Modify Levenshtein-Distance to ignore order

I'm looking to compute the the Levenshtein-distance between sequences containing up to 6 values. The order of these values should not affect the distance.

How would I implement this into the iterative or recursive algorithm?

Example:

# Currently 
>>> LDistance('dog', 'god')
2

# Sorted
>>> LDistance('dgo', 'dgo')
0

# Proposed
>>> newLDistance('dog', 'god')
0

'dog' and 'god' have the exact same letters, sorting the strings before hand will return the desired result. However this doesn't work all the time:

# Currently 
>>> LDistance('doge', 'gold')
3

# Sorted
>>> LDistance('dego', 'dglo')
2

# Proposed
>>> newLDistance('doge', 'gold')
1

'doge' and 'gold' have 3/4 matching letters and so should return a distance of 1. Here is my current recursive code:

def mLD(s, t):
    memo = {}
    def ld(s, t):
        if not s: return len(t)
        if not t: return len(s)
        if s[0] == t[0]: return ld(s[1:], t[1:])
        if (s, t) not in memo:
            l1 = ld(s, t[1:])
            l2 = ld(s[1:], t)
            l3 = ld(s[1:], t[1:])
            memo[(s,t)] = 1 + min(l1, l2, l3)
        return memo[(s,t)]
    return ld(s, t)

EDIT: Followup question: Adding exceptions to Levenshtein-Distance-like algorithm

You don't need the Levenshtein machinery for this.

import collections
def distance(s1, s2):
    cnt = collections.Counter()
    for c in s1:
        cnt[c] += 1
    for c in s2:
        cnt[c] -= 1
    return sum(abs(diff) for diff in cnt.values()) // 2 + \
        (abs(sum(cnt.values())) + 1) // 2   # can be omitted if len(s1) == len(s2)