On the probability of a collision in RFO-rCTR My slides show that assuming a polynomial number $q(n)$ of blocks have been encrypted by a random function oracle in the context of a CPA-game, the probability of having a collision in the input to the oracle is bounded by $\frac{q(n)^2}{2^{l(n)}}$ (cfr. Katz and Lidell, 2nd edition pages 93-94).

Then I am proposed the following: assuming that in total $e(n)$ messages are encrypted each consisting of at most $s(n)$ blocks then the probability of a collission in the input is bounded by $\frac{e(n)^2s(n)}{2^{l(n)}}$.

Is this a typo? Why are we dropping the square from $s(n)$?

Here you may see the slides I refer to (see slide 18 and 19)
The key idea here is the blocks within the same message do not collide—message $1$ will have no possible block collisions, message $2$ can collide with the message $1$'s $s(n)$ blocks, message $i$ can collide with message $i-1$'s $s(n)$ blocks or message $i-2$'s $s(n)$ blocks, etc.

In other words, the number of possibly colliding block combinations is given by
$$
\sum_{i=1}^{e(n)} (i - 1)s(n)\,,
$$
which simplifies to $s(n) {e(n) \choose 2}$. We turn this into a probability $$ \frac{s(n){e(n) \choose 2}}{2^{l(n)}} \le \frac{s(n)e(n)^2}{2^{l(n)}}\,.$$