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How do you calculate the total amount of combinations in the standard 12-word seed from a 2048 word dictionary?

$log_2(2048^{12})$ which equals $2^{132}$? Is this the right way?

And what is the "minimum" sort of recommended combinations for the next 5-10 years, $2^{110}$? $2^{100}$?

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The total number of possible 12-word "seeds" is $2048^{12}$ as you already noticed.

The entropy of such a seed is $\log_2{(2048^{12})} = \log_2{((2^{11})^{12})} = \log_2{(2^{132})} = 132$ bits.

Therefore, remembering such a 12-word seed is equivalent to remembering a 132-bit key.

If you want to play it safe, pick a seed with 256 bits of entropy or with 24 words. (Because $\log_2{2^{11\lambda}} \ge 256 \Rightarrow \lambda > 23.3.)$

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  • $\begingroup$ Ok but is 120 bit safe the next 5-10 years? $\endgroup$ – NetCoder Feb 3 '18 at 11:11
  • $\begingroup$ I would say so, yes. $\endgroup$ – Alin Tomescu Feb 3 '18 at 11:21

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