A thought / question on seed storage devices with pre printed letters

[DaveF]

So something like this:


With the letters that come like this:



So, if you create a 12 word seed and only take out the 48 punched letters you need to save it.
AND someone gets a hold of those metal sheets AND they know how many of each letter was supposed to be there.

The question I have is how much security (if any) did you loose if someone was trying to brute force your seed.

I can't be the only person who thought of this, but I did a fair amount of googling and could not find an answer.
There has to be someone with the math skills that ran the numbers for a situation like this.

-Dave

Note: Self mod thread to keep out spammers and AI posters and such.

[satscraper]

Roughly.

n=48 (number of empty places in plates, each place is relevant to English character.

Number of permutations:

P_n = n! = 48!

The probability that each 4-letter block in every permutations corresponds to one word from BIP39 = 2048/26⁴ (2048 - number of words in BIP39 list, 26⁴-number of  permutation m out of k with reiteration, m=26 - number of English letters in alphabet, k=4

As 48-letter string has 12 4-characters blocks and probability of each block to correspond to BIP39 list doesn't depend on the relevant probability of any other block the whole probability to all 48-letters string to correspond BIP 39 lisst equals to 2048¹²/26^4x12 . (The probabilities of independent events are multiplied).

So the probability to find correct phrase would be (1/48!) x (2048^12//26^4x12) (without taking into account that the last word is the checksum.).

Taking the checksum factor into account reduces the resulting probability by roughly 16

Resulting number is still very and very small  to take the threat seriously.

P.S. I said roughly at the start since m‑out‑of‑k calculation ignores the restriction that 4 ^{and probably even 3} letters in 4-letter block cannot be the same.  So anyone is free to refine result by considering this factor.  

[Mrbluntzy]

The question I have is how much security (if any) did you loose if someone was trying to brute force your seed.

You have not lose so much security of your seed phrase but you just need to be aware that your security is not as strong as before, if I'm to rate your risk exposure, it's <1%, just imagine being asked to locate a specific grain of sand in a deserts.

I have seen this metal seed phrase storer before and a similar question has pop in my head then but when I asked about the possibility of brutforcing those seeds if the owner lose the metal sheet, I was told that my security is still very strong but not as strong as before. Your security is still strong because that metal sheet has the 48 missing punched holes and the attacker doesn't have the letters to those holes but he or she only have the sheet and only knows the position of those letters so it will be difficult to brutforce those seeds, meanwhile about 10 to 20 different letters can match each of the punched holes, so the computational power will have to start matching up 10 to 20 letters for each of the holes and we have 48 holes, so it will need to do those matching for the 48 holes  

So that will be 10letters multiplied by 48 holes, meanwhile a modern GPU (like I read) can compute only 10^9 hash/s, so doing 10^48 is really massive, the task is not going to be feasible.

Therefore your risk of exposure is still not too much, but just have it at the back of your mind that it's not 100% strong as before.

[DaveF]

Roughly.

n=48 (number of empty places in plates, each place is relevant to English character.

Number of permutations:

P_n = n! = 48!

The probability that each 4-letter block in every permutations corresponds to one word from BIP39 = 2048/26⁴ (2048 - number of words in BIP39 list, 26⁴-number of  permutation m out of k with reiteration, m=26 - number of English letters in alphabet, k=4

As 48-letter string has 12 4-characters blocks and probability of each block to correspond to BIP39 list doesn't depend on the relevant probability of any other block the whole probability to all 48-letters string to correspond BIP 39 lisst equals to 2048¹²/26^4x12 . (The probabilities of independent events are multiplied).

So the probability to find correct phrase would be (1/48!) x (2048^12//26^4x12) (without taking into account that the last word is the checksum.).

Taking the checksum factor into account reduces the resulting probability by roughly 16

Resulting number is still very and very small  to take the threat seriously.

P.S. I said roughly at the start since m‑out‑of‑k calculation ignores the restriction that 4 ^{and probably even 3} letters in 4-letter block cannot be the same.  So anyone is free to refine result by considering this factor.  

Figured as much. Just could not get the paranoid voice out of my head saying that it could in theory weaken security by a noticeable amount. Was wondering what the worst case scenario was.

It's also super simple to pop out another couple of dozen other letters and put in them in the trash someplace else so even if someone did find the plates they would have no idea what letters were used.


-Dave

[satscraper]

It's also super simple to pop out another couple of dozen other letters and put in them in the trash someplace else so even if someone did find the plates they would have no idea what letters were used.


-Dave

Yeah, it is exactly what I thought after reading OP. To calm yourself, you might want to make the task harder at any time by removing some extra letters engraved on the top of squares held by plates. But even if they are not removed, and someone is lucky enough to find the plates with 48 holes, the maximum they can do is use the remaining letters for their own needs.

[Cricktor]

It's just that this particular metal backup design is terrible when exposed to harsh conditions (and that's why we want to have a resilient metal backup to withstand such). See Jameson Lopp's test of this (expensive) crap.

When it warps due to high temperatures and maybe some pressure (fire, collapsing building parts) your small character plates will go all over the place. Good luck to reassemble your mnemonic recovery word fragments (you have a redundant backup at some other place, haven't you?).

Your metal backup should've a minimal number of movable parts. If you think you're safe because movable parts are fixated by some mechanism which can break under harsh conditions, I'd say you're screwed already from the start.

[dkbit98]

So something like this

Don't use this as a backup, it's garbage compared to all other metal backup solutions, because it has to many moving parts.
I actually have something similar that I personally tested and I would not recommend it to anyone.
What I would suggest instead is going with stainless seed washers, or solid plate from titanium or stainless steel.

Jameson Lopp tested most metal backups and this type of backup always had lower ratings in his reviews:
https://jlopp.github.io/metal-bitcoin-storage-reviews/

[Pmalek]

You self-moderate the thread to keep out AI posters, but this calculation and resulting estimate sounds like the perfect job for AI to perform. I wonder how similar/different the answer would be to what satscraper wrote in post #2. If you give an AI a detailed-enough prompt so it doesn't misunderstand what you want it to do, I am sure it would spit out a useful reply.