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 $p &gt; 1$ 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 $p ( X=1)$.


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