probability of a certain event in a repeated sampling with replacement (without ordering) I have a problem that is bugging me for a couple of weeks now. I have asked some friends etc but the answers were not satisfying at all.
So here we go.
Suppose we have a set S = {1,2,3,4,5,6,7,8,9,10} and each time we pick a random number from this set. After $N$ pickups suppose that the number, lets say 6, has never been picked. 
The question is, what is the probability of picking number 6 in the $N + 1$ pickup knowing that it has not been picked for the previous $N$ pickups?
I know that people will rush and say, its just $\frac 1 {10}$ but if we consider the previous pickups, meaning that we have a "memory" in the "system" doesn't it change everything? 
Imagine a representation of the distribution of # of picks on X axis and # of actually possible values to be picked in Y axis. Law of Large Numbers says that approximately every possible value will have the same number of pickups after a very large number of repetitive samplings N. Supposing that we have rolled for a large number of, lets say dice, and we see that a specific number has not been rolled yet, lets say the 6. Doesn't this implies that it is almost certain that the roll 6 has to come, it was just too unlikely to not have been rolled up to now, given that the dice is "fair"?
I tried to solve this as binomial with Conditional Probability but i think its wrong cause the set is dynamic while Conditional Probability needs a constant set.
My wrong answer is here:
$N$ is known
Number $k$ $\in$ $S$ has $P_k$ = $\frac 1 {10}$ and $Q_k$ = $\frac 9 {10}$
In $N$ pickups no success:  $A$
In $N + 1$ pickups $1$ success: $B$
Binomial:
$p ( X=0)=$ $\binom{N}{0} P^0 Q^N =(\frac {9} {10})^N$
$p ( X=1)=$ $\binom{N+1}{1} P^1 Q^{N} ={\frac {(N+1)!} {N!}} (\frac {1} {10})(\frac {9} {10})^{N}=(N+1)(\frac {1} {10})(\frac {9} {10})^{N}$
And so we ask for the probability of B, given that A has been fulfilled:
$p(B/A) = \frac {P(B\bigcap A)} {P(A)} = \frac {\frac {1} {10}} {(\frac {9}{10})^N} =(\frac {10}{9})^N (\frac {1}{9})$
This seems to be wrong since $p > 1$ for large N.
Thanks in advance and sorry if i did not make the question clear enough. I do not know well the mathematics' definitions in English.
EDIT: I have corrected $P(B\bigcap A)$ and $p ( X=1)$.
 A: 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 $$\frac{9^N}{9^N\cdot 10}=\frac1{10}.$$
