generating random numbers in python with percentage function on what selected value in python

Question

Lets say I wanted to generate a random number in python using something like random.randint(1,100) but how would I go about making Python want to tend toward selecting higher random numbers? So like if i wanted 80% of numbers above 50 to be chosen with 20% of numbers below 50 to be chosen. like a casino slot machine probability. (we all know there not truly random but are set with probability).

I have tried

import random

random.randint(1,100)
select % above 50

Answer 1

Try using a weighted list, like this:

import random
my_list = ['A'] * 5 + ['B'] * 5 + ['C'] * 90
random.choice(my_list)

See Python Weighted Random. To make it twice as likely that a number from 51 to 100 will be chosen instead a number from 1 to 50, do this:

import random

my_list = []

for i in range(1, 100):
    my_list.append(i)
    if i > 50:
        my_list.append(i)

print my_list
print 'Random (sort of) choice: %s' % random.choice(my_list)

Answer 2

Use a non-uniform distribution such as the triangular distribution, set the mode equal to the range max, and cast to int or round it.