How can I speed up my calculations with loops? Python

Question

I wrote this code. But it works very slowly.

I'm figuring out how many times I have to run the case generator to find numbers less than or equal to inv, in this case six. I count the number of attempts until a digit <= 6 is generated. I find inv equal to 1 and repeat the loop. Until inv is 0. I will keep trying to generate six digits <= 6.

And I will repeat all this 10 ** 4 degrees again to find the arithmetic mean.

Help me speed up this code. Works extremely slowly. The solution should be without third-party modules. I would be immensely grateful. Thank!

import random

inv = 6

    def math_count(inv):
        n = 10**4
        counter = 0
        while n != 0:
            invers = inv
            count = 0
            while invers > 0:
                count += 1
                random_digit = random.randint(1, 45)
                if random_digit <= invers:
                    invers -= 1
                    counter += count
                    count = 0
    
            if invers == 0:
                n -= 1
                invers = inv
        
        return print(counter/10**4)

math_count(inv)

Answer 1

The code is very hard to follow. Here's a possible refactor to make its intent clearer. Some key ideas in the simplification: (1) use loops to manage counters rather than manually incrementing/decrementing; (2) use more meaningful variable names when context isn't clear; and (3) delegate some of the complexity to a helper function. The third idea is the most impactful: the helper function is simple because it does so little; and the main function, having been relieved of a burden, is simpler as well -- so simple, in fact, that it boils down to computing a sum over two loops.

def math_count(inv, trials = 10000):
    total_attempts = sum(
        attempts_to_get_lte(x)
        for _ in range(trials, 0, -1)
        for x in range(inv, 0, -1)
    )
    return total_attempts / trials

def attempts_to_get_lte(x):
    attempts = 0
    while True:
        attempts += 1
        if random.randint(1, 45) <= x:
            return attempts

But attempts_to_get_lte() is an easily solved probability problem: rather than have Python simulate the manual flipping of coins, we can just do a little math. If you prefer that approach, I believe the following is correct (probability experts should chime in if I've gone astray):

def math_count_exact(inv):
    return sum(45 / x for x in range(inv, 0, -1))