text: user2862: Suppose my genome is 3 million bases and that my reads are 100 nucleotide long. I need to know how many reads I need to cover the entire genome.
  I start from using the equation $C = \frac{N \cdot L}{G}$ where C is the coverage, N the number of reads, L the length of a read and G the length of the haploid genome. I also know that the average coverage follows a Poisson distribution, so I need this for C.
  On my slides, in fact, I have the following:
  $$
  C = \frac{N \cdot L}{G} \approx ln \left( \frac{G}{L \cdot \epsilon}\right),
  $$
  if we assume that this is the average coverage needed to cover the entire genome with probability $1-\epsilon$.
  The problem is that I do not understand how to arrive at the expression on the right. I thought that if we want to be sure to have a coverage of at least 1 on average, we would have had to compute $1 - P(X=0) = 1 - e^{-\frac{N \cdot L}{G}} = 0.99$, if we take $\epsilon=0.01$. But I do not get the same result as with the equation above.
  
  user3043: 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 $|e^{-C}| \times \frac{G}{L} << 1$. 
  In your case, $\frac{G}{L} = 30,000$, so the approximation only really starts getting accurate when $C > 15$ or so.
  Following through with our approximation, we get
  $$ \epsilon \approx 1 - (1 - (G/L) e^{-C}) $$
  $$ \epsilon \approx \frac{G}{L} \times e^{-C}. $$
  
original_text: 2862: <p>Suppose my genome is 3 million bases and that my reads are 100 nucleotide long. I need to know how many reads I need to cover the entire genome.</p>
  
  <p>I start from using the equation $C = \frac{N \cdot L}{G}$ where C is the coverage, N the number of reads, L the length of a read and G the length of the haploid genome. I also know that the average coverage follows a Poisson distribution, so I need this for C.</p>
  
  <p>On my slides, in fact, I have the following:</p>
  
  <p>$$
  C = \frac{N \cdot L}{G} \approx ln \left( \frac{G}{L \cdot \epsilon}\right),
  $$</p>
  
  <p>if we assume that this is the average coverage needed to cover the entire genome with probability $1-\epsilon$.</p>
  
  <p>The problem is that I do not understand how to arrive at the expression on the right. I thought that if we want to be sure to have a coverage of at least 1 on average, we would have had to compute $1 - P(X=0) = 1 - e^{-\frac{N \cdot L}{G}} = 0.99$, if we take $\epsilon=0.01$. But I do not get the same result as with the equation above.</p>
   3043: <p>Let's see what your slides are claiming $\epsilon$ to be, using:</p>
  
  <p>$$ C = \frac{NL}{G} \approx \ln \left( \frac{G}{L\epsilon} \right) $$</p>
  
  <p>We can rearrange to get</p>
  
  <p>$$ \epsilon \approx \frac{G}{L} \times e^{-C} $$</p>
  
  <p>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.</p>
  
  <p>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.</p>
  
  <p>From the Poisson distribution, we know that P(a bin has 0 balls) = $e^{-C}$.</p>
  
  <p>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 </p>
  
  <p>$$ \epsilon = 1 - (1 - e^{-C})^{G/L}. $$</p>
  
  <p>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.</p>
  
  <p>The mismatch in the number of exponential is a hint that another mathematical approximation is at play. We can use a <a href="https://en.wikipedia.org/wiki/Binomial_series" rel="nofollow noreferrer">first-order approximation of the binomial series</a> to obtain</p>
  
  <p>$$ (1-e^{-C})^{G/L} \approx 1 - (G/L)e^{-C} $$ which holds particularly when $|e^{-C}| \times \frac{G}{L} &lt;&lt; 1$. </p>
  
  <p>In your case, $\frac{G}{L} = 30,000$, so the approximation only really starts getting accurate when $C &gt; 15$ or so.</p>
  
  <p>Following through with our approximation, we get</p>
  
  <p>$$ \epsilon \approx 1 - (1 - (G/L) e^{-C}) $$</p>
  
  <p>$$ \epsilon \approx \frac{G}{L} \times e^{-C}. $$</p>
  