Probability of hitting X shots in N tries knowing that the P(hit) is the ratio of previous hits Let $X$ be the number of hits in $N$ tries, I know that the probability of the next hit is $P(\text{Hit}) =X/N$.
How can I get the generic expression for the probability distribution function of hitting $x$ hits in $N$ tries, knowing that the probability of hitting the first shot is $p_1$?
I've been trying to find the solution to this, I know from brute forcing that for $p_1 = 0.5$ the distribution is a constant ( $1/(N+1)$ I believe), when it's higher than 0.5 it's a positive slope line and lower than 0.5 a negative slope line. I just don't know how to mathematically reach a result.
Edit: Clarification.
Let's assume the first two 6 shots are known. There were 3 hits and 3 misses, and so the probability of the 7th shot hitting is 0.5. Should that hit, the probability of hitting the 8th shot would be $4/7$. Should that miss, the probability of hitting the 8th shot would be $3/7$.
I forgot to say that when I was brute forcing I assumed the first two shots to be a hit and a miss, but knowing any sequence of two or more previous shots would be necessary for the problem to make sense.
 A: "There were 3 hits and 3 misses, and so the probability of the 7th shot hitting is 0.5. " Do you mean: "There were 3 hits and 3 misses, and so my estimate of the probability of the 7th shot hitting is 0.5. "
Some more information is needed before your question can be answered. Firstly, is the process stable? i.e. is the real probability of a hit constant, or does it drift over time?
Secondly, what prior knowledge do you have of the process? If you can encode your prior knowledge as a probability distribution of the unknown P then bayesian methods provide a precise answer to your question.
In the absence of prior knowledge your question has no answer. Put it this way: suppose I have already planned the infinite sequence miss, miss, hit, hit, miss, hit, miss ... according to some whim, then the first $n$ results don't really tell you anything about the $n+1^{st}$
A: Let $P(x, N)$ denote the probability given $N$ throws of making exactly $x$ of them.
We have
$$P(x, N+1) = P(x-1, N)\cdot\frac{x-1}{N} + P(x, N)\cdot\frac{N-x}{N}.$$

Starting from $P(x, 6)=\mathbb{I}(x=3)$ as you suggest:
Note that $P(2, N)=0$ for all $N$.  Using the recurrence,
$P(3, N+1)=0+P(3,N)\cdot\frac{N-3}{N}$, and so
$P(3, N) = 1 \frac36 \frac47\frac58\frac69\dotsb\frac{N-4}{N-1}=\frac{60}{(N-3)(N-2)(N-1)}$.  I don't particularly imagine there's a great closed-form for the general term.

Starting from $P(x, 2)=\begin{cases}a,&\text{x=0}\\b,&\text{x=1}\\c,&\text{x=2}\end{cases}$:
I didn't carefully state the recurrence for small values, but it's not hard to see that $P(0, N)=a$ and $P(N, N)=c$ for all $N$.  Then it turns out that $P(x, N)=b/(N-1)$ for all $x\in(0, N)$, the uniform distribution you suggested earlier (especially taking $a=c=0$).  This is nice enough that there ought to be a slick proof, but I don't have it yet.  It can be proved in a not-so-slick way by doing the computation for $P(1, N)$, $P(N-1, N)$, and the rest of the cases separately.
