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One of the popular cold storage wallets provides double-sided tiles for users to put in the first four letters of each of their 24 words (96 characters total).
They include ~245 double sided tiles, plus some blanks.
By my math on the BIP39 wordlist, they should need to provide ~485 double sided tiles to provide for every valid 24 word mnemonic.
I simply calculated, for each letter, the worst case 24 word combination, then added up how many of each letter I would need, then divided by 2 since the tiles are double sided.
Is there something I'm missing here?
Thanks!
EDIT I have found at least one flaw in my reasoning. The last word is a checksum, so not entirely random. But accounting for this only saves us 33 letter tiles!
Perhaps you'd like to explain how you calculated the "worst case" combination - it would be easier to check your calculation than to try and redo it. There may also be an issue with double sided tiles, since not all combinations can be used together. – Nate Eldredge – 2018-02-12T03:54:14.297
But for practical purposes, you don't need to be able to handle all possible mnemonics, so long as you can handle a pretty large percentage (say 99.99%). I would bet the 245 provided tiles can do that. If you are unlucky and get one that doesn't work, you can have your wallet generate a different one and send all your coins there. – Nate Eldredge – 2018-02-12T03:56:17.190
Worst case combo was calculated by, with the letter A for example, finding the 24 words in the list with the most As in the first four letters (since one only needs to store the first four letters). There are 37 words with 2 As in the first four letters, and none with 3 As, so the worst case for A is 48 As in a seed. – LoyQrtlomd – 2018-02-12T04:00:05.263
Cutting off the long tails by only providing for some large majority is interesting, but I can't help but think that there is something more I'm missing. – LoyQrtlomd – 2018-02-12T04:01:38.827
No, I think your logic is fine. It's clear that 245 tiles can't account for all possible mnemonics. – Nate Eldredge – 2018-02-12T04:12:32.333
I have found at least one flaw in my reasoning. The last word is a checksum, so not entirely random. But accounting for this only saves us 33 letter tiles!
(also added as an edit to original post) – LoyQrtlomd – 2018-02-12T19:06:46.733