How to check if all elements of 1 list are in the *same quantity* and in any order, in the list2?

Use collections.Counter to convert to a dict_items view Set of (value, count) pairs. Then you can use normal set operations.

from collections import Counter

def a_all_in_b(a, b):
    """True only if all elements of `a` are in `b` in the *same quantity* (in any order)."""
    return Counter(a).items() <= Counter(b).items()

Note that Counter works on hashable elements only, because it's a subclass of dict.

Modify this answer to Checking if list is a sublist to check for equality of occurences:

from collections import Counter 

list1 = [2,2,2,6]    
list2 =[2,6,2,5,2,4]

def same_amount(a,b):
    c1 = Counter(a)
    c2 = Counter(b)

    for key,value in c1.items():
        if c2[key] != value:
            return False
    return True


print(same_amount(list1,list2))
print(same_amount(list1 + [2],list2))

Output:

True
False

There is almost no transfere-knowledge needed to create this answer, thats why I suggested it as dupe. This question is simply a more specific case of what Checking if list is a sublist discussed.