The probability of choosing $6$ at any given time is independent of whether it has been chosen before, which is why the answer is simply $\frac1{10}.$

You can still calculate it as a conditional probability, of course, but you've miscalculated your $P(B\cap A),$ which is why you get a "probability" greater than $1$ eventually. Note that $B\cap A$ is the event in which $6$ is not chosen in the first $N$ picks, but is chosen once in $N+1$ picks (so necessarily on pick $N+1$). Equivalently, $B\cap A$ is the event of choosing $6$ for the first time on pick $N+1$. Part of the issue is that your $A$ and $B$ are not independent, so that makes calculation more cumbersome. Instead, let $C$ be the event of choosing $6$ on pick $N+1,$ and note that $C\cap A=B\cap A,$ and that the events $A,C$ ("never having chosen $6$ before" and "choosing $6$ now") are independent. Thus, $$p(B\cap A)=p(C\cap A)=p(C)p(A)=\frac1{10}\cdot\left(\frac9{10}\right)^N.$$ Put another way: 


In $\left(\frac9{10}\right)^N$ of our possible outcomes, we have not chosen $6$ in the first $N$ picks.
In $\frac1{10}$ of those outcomes, we'll choose $6$ at pick $N+1$--specifically, when we choose $6$ next, rather than some other number.
$\frac1{10}$ of $\left(\frac9{10}\right)^N$ is $\frac1{10}\cdot\left(\frac9{10}\right)^N,$ the fraction of our possible outcomes in which we choose $6$ for the first time at pick $N+1$.


Thus, the probability is $$p(B\mid A)=\frac{p(B\cap A)}{p(A)}=\cfrac{\frac1{10}\cdot\left(\frac9{10}\right)^N}{\left(\frac9{10}\right)^N}=\frac1{10}.$$



Suppose we approach it combinatorically, instead?

There are $10^{N+1}$ total ways to make a sequence of $N+1$ picks from $S,$ but we already know that $6$ was not chosen in the first $N$ picks, so our sample space is narrowed (perhaps substantially, depending on how large $N$ is).

There are $9^N$ possible ways to make a sequence of $N$ choices from $S$ without choosing $6$. From this, it follows that:


There are $9^N\cdot 10$ possible ways to make a sequence of $N+1$ choices from $S$ without choosing $6$ except possibly at pick $N+1$. (This is the cardinality of our "narrowed" sample space.)
There are $9^N$ possible ways to make a sequence of $N+1$ choices from $S$ where $6$ is chosen for the first time at pick $N+1$. (This is the cardinality of the event we're interested in.)


Hence, the probability that we choose a $6$ on pick $N+1$, given that we haven't chosen $6$ in any of the previous $N$ picks, is $$\frac{9^N}{9^N\cdot 10}=\frac1{10}.$$

probability of a certain event in a repeated sampling with replacement (without ordering)
I read somewhere that the Talmud says that all prayers from around the world ascend to Heaven from the Western Wall. (i.e. the kosel)

Does anyone know where in the Talmud that it says this?
