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How long time would it realistically take to crack each of following scenarios and how many combinations does each scenario consist of?

Mnemonic Phrase, it is not known if we are looking for a pass-phrase of 12, 13, 24 or 25 words

Scenario 1: There are total 90 words shown on a single sheet of paper.

Scenario 2: There are total 180 words shown on two sheets of paper (90 each). It is not known how many words (of pass-phrase) are on each sheet of paper (meaning one sheet could have 0 and the other all)

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You can calculate the number of word combinations for each possible pass phrase length and then add them together. For mnemonic phrases any order is valid, so it is a question of calculating the exponentiation of the number of words on the sheet with the number of words in the phrase.

Finally, there are two scenarios, which should be considered separately. For the first scenario there is just a single sheet of 90 words. For the second there are two sheets, but as the words do not need to be distributed over the sheets you can simply think of it as one sheet of 180 words.

Finally, if you want to know the number of bits of security you can simply count the number of digits, divide by 3 and then multiply with 10 to get a rough estimate (or perform $\log(c) / \log(2)$ where $c$ it the total number of combinations).

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  • $\begingroup$ Thank you very much Maarten, I will give this a go and see what it looks like. Cheers and wish you a nice evening. $\endgroup$
    – Anders
    Commented Mar 31, 2018 at 21:44
  • $\begingroup$ For the first scenario you should already go over 160 bits of security, which is way more than is possible to brute force. And just the 12 word version is just out of reach with 78 bits. $\endgroup$
    – Maarten Bodewes ♦
    Commented Mar 31, 2018 at 21:52
  • $\begingroup$ I'm not sure that I understand all the things you wrote down there. I presumed a phrase must always be entered in a specific order, and that words may be repeated. Neither PERMUT or COMBIN is exponentiation, e.g. 90 to the power of XX $\endgroup$
    – Maarten Bodewes ♦
    Commented Apr 7, 2018 at 18:41
  • $\begingroup$ Sorry, for confusion and now understand why - my mistake and mixed up. Now got same verified and came to the 77,9 bits for 12 word version and 162,2 bits for the 25 word version - just like u initially did. Wish you a nice weekend and appreciate all your kind help. Cheers, Anders $\endgroup$
    – Anders
    Commented Apr 8, 2018 at 11:49
  • $\begingroup$ Hey, no problem, glad you got it solved. If you like the answer you can accept it by clicking the V mark to the left hand side of it; it's the best way to say thanks on StackExchange. $\endgroup$
    – Maarten Bodewes ♦
    Commented Apr 8, 2018 at 20:11

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