Question: On a site that I operate I use several layers of protection for our subscription content. Among the other protections, each content item is placed into a folder with a randomly generated 19-character name. That name can include numbers, lowercase letters, and uppercase letters. 

While it will not really lower the benefit provided by this layer of protection for a second file to end up in a folder, I am curious about the chances of that happening. That leads me to two related questions: 

1. Assuming that we have 2500 files/folders, what is the chance that the next folder we generate will be a repeat?
2. How many folders must exist for the chance of repetition to rise above 1%?

Answer: 
Let's say you generate $K$-character folder names from an alphabet consisting of $A$ different symbols, and you currently have $N$ *distinctly named* folders.

The chance of a particular folder name being generated is $1 / A^K$. The chance that your next folder name matches one that already exists therefore $N/A^K$.

For this to be above a particular value $p$, you need $N > pA^K$.

For you, you have $K=19$, $A=62$ and $N=2500$, so the chance of a repeat is $2500/62^{19}$, whcih is around 1 part in $5\times 10^{30}$, i.e. vanishingly small.

For there to be a greater than 1% chance of a collision, you need $N > 0.01 \times 62^{19} = 10^{32}$ folders.

Note that since each folder name takes up around 20 bytes (19 characters plus a null character), you would need in excess of $10^{24}$ GB of disk space just to store the names of this many folders, never mind their contents.
There are a total of $62^{19} \approx 1.136 \times 10^{34}$ possible 19-character case-sensitive alphanumeric strings.

1. Thus, assuming you have 2500 distinct folders already, the chance of the next folder generated being a repeat is $\frac{2500}{1.136 \times 10^{34}} \approx 2.200 \times 10^{-31} $.
2. (a) If you are asking what the probability of *any* two folder names colliding, then
this is a classic [birthday problem](http://en.wikipedia.org/wiki/Birthday_problem). The probability that given $n$ folders, there is at least one repeat, is given by $$ p(n) = 1 - \frac{n! \cdot {62^{19} \choose n}}{62^{19n}} ,$$
but this is expensive to compute, so we approximate by $$ p(n) \approx 1 - e^{-n(n-1)/(2 \times 62^{19})}.$$
Solving for $p(n)=0.01$ gives $n \approx 1.5 \times 10^{16}$ folders required.

(b) If, on the other hand, you are asking how many folders are needed for the probability of the *next* folder colliding to be at least 0.01, then Chris Taylor's answer covers that superbly.