Suppose that there is a urn with $n$ balls, $pn$ red balls and $(1-p)n$ green balls, where $0< p < 1$.
I pick from this urn balls in a sequential order, uniformly without replacement. For $1 \leq k \leq n$, let $R_k$ be the total number of red balls I have at the $k$'th step. I am interested in the distribution of $R_k$, for every $k$.
My attempt. First, we know that $\Pr[R_1=1] = p$, and that $\Pr[R_n = pn] = 1$. We also know that $$\Pr[R_{k+1} = R_k+1 | R_k] = \frac{pn - R_k}{n-k},$$ since there are $pn-R_k$ red balls left out of $n-k$ (if this value is below $0$ (or above $1$), I assume it is $0$ (or $1$)). Thus, for some value $m$, $$\Pr[R_{k+1} = m] = \Pr[R_{k+1} = R_{k}+1 | R_{k} = m-1]\times \Pr[R_{k} = m-1] \\ + \Pr[R_{k+1} = R_{k} | R_{k} = m]\times \Pr[R_{k} = m], $$
which continues in a recursive manner. I am not sure how to extract from this recursion a closed formula of the distribution of $R_k$, or if it even possible.
The distribution of $R_k$ is hypergeometric and can be found directly without any appeal to recursion.
Picking $k$ balls one by one in a specific order does not essentially differ from picking $k$ balls simultaneously if at stake is the finding of the distribution of $R_k$.
Recursion might come in sight if you are after the distribution of random vector $(R_1,…,R_k)$.