Probability approximation of no collisions I am comparing the actual probability of no collisions to the probability approximation formula of no collisions from the Understanding Cryptography text. 

The approximation is as follows:

$$P(\text{no collisions}) = \frac{e^{-t(t-1)}}{2 \cdot 2^n}$$

I know that t represents the number of elements but I am confused about the denominator. The text says that $n$ is the output width of $h()$. I don't really understand this. Does this mean that for lets say the Birthday Paradox, $n  = 365$? So the denominator would be $2 \cdot 2^{365}$?
For a hash function with a 128 bit output, n=128. Such a function has $2^{128}$ possible outputs.

For the birthday paradox you have 365 different outputs, i.e. $2^n = 365$ or $n = \log_2 {365} \approx 8.5$.

You'll also find alternative versions of this formula, which omit the 2^ part in the formula: $\frac{e^{-t(t-1)}}{2 \cdot N}$. In that case you'd use $N=365$ for the birthday paradox and $N=2^{128}$ for the 128 bit hash.

If $t$ is small compared to $N=2^n$, the formula can be approximated as:

$$1-\frac{t(t-1)}{2}\frac{1}{
N}=1-\frac{1+2+3+...+(t-1)}{N}=1-(\frac{1}{N}+\frac{2}{N}+\frac{3}{N}+...+\frac{t-1}{N})$$

Which is intuitive, since the first value has no chance of collision, the second can collide with 1 existing value with probability $\frac{1}{N}$, the third can collide with 2 existing values with probability $\frac{2}{N}$, etc.