Create random numbers with left skewed probability distribution

This is the answer you are looking for using the SciPy function 'skewnorm'. It can make any positive set of integers either left or rightward skewed.

from scipy.stats import skewnorm
import matplotlib.pyplot as plt

numValues = 10000
maxValue = 100
skewness = -5   #Negative values are left skewed, positive values are right skewed.

random = skewnorm.rvs(a = skewness,loc=maxValue, size=numValues)  #Skewnorm function

random = random - min(random)      #Shift the set so the minimum value is equal to zero.
random = random / max(random)      #Standadize all the vlues between 0 and 1. 
random = random * maxValue         #Multiply the standardized values by the maximum value.

#Plot histogram to check skewness
plt.hist(random,30,density=True, color = 'red', alpha=0.1)
plt.show()

Please reference the documentation here: https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.skewnorm.html

Histogram of left-skewed distribution

The code generates the following plot.

Like you described, just make sure your skewed-distribution adds up to 1.0:

pers = np.arange(1,101,1)

# Make each of the last 41 elements 5x more likely
prob = [1.0]*(len(pers)-41) + [5.0]*41

# Normalising to 1.0
prob /= np.sum(prob)

number = np.random.choice(pers, 1, p=prob)

The p argument of np.random.choice is the probability associated with each element in the array in the first argument. So something like:

    np.random.choice(pers, 1, p=[0.01, 0.01, 0.01, 0.01, ..... , 0.02, 0.02])

Where 0.01 is the lower probability for 1-59 and 0.02 is the higher probability for 60-100.

The SciPy documentation has some useful examples.

http://docs.scipy.org/doc/numpy-dev/reference/generated/numpy.random.choice.html

EDIT: You might also try this link and look for a distribution (about half way down the page) that fits the model you are looking for.

http://docs.scipy.org/doc/scipy/reference/stats.html