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I saw in another thread: Is using 7-8 random words from all words of a language as password a good idea?

These calculations:

If we assume that English has 171,476 words. Then with 8 words the entropy is:

$$P(171476,8) \approx 7.474 \times 10^{41} \approx 2^{140}$$

Therefore you will have lower entropy than Bip-39, again. And with 7 words:

$$P(171476,7)≈4.358×10^{36}≈2^{122}$$

I like to use this calculation method myself. I have the following related questions:

  1. What is the name of this mathematical formula called $P(x,y)$?

  2. Any online calculator available? Or even better, a formula for MS Excel or python?

  3. Is this method commonly used to calculate entropy? If not please hint me to a more widely used formula.

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  • $\begingroup$ crypto.stackexchange.com/q/374/18298 $\endgroup$ – kelalaka Jan 15 at 14:14
  • $\begingroup$ thx! and do u know if there is ready Excel formula or python formula available? $\endgroup$ – johnsmiththelird Jan 15 at 14:18
  • $\begingroup$ I did not read, however, first with google search password entripy calculator excel $\endgroup$ – kelalaka Jan 15 at 15:34
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    $\begingroup$ @kelalaka: Right. And, checking, the numbers are closer to your hypothesis than mine, and could have been obtained (checking the source: by you!) using a rounding down of the result with permutation. However the question is about 7 or 8 random words, not 7 or 8 random distinct words, therefore the mathematically correct function is power, and there is no reason to invoke permutation. With the numbers at hand the difference is <0.00024 bit of entropy. $\endgroup$ – fgrieu Jan 15 at 20:25
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    $\begingroup$ @fgrieu I've updated. $\endgroup$ – kelalaka Jan 15 at 23:00