I want to generate a random integer within a range but on average I want it to be biased toward a particular number. For example, I want to pick a number from 0-50 N times and the average of all the chosen random numbers should be around 20. I tried to use the random.normalvariate function in python and casting it from float to int but I'm not sure if it's a proper solution. I set the mean to 20 but I am not sure what the standard deviation should be in order to make sure that on average we pick 20 from the range.

r = int(random.normalvariate(20, st_dev???)) 
r = 50 if r > 50 else r # clamp

CodePudding user response：

There are several choices, but one discrete random distribution that satisfies the properties is binomial distribution. In particular, a binomial distribution with parameters n=50 and p=0.4 is distributed on {0, ..., 50} and has mean 20 (because n * p = 20). Also its distribution is close to a normal distribution (loosely by central limit theorem).

from scipy.stats import binom

r = binom.rvs(50, 0.4, size=1000)
print(r) # [20 16 24 24 18 21 21 25 27 22 22 22 20 18 23 ... (a sample)
print(sum(r) / len(r)) # 20.022 (sample mean)

CodePudding user response：

If a mean of 20 is the only requirement for pseudo-random integers generated from 1 to 50, this simple generator based on random.random() will meet it:

        import random
        def biased_uniform(n, invert_threshold):
            u = random.random()
            if u > 0.5 and random.random() > invert_threshold:
                u = 1 - u
            v = int(50 * u   0.5)   1
            v = 1 if v == 0 else v
            return v
        total, n = 0, 1000000
        for i in range(n):
            v = biased_uniform(50, 0.52)
            total  = v
        mean = total / n
        print(f"mean {mean}")

You can tune the mean by adjusting the invert_threshold in the semi-open interval [0.0, 1.0).