How many reads do I need to cover the entire genome?
Let's see what your slides are claiming $\epsilon$ to be, using:

$$ C = \frac{NL}{G} \approx \ln \left( \frac{G}{L\epsilon} \right) $$

We can rearrange to get

$$ \epsilon \approx \frac{G}{L} \times e^{-C} $$

Now, what does it mean for an entire genome to be covered? Having a clear understanding of this definition is crucial, and your misunderstanding seems to partially stem from this.

I claim it's natural to consider a genome as consisting of $G/L$ bins. In this setting, we have $N$ balls that each randomly goes into a bin. Then, the entire genome is covered when all bins contain at least $1$ ball/read.

From the Poisson distribution, we know that P(a bin has 0 balls) = $e^{-C}$.

Then, we define $\epsilon$ to be the probability that at least one of the $G/L$ bins has 0 balls. Given the way we've modeled the problem, this is exactly equal to 

$$ \epsilon = 1 - (1 - e^{-C})^{G/L}. $$

From above, your slides are claiming $ \epsilon \approx \frac{G}{L} \times e^{-C}$. This might be strange, as it appears to be equal to the expected number of bins with 0 balls, instead of the probability that there is at least 1 bin with 0 balls. Furthermore, it's strange because it contains only one exponential, compared to our exact derivation which has a double exponential.

The mismatch in the number of exponential is a hint that another mathematical approximation is at play. We can use a first-order approximation of the binomial series to obtain

$$ (1-e^{-C})^{G/L} \approx 1 - (G/L)e^{-C} $$ which holds particularly when $|e^{-C}| \times \frac{G}{L} &lt;&lt; 1$. 

In your case, $\frac{G}{L} = 30,000$, so the approximation only really starts getting accurate when $C &gt; 15$ or so.

Following through with our approximation, we get

$$ \epsilon \approx 1 - (1 - (G/L) e^{-C}) $$

$$ \epsilon \approx \frac{G}{L} \times e^{-C}. $$