$E$(hits) if for every fail, you divide the probability of hitting by $C$, and for each hit, you divide the probability of failing by $C$? Suppose a player is playing darts, and they can hit or miss a certain target. With each miss they get less confident, and with each hit their aim gets better.
Let $p_0$ be the initial probability of them hitting the target. Let $p_n$ be the probability of them hitting the target after the $n$th attempt, and $f_n$ the probability of failing the target after the $n$th attempt. Of course, $f_n + p_n = 1.$
Let:
\begin{align*}
p_n = \dfrac {p_{n-1}}{C} \ \ , \ \ \text{if he fails the $n$th throw},  \\
f_n = \dfrac {f_{n-1}}{C} \ \ , \ \ \text{if he hits the $n$th throw},
\end{align*}
where $C$ is a real constant bigger than $1$, that symbolizes their change in confidence.
My question is: what's the average number of hits after $N$ throws, for a certain $p_0$ and $C$?

Answer
I know the answer: $E(N,p_0,C) = $$Np_0$, the same as if $C$ was $1$ and there was no change in the game. I did some computations that showed this result, and then proceeded to prove it by induction. The proof is as follows:
For $N=1$ this is obvious. Assume it for the expected values of $N-1$ throws. Then, on the first throw, we have that:
$$E(N,p_0,C) = p_0 \left (1 + E(N-1,p_0 + \frac{(1-p_0)(C-1)}{C}  ,C) \right) + (1-p_0) \left(E(N-1,\frac {p_0}{C}, C) \right) $$
Where the big expressions for probability are the probability of hitting the second throw after hitting and missing on the first, respectively.  If we replace the $E(N-1,p, C)$s for $(N-1)p$ we get the desired result that $E(N,p_0,C) = Np_0$, after some cancelation. But why?
What I am looking after
I am seeking a solution to this question that doesn't require a priori knowledge about the solution. I don't see why the special kind of symmetry proposed in the problem would imply that the expected value stays the same regardless of $C$: the latter seems stronger at first sight. Also, I originally intended to submit this question to a (now canceled) student contest, and it didn't seem fair that they would have to find the answer a priori.
I've been able to prove that $E(p_0,N,C) + E(1-p_0,N,C) = N$, and so, for a given $N$ and $C$, the solution has symmetry across $p_0=\frac 1 2$, and that implies $E(1/2,N,C) = \frac N 2$ for any $C$. Also, the equation from before is still valid. If one could show that $E(1/2,N,C)$ doesn't depend on $C$, that alone would be enough to prove the result as well.
 A: A simple and natural approach
A natural approach is to try computing the probability of the second throw being a hit.
Let $q_1 $ be the probability of the second throw being a hit. Note $p_0$ is the probability of the first throw being a hit.
$$\begin{aligned}q_1&=p_0P[\text{second throw is a hit}\mid\text{the first throw is a hit}]\\
&\quad\quad+(1-p_0)P[\text{second throw is a hit}\mid\text{the first throw is a miss}]\\
&=p_0(1-(1-p_0)/C) + (1-p_0)(p_0/C)\\
&=p_0
\end{aligned}$$
The equality says the probability of hitting the target does not change from the first throw to the second throw. So the probability of hitting the target  stays at $p_0$ forever.
$$E[N, p_0,C]=np_0.$$
A trivial generalization
The computation above shows that $C$ can vary at each attempt.
That is, the expected number of hits after $N$ throws is still $np_0$ if we define $$\begin{align*}
p_n = \dfrac {p_{n-1}}{C_{n-1}} \ \ , \ \ \text{if he fails the $n$th throw},  \\
f_n = \dfrac {f_{n-1}}{C_{n-1}} \ \ , \ \ \text{if he hits the $n$th throw},
\end{align*}$$
where $C_0, C_1, C_2,\cdots$ is a sequence of numbers that are $\ge1$.
In fact, we do not need $C_i\ge1$ as long as $p_n,f_n\le 1$.
