Creating a random list of 4 letters in which each letter appears 15 times and does not repeat twice in a row

I would like to create a random list of 4 letters, with the letters 'A', 'B', 'C' and 'D', where each letter appears 15 times and the same letter is not repeated twice in a row.

I tried the code below but the letters repeated twice in a row appear in the result:

import random

def create_random_list():
    try:
        letters = ['A', 'B', 'C', 'D']
        freq = 15
        
        lst = []
        for letter in letters:
            lst += [letter] * freq
        
        random.shuffle(lst)
        
        # Check that no letter appears twice in a row
        for i in range(1, len(lst)):
            if lst[i] == lst[i-1]:

                # If a letter appears twice in a row, shuffle the list again
                random.shuffle(lst)
                break
        
        return lst
    except Exception as e:
        print(f"Error: {e}")
        return []

print(create_random_list())

In principle, you can fix your code like this. You don't know that the first reshuffle will produce a correct result. So you need to keep shuffling:

shuffle = True
while shuffle:
    random.shuffle(lst)
    
    # Check that no letter appears twice in a row
    for i in range(1, len(lst)):
        if lst[i] == lst[i-1]:
            break
    else:  # yes: this is correct. Check for-else in Python
        shuffle = False

But my guess is that this might take forever given your parameters. Probably only a tiny fraction of possible permutations has no repetitions. So you probably need some generative approach, most likely some backtracking algorithm that picks random next elements based on the previous ones.

The issue with your current code is that you're checking for consecutive letters that are the same after shuffling the list, but you're not guaranteeing that the initial shuffled list satisfies the condition. To solve this, you can modify your code as follows:

import random

def create_random_list():
    try:
        letters = ['A', 'B', 'C', 'D']
        freq = 15
    
        lst = []
        for letter in letters:
            lst += [letter] * freq
    
        while True:
            random.shuffle(lst)
        
        # Check that no letter appears twice in a row
            for i in range(1, len(lst)):
                if lst[i] == lst[i-1]:
                    break
            else:
                # No consecutive letters found, so the list is valid
                return lst
    
    except Exception as e:
        print(f"Error: {e}")
        return []

print(create_random_list())

This modified code uses a while loop to keep shuffling the list until a valid arrangement is found where no letter appears twice in a row. The for loop checks for consecutive letters, and if it finds any, it breaks out of the loop and shuffles the list again. If the loop completes without finding any consecutive letters, it means the list is valid, and it is returned

You'd need a pool of letters and take a random letter from this pool. If it is different than the last taken letter, then add it to the result. If it is the same, then re-add the letter back to the pool.

You have to make sure though, that never there's a letter that's more than a number of other letters + 1 in the pool (e.g., count(A)=15, but count(BCD)=13), or you will get an infinite loop.