text: user6787: I am reading a book about RNAseq analysis and it says
  
  "To calculate the probability that a read will map to a specific gene, we can assume an average gene size of 4000 nt (100 M nt divided by 25,000 genes). At 30 M reads equivalent to 30× coverage, at single read 100 nt (or paired-end read 50 nt) length, we can expect a single read to map to the average expressed and length gene 4000 nt× 30 coverage/100 nt 1200 times. Thus, if the gene is expressed at a level of 1/1200 compared to the average gene, then we have a 50:50 probability to have a read map to it."
  
  I understand that a single read would map to the average expressed 1200 times but I don't understand how the 50:50 probability is calculated. Please could you explain how this probability is determined.
  
  user5077: I actually disagree with that.. I guess I write this down as a discussion.
  The expected value for an average gene is 1200. For this gene at 1/1200 expression, you expect 1 read.
  However, because of sampling, sometimes you get 0, 1 or 2. If we assume the sampling follows a poisson distribution, we can calculate the probability of getting 0, 1 , 2 ... reads:
  plot(dpois(0:10,lambda=1),ylab="Probability",xlab="No of reads")
  
  
  The probabilities are like:
  data.frame(n=0:5,p=dpois(0:5,lambda=1))
  
    n           p
  1 0 0.367879441
  2 1 0.367879441
  3 2 0.183939721
  4 3 0.061313240
  5 4 0.015328310
  6 5 0.003065662
  
  The probability of getting exactly 0 or 1 read should be ~0.37. The probability of getting at least 1 read would be 1 - P(n=0) = 1- 0.367879441 = 0.6321206
  
original_text: 6787: <p>I am reading a book about RNAseq analysis and it says</p>
  <blockquote>
  <p>&quot;To calculate the probability that a read will map to a specific gene, we can assume an average gene size of 4000 nt (100 M nt divided by 25,000 genes). At 30 M reads equivalent to 30× coverage, at single read 100 nt (or paired-end read 50 nt) length, we can expect a single read to map to the average expressed and length gene 4000 nt× 30 coverage/100 nt 1200 times. Thus, if the gene is expressed at a level of 1/1200 compared to the average gene, then we have a 50:50 probability to have a read map to it.&quot;</p>
  </blockquote>
  <p>I understand that a single read would map to the average expressed 1200 times but I don't understand how the 50:50 probability is calculated. Please could you explain how this probability is determined.</p>
   5077: <p>I actually disagree with that.. I guess I write this down as a discussion.</p>
  <p>The expected value for an average gene is 1200. For this gene at 1/1200 expression, you expect 1 read.</p>
  <p>However, because of sampling, sometimes you get 0, 1 or 2. If we assume the sampling follows a poisson distribution, we can calculate the probability of getting 0, 1 , 2 ... reads:</p>
  <pre><code>plot(dpois(0:10,lambda=1),ylab=&quot;Probability&quot;,xlab=&quot;No of reads&quot;)
  </code></pre>
  <p><a href="https://i.stack.imgur.com/R3q8V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R3q8V.png" alt="enter image description here" /></a></p>
  <p>The probabilities are like:</p>
  <pre><code>data.frame(n=0:5,p=dpois(0:5,lambda=1))
  
    n           p
  1 0 0.367879441
  2 1 0.367879441
  3 2 0.183939721
  4 3 0.061313240
  5 4 0.015328310
  6 5 0.003065662
  </code></pre>
  <p>The probability of getting exactly 0 or 1 read should be ~0.37. The probability of getting at least 1 read would be 1 - P(n=0) = 1- 0.367879441 = 0.6321206</p>
  