Does not enter the loop even though the conditions are met. Where is the mistake?

Question

Python

import random
import sys


def print_mimic(mimic_dict, word):
    bred = []
    x = 0
    while x < 200:
        if x == 0:
            first = word                                                          ### В это части цикла всё хорошо
            dobav = random.choice(mimic_dict[first]) 
            bred.append(dobav)
            x += 1
        else:
            print (dobav in mimic_dict)
            if dobav in mimic_dict == True:                              ### А вот тут проблема. Строчка 
                dobav = random.choice(mimic_dict[dobav])      ### выше выдает True но почему  то в цикл if  
                bred.append(dobav)                                          ### не переходит, а соскакивает 
                x +=1                                                                 ### в else. Как такое 
            else:                                                                ### может быть? Спасибо
                dobav = random.choice(mimic_dict[" "])
                bred.append(dobav)
                x += 1



    print (bred)
    return bred
    
mimic_dict = {" " : ["A", "B" , "C"], 
                      "A" : ["1", "2" "3"],
                      "B" : ["4", "5" "6"], 
                      "C" : ["7", "8" "9"]   }
print_mimic(mimic_dict, " ")

Answer 1

The priority of operations must be taken into account, apparently:
it will give False, since (1, 2, 3) == True will be executed first, which will return False, and then the check 1 in False will turn out, which is False.
And this is correct: it
will give True.
And this is even more correct:

if 1 in (1, 2, 3):
    pass

Answer 2

There are a lot of strange places in the source code. For example, the commas in the dictionary definition are omitted in a strange and dangerous way, the dangerous use of the dobav variable in the else branch... This all leads to unsafe code in terms of understanding and future modifications. There is an obvious lack of experience.
You have already been answered about the precedence of operations, but let's try to refactor your code step by step for clarity.
This is what your function body is equivalent to:

def print_mimic(mimic_dict, word):
    bred = []
    x = 0
    dobav = word
    while x < 200:
        if x == 0 or dobav in mimic_dict:
            dobav = random.choice(mimic_dict[dobav])
        else:
            dobav = random.choice(mimic_dict[" "])
        bred.append(dobav)
        x += 1
    print (bred)
    return bred

I have no idea what this code is and what it is for. I made the previous refactoring leaving the code equivalent to the original (with the exception of fixing obvious errors). It can be assumed that there are a number of logical errors in the code. Perhaps the behavior on the "x == 0" branch should not differ from the behavior on the else branch of the "add in mimic_dict" condition, and word should have been put instead of the constant " ". Then the code becomes much simpler, clearer and more elegant.

def print_mimic(mimic_dict, word):
    bred = []
    dobav = word
    for x in range(200):
        dobav = random.choice(mimic_dict.get(dobav, mimic_dict[" "])) 
        bred.append(dobav)

    print (bred)
    return bred

Answer 3

if dobav in mimic_dict == True:
>>> False
if (dobav in mimic_dict) == True:
>>> True

Answer 4

if dobav in mimic_dict == True:

And that's it.