}^\infty \left( \frac{1-p}{2}\right)^n.$$
This can now be solved using the identity for the geometric series: $\sum_{n=0}^\infty x^n = \frac1{1-x}$, when $ 0 \leq x < 1$, so we have
$$ \mathbf{E}[W] = 2p \frac{1}{1 - \frac{1 - p}{2}} = 2p \frac{2}{1+p} = \frac{4p}{1+p}.$$
When $p = 1/3$ we see that $\mathbf{E}[W] = 1$, so if the game costs 1 dollar to enter, then we would expect neither to make a gain or loss.
