Urn problem with multiple drawings and conditional replacement Problem description
Imagine a urn, initially completely filled with $N$ marbles of color green. Define that 1) each time that a green colored marble is drawn from the urn another marble of color red is put back instead of the green marble; and 2) if a red colored marble is drawn it is simply put back into the urn. Under the above conditions, is there a probability distribution for the probability of drawing a green (or red) marble after $k$ draws?
Additional information
I suspect that the distribution is related to the binomial distribution, or to the Poisson binomial distribution, given that the probability of drawing a green or red marble will change throughout the consecutive $k$ draws.
I have calculated the probabilities of obtaining a red marble for $k=1,2,3$ as
\begin{equation}
\textrm{Pr}(R,k=1)=0
\end{equation}
\begin{equation}
\textrm{Pr}(R,k=2)=\frac{1}{N}
\end{equation}
\begin{equation}
\textrm{Pr}(R,k=3)=\frac{1}{N}\times \frac{1}{N} + \frac{2}{N}\times\frac{(N-1)}{N}=\frac{2}{N}-\frac{1}{N^2}
\end{equation}
Note: I am trying to solve a continuous (rather than discrete) problem equivalent to the above and hope to gain some insight from the current question. Thank you in advance.
 A: We view the process as an $(n+1)$-state Markov chain, the state being the number of red balls in the urn.  If we are in state $0$, then we must transition to state $1$ on the next draw, and if we are in state $n$, we must stay there.  If we are in state $0<i<n$ then we stay in the same state if we draw a red ball (probability $\frac in$), and transition to state $i+1$ with probability $\frac{n-i}{n}$.  The transition matrix is
$$
\begin{equation}   P=\begin{bmatrix}   0&1              & 0           &0&0&\ldots&0\\   0&\frac{1}{N}& \frac{N-1}{N} &0&0&\ldots&0\\   0&0            & \frac{2}{N}&\frac{N-2}{N} &0&\ldots&0\\   0&0&0&\frac{3}{N}&\frac{N-3}{N}&\ldots&0\\            0 & 0&0&0&\ddots&\ddots&0\\   \vdots & \vdots&\vdots&\vdots&\vdots&\frac{N-1}{N}&\frac{1}{N}\\   0 & 0&0&0&0&0&1\\ \end{bmatrix} \end{equation}
$$
I claim that $$P=TDT^{-1}$$ where
$$
D=\operatorname{diag}\left(0,\frac1n,\frac2n,\dots,\frac{n-1}n,1\right)
$$
and $T=(t_{ij})$ with
$$
t_{ij}=\binom{n-i}{j-i},\ 0\leq i,j\leq n 
$$
and as usual, we take $\binom{n-i}{j-i}=0$ when $j-i<0$.
To prove the claim, we must show that $PT=TD$.
Let $A=(a_{ij})=PT$.  Then
$$
\begin{align}
a_{ij}&=\sum_{k=0}^np_{ik}t_{kj}\\
&=p_{ii}t_{ij}+p_{i,i+1}t_{i+1,j}\\
&=\frac in\binom{n-i}{j-i}+\frac{n-i}{n}\binom{n-i-1}{j-i-1}\\
&=\binom{n-i-1}{j-i-1}+\frac in\left(\binom{n-i}{j-i}-\binom{n-i-1}{j-i-1}\right)\\
&=\binom{n-i-1}{j-i-1}+\frac in\binom{n-i-1}{j-i}\\ 
&=\frac{(n-i-1)!}{(j-i-1)!(n-j)!}+\frac in\cdot\frac{(n-i-1)!}{(j-i)!(n-j-1)!}\\
&=\frac{(n-i-1)!}{(j-i-1)!(n-j-1)!}\left(\frac1{n-j}+\frac in\cdot\frac1{j-i}\right)\\
&=\frac{(n-i-1)!}{(j-i-1)!(n-j-1)!}\left(\frac{n(j-i)+i(n-j)}{(n-j)n(j-i)}\right)\\
&=\frac{(n-i-1)!}{(j-i-1)!(n-j-1)!}\frac{j(n-i)}{(n-j)n(j-i)}\\
&=\frac jn\binom{n-i}{j-i}
\end{align}
$$
On the other hand, if
$$B=(b_{ij})=TD$$ then $$b_{ij}=\sum_{k=0}^n t_{ik}d_{kj}=t_{ij}d_{jj}=\binom{n-i}{j-i}\frac jn$$ and we see that $A=B$ as claimed.
Now we have to compute $T^{-1}$.  I claim that this is the same as $T$, but with alternating signs.  That is if $T^{-1}=(s_{ij})$, then $$s_{i,j}=(-1)^{i-j}\binom{n-i}{j-i}$$  This is probably a well-known fact, but I haven't been able to find it online, so we slog through the proof.
We must show
$$
\sum_{k=0}^n\binom{n-i}{k-i}(-1)^{j-k}\binom{n-k}{j-k}=\delta_{ij}
$$  The first binomial is $0$ if $i>k$ and the second is $0$ if $k>j$, so the sum is $0$ for $i>j$ and for $i\leq j$, we must show
$$
\sum_{k=i}^j(-1)^{j-k}\binom{n-i}{k-i}\binom{n-k}{j-k}=\delta_{ij}
$$
Now
$$
\begin{align}
\sum_{k=i}^j(-1)^{j-k}\binom{n-i}{k-i}\binom{n-k}{j-k}&=
\sum_{k=i}^j(-1)^{j-k}\frac{(n-i)!}{(k-i)!(n-k)!}\frac{(n-k)!}{(j-k)!(n-j)!}\\
&=\sum_{k=i}^j(-1)^{j-k}\frac{(n-i)!}{(k-i)!(j-k)!(n-j)!}
\end{align}
$$
If we make the change of variables $$
t=k-i, m=j-i, K=n-i$$ then the last sum becomes $$
\begin{align}
\sum_{t=0}^m(-1)^{m-t}\frac{K!}{t!(m-t)!(K-m)!}&=(-1)^m\frac{K(K-1)\cdots(m+1)}{(K-m)!}\sum_{t=0}^m(-1)^t\frac{m!}{t!(m-t)!}\\
&=(-1)^m\binom{K}{m}\sum_{t=0}(-1)^t\binom mt\\
&= \delta_{m0}
\end{align}
$$
By the definition of $m$, this means that the sum equals $\delta_{ij}$.
Now the probability of getting a red marble on draw number $k$ is
$$
XTD^{k-1}T^{-1}Y\tag1
$$
where $$X=\begin{bmatrix}1&0&0&\cdots&0\end{bmatrix}$$ and
$$Y=\begin{bmatrix}0\\\frac1n\\\frac2n\\\vdots\\\frac nn\end{bmatrix}$$
We can rewrite $(1)$ as
$$n^{-k}XTET^{-1}Z\tag2$$ where
$$
E=\operatorname{diag}\left(0^{k-1},1^{k-1},2^{k-1},\dots n^{k-1}\right)
$$ and
$$ Z=\begin{bmatrix}0\\1\\2\\\vdots\\ n\end{bmatrix}$$
Now easily,$$
XTE = \begin{bmatrix}0^{k-1}\binom n0&1^{k-1}\binom n1&2^{k-1}\binom n2&\cdots&n^{k-1}\binom nn\end{bmatrix}\tag3$$
If $C=(c_{ij})=T^{-1}Z$, then
$$
\begin{align}
c_{ij} &= \sum_{k=0}^n(-1)^{k-i}\binom{n-i}{k-i}k\\
&= \sum_{k=i}^n(-1)^{k-i}\binom{n-i}{k-i}k\\
&= \sum_{t=0}^m(-1)^t\binom mt(i+t)\\
&= i\sum_{t=0}^m(-1)^t\binom mt+\sum_{t=0}^m(-1)^t\binom mtt\\
&= i\delta_{m0}-\delta_{m1} 
\end{align}
$$
so that
$$ 
T^{-1}Z=\begin{bmatrix}0\\0\\0\\\vdots\\0\\-1\\n\end{bmatrix}\tag4
$$
Multiplying $(3)$ and $(4)$ give that the probability of drawing a  red ball on draw $k$ is
$$n^{-k}\left(-(n-1)^{k-1}\binom n{n-1}+n^k\binom nn \right)=\boxed{1-\left(\frac{n-1}{n}\right)^{k-1}}$$
I didn't expect such a pleasant answer.  Now that I see it, I wonder if there isn't a way to prove it without all this calculation.  Perhaps the powers of the transition matrix $P$ have a simple form.
