Determine probability of counting problem At time $0$, an urn contains $1$ black and $1$ white ball. At each time $1,2,3, \ldots $ a ball is chose at random from the urn and is replaced together with a new ball of the same colour. Just after time $n$, there are therefore $n+2$ balls in the urn, of which $\beta_n +1$ are black, where $\beta_n$ is the number of black balls chosen by time $n$. 
Let $\mu_n = \frac{\beta_n+1}{n+2}$, the proportion of black balls in the urn after time $n$. I showed that $(\mu_n)_{n \geq 1}$ is a martingale. So
\begin{align}
\mathbb{E}[\mu_{n+1} \mid \mu_0 \ldots \mu_n] = \mu_n.
\end{align}
Now I want to prove that $P(\beta_n = k) = \frac{1}{n+1}$ for $0 \leq k \leq n$. Using the property that $(\mu_n)_{n \geq 1}$ is a martingale, however I do not know in which way I can use this information? 
EDIT:
See answer.
 A: The fact that $(\mu_n)_{n \geq 1}$ is a martingale is not useful at all to solve this problem. However, by induction you can say that:
•$n=1, \qquad P(B_1 = k ) = \frac{1}{2} \qquad \text{for } 0 \leq k \leq 1$ 
•$n=p, \qquad P(B_p = k ) = \frac{1}{p+1} \qquad \text{for } 0 \leq k \leq p$
•$n=p+1 \ \ \ \ P(B_{p+1} = k ) \qquad \qquad \text{for } 0 \leq k \leq p+1$
\begin{align}
&= P(B_p = k-1, \text{ball on $p+1$ is black})\\
&+ P(B_p = k, \text{ball on $p+1$ is white}) \\
&= P(\text{ball on $p+1$ is black}\mid B_p = k-1)\cdot P(B_p = k-1)\\
&+ P(\text{ball on $p+1$ is white}\mid B_p = k)\cdot P(B_p = k)\\
&=\frac{k}{p+2} P(B_p = k-1) + \frac{p+2-(k+1)}{p+2}P(B_p = k) \\
(*)&= \frac{k}{p+2} P(B_p = k-1) + \frac{p+2-(k+1)}{p+2}\cdot \frac{1}{p+1}.
\end{align}
But what is $P(B_p = k-1) $? For me this was not immediately clear.
Well, ideally $P(B_{p+1} = k ) = \frac{1}{p+2}$. From $(*)$ we can determine that $P(B_p = k-1) = \frac{1}{p+1}$. Which is obvious since $P(B_p = k )$ for all $0 \leq k \leq p$. From this it follows that 
\begin{align}
P(B_{p+1} = k ) = \frac{1}{p+2}.
\end{align}
Which completes the proof.
