Review

In class, we were computing the probability of a birthday collision for a group of 30 people.

Consider the following experiment: Extract one person from the population and insert in the group. That person will have some birthday.

Then insert a second person. The chance of a birthday collision is $1/365$ (the second person would have to have the exact same b'date as the first). The chance of no-collision is $364$. In other terms, there are $365$ pairs of people with same birthday, and $365\cdot 364$ pairs of people with different birthdays. (The second person must have a birthday different from the first).

Insert a third person. Now, there are three people. Of all possible b'date configurations of three people taken from the general population, $365 \cdot 364 \cdot 363$ would have different birthdays (the second would have to be different from the first AND the third from the other two).

The ratio between configurations without collisions and all configurations for 30 people is therefore:

$$\frac{365\cdot 364\cdot 363\cdot\ldots\cdot336}{365^{30}} = \frac{365!}{(365-30)! 365^{30}}$$

Now, let's generalize it for hash functions of hash-length $\log t$. After computing $k$ hashes, the possibility of having no collisions is

$$\frac{t!}{(t-k)! t^k}$$ (1)