Distributed probability random number generator

c# random probability probability-theory
Mark Conway picture Mark Conway · Mar 31, 2012 · Viewed 19.2k times · Source

I want to generate a number based on a distributed probability. For example, just say there are the following occurences of each numbers:

Number| Count           
1    |  150                
2    |  40          
3    |  15          
4    |  3  

with a total of (150+40+15+3) = 208     
then the probability of a 1 is 150/208= 0.72    
and the probability of a 2 is 40/208 = 0.192

How do I make a random number generator that returns be numbers based on this probability distribution?

I'm happy for this to be based on a static, hardcoded set for now but I eventually want it to derive the probability distribution from a database query.

I've seen similar examples like this one but they are not very generic. Any suggestions?

Answer

Adam Zalcman picture Adam Zalcman · Mar 31, 2012

The general approach is to feed uniformly distributed random numbers from 0..1 interval into the inverse of the cumulative distribution function of your desired distribution.

Thus in your case, just draw a random number x from 0..1 (for example with Random.NextDouble()) and based on its value return

  • 1 if 0 <= x < 150/208,
  • 2 if 150/208 <= x < 190/208,
  • 3 if 190/208 <= x < 205/208 and
  • 4 otherwise.

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