Probability for collisions of a one-way compression function Given a one-way compression function $h:\{0,1\}^n \rightarrow \{0,1\}^m$ and an attacker that picks $x_1 \ldots x_t \in \{0,1\}^n$ (uniformly distributed), I have to show that the probability to find a collision in the picked elements is smaller than $\epsilon$ if $m &gt; log(\frac{t^2}{2 \epsilon})$.
Assumption: $h$ is surjective and has the same number of inverse images for each image.

What I found out already:
Because a binary alphabet is used, I have to show that the probability is smaller than $\epsilon$ if there are at least $\frac{t^2}{2 \epsilon}$ elements that the compression function can map to. And because I want to find a collision between two elements, there are ${t \choose 2} \le \frac{t^2}{2}$ possible combinations for a collision pair in the set of picked elements.

Question:
How can I go on from here and how do I handle the $\epsilon$?
$\left|\hspace{.01 in}\operatorname{Range}(h)\hspace{.01 in}\right| \:$ is the number of elements that the compression function can map to.

If $\: m &gt; \operatorname{log}_{\hspace{.01 in}2}\left(\hspace{-0.03 in}\frac{t^2}{2\cdot \epsilon}\hspace{-0.04 in}\right) \:$ then $\;\; \left|\hspace{.01 in}\operatorname{Range}(h)\hspace{.01 in}\right| \: = \: 2^m \: &gt; \: 2^\left(\operatorname{log}_{\hspace{.01 in}2}\left(\hspace{-0.03 in}\frac{t^2}{2\cdot \epsilon}\hspace{-0.04 in}\right)\right) \: = \: \frac{t^2}{2\cdot \epsilon} \;\;\;\;$.