Problem
An urn contains $B$ blue balls and $R$ red balls. Suppose that one extracts successively $n$ balls at random such that when a ball is chosen, it is returned to the urn again along with $c$ extra balls of the same color. For each $n \in \mathbb N$, we define $R_n=\{\text{the n-th ball extracted is red}\}$, and $B_n=\{\text{the n-th ball extracted is blue}\}.$
Prove that $P(R_n)=\dfrac{R}{R+B}$.
I thought of trying to condition the event $R_n$ to another event in order to use induction. For example, if $n=2$, I can express $$P(R_2)=P(R_2|R_1)P(R_1)+P(R_2|B_1)P(B_1)$$$$=\dfrac{R+c}{R+B+c}\dfrac{R}{R+B}+\dfrac{R}{R+B+c}\dfrac{B}{R+B}$$$$=\dfrac{R}{R+B}.$$
Now, suppose the formula is true for $n$, I want to show it is true for $n+1$.
So, $P(R_{n+1}=P(R_{n+1}|R_n)P(R_n)+P(R_{n+1}|B_n)P(B_n)$$$=P(R_{n+1}|R_n)P(R_n)+P(R_{n+1}|B_n)(1-P(R_n)$$$$=P(R_{n+1}|R_n)\dfrac{R}{R+B}+P(R_{n+1}|B_n)(1-\dfrac{R}{R+B}).$$
I am having some difficulty trying to calculate $P(R_{n+1}|R_n)$ and $P(R_{n+1}|B_n)$. I would appreciate if someone could complete my answer or suggest me how can I finish the proof if what I've done up to now is correct.
Best Answer
The key is to condition by the composition of the urn at time $n$, say $X_n$ red balls and $Y_n$ blue balls, since $$P(R_{n+1}\mid X_n,Y_n)=Z_n,\qquad Z_n=X_n/(X_n+Y_n).$$ Obviously, $X_0=R$, $Y_0=B$, and, for every $n$, $$X_n+Y_n=R+B+nc,$$ which is deterministic. Conditionally on $(X_n,Y_n)$, one adds $c$ red balls with probability $Z_n$ and zero otherwise, hence $$E(X_{n+1}\mid X_n,Y_n)=X_n+cZ_n=(X_{n+1}+Y_{n+1})Z_n,$$ which implies $$E(Z_{n+1}\mid X_n,Y_n)=Z_n.$$ In particular, for every $n$, $$P(R_{n+1})=E(Z_n)=Z_0=R/(R+B).$$