Theorem: Choose $Q$ random natural numbers in the set $\{1,2, \dots, M\}.$ The probability of getting at least one collision is
$$P_C(Q) = 1 - \frac{M - (Q - 1)}{M} P_{\neg C}(Q-1).$$
Notation: By $P_C$, I mean the probability of getting a collision. By $P_{\neg C}$ I mean the probability of not getting a collision.
Remark: This is the birthday problem.
Remark: So $P_C(Q)$ is just being computed by using its complement. The reason I express the theorem this way is because its induction proof relates directly to it being enunciated this way.
Theorem: $$P_C(Q) \approx 1 - e^{-\tfrac{(Q-1)Q}{2M}}.$$
Proof: We know $$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \ldots$$
If we take the two terms of this expansion, we get $e^{-x} \approx 1 - x$. Then
\begin{align} P_{\neg C}(Q)&= \prod_{i=1}^{Q-1} \left(1 - \dfrac{i}{M}\right)\\ &\approx \prod_{i=1}^{Q-1} e^{-i/M} \\ &= e^{-1/M} e^{-2/M} \dotsc\ e^{-(Q - 1)/M} \\ &= e^{-\sum_{i=1}^{Q-1} i/M} \\ &= e^{-\dfrac{1}{M} (Q-1)Q/2}\\ &= e^{-\dfrac{(Q-1)Q}{2M}}, \end{align}
So $P_C(Q) \approx 1 - P_{\neg C}(Q),$ as desired.
Question: How can I (at least) have some notion of how off the right number I am if I use this estimation to compute the probability of getting a collision in a concrete case?