Question

I want to generate a number based on a distributed probability. For example, just say there are the following occurrences 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 random 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 1

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.