[Proposal] Diceware multilingual mixing

[DonaldTsang]

Have an option that allows mixing Japanese words and English words etc.

[grempe]

My intuition is that this does not help that much, and you'd be better served adding one additional word or a few symbols to gain the same entropy advantage. Here's what claude says on the topic, which matches my thoughts.

Good question. It depends on what the attacker knows.

  If the attacker knows which list each word comes from (e.g., the pattern is always EN, JP, EN, JP...):

  No entropy change. Each word is still drawn from a pool of 7,776, so it's still log2(7776) ≈ 12.925 bits/word. The total entropy for N words is still N ×
  12.925.

  If the attacker doesn't know which list each word came from, and the lists have no overlapping words:

  Each word position could be from any of K lists, so the effective pool is K × 7,776. Entropy per word becomes:

  log2(K × 7776) = log2(7776) + log2(K)

  With 2 lists: 12.925 + 1 = 13.925 bits/word (+1 bit)
  With 4 lists: 12.925 + 2 = 14.925 bits/word (+2 bits)
  With 22 lists (all of ours): 12.925 + 4.46 = ~17.4 bits/word

  The practical problem: this only works if the attacker can't identify which list a word came from. But "correct sushi mountain 桜" makes it obvious which
  words are English and which are Japanese. Even within the same language, word frequency and style differ between lists. So in practice the attacker can
  usually narrow each word to its source list, and the entropy gain collapses back to zero.

  The honest answer: using multiple lists doesn't meaningfully increase entropy. Adding one more word to a single list gives you a full 12.925 bits — far more
  than the theoretical best case of mixing lists.

So, if you assume the attacker has access to all of the word lists, just looking at the words can tell you which list they probably came from, removing any advantage. A six/seven passphrase from a single word list will give you all the entropy you likely need against all attackers.