I had a hell of a time figuring out how to cache dynamic pages accessed by a URI scheme like this:

domain.com/admin/page/1
domain.com/admin/page/2
domain.com/admin/page/3


Now the problem is that the appcache won't cache each individual admin/page/... unless you visit it.

What I did was use the offline page to represent these pages that you may want to allow a user to access offline.

The JS in the offline page looks up the URI and scrapes it to find out which page it should show and fetches the data from localStorage which was populated with all the page data when the user visited the admin dashboard before being presented with the links to each individual page.

I'm open to other solutions but this is all I could figure out to bring a bunch of separate pages offline with only visiting the single admin page.

Appcache for dynamic site
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 $p ( X=1)$.
