A box has red balls and green balls. To each step, we take a ball and we put it back in the box with an other ball of the same color. At the beginning, the box has exactly one ball red and one ball red. Let $R_n$ the number of red balls after $n$ step. Show that $$\left(S_n=\frac{R_n}{n+2}\right) $$ is a martingale refer to $(R_n)$.
Q1) I just have a problem to understand why do we have that $$\mathbb E[R_{n+1}\mid R_0,...,R_n]=\mathbb E[R_{n+1}\mid R_n]\ \ ?$$
Is $R_{n+1}$ and $R_0,...,R_{n-1}$ are independent ? And if yes, why ?
Q2) Moreover, if $$\mathbb E[R_{n+1}\mid R_n=i]=\frac{n+3}{n+2}i$$ with $i\in\{1,...,n+1\}$, why do we have that $$\mathbb E[R_{n+1}\mid R_n]=\frac{n+3}{n+2}R_n\ \ \ ?$$
Note that $R_{n+1}=R_n$, if a green ball is chosen at step $n+1$ and $R_{n+1}=R_n+1$ if a red ball is chosen at step $n+1$. So, given $R_0,\cdots,\ R_{n-1},R_n$, we have $R_{n+1}=Z_n R_n+(1-Z_n)(R_n+1)$ where $Z_n\sim Bin(1-S_n)$ which means that $R_{n+1}$ is independent of $R_0,\cdots,\ R_{n-1}$ given $R_n$.
So $$\mathbb{E}(R_{n+1}\mid R_0,\cdots,\ R_{n-1},R_n)=R_n(1-S_n)+(1+R_n)S_n=R_n+S_n=\frac{n+3}{n+2}R_n$$