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I have the following password with a prescribed format: it starts with a fixed value of 111, followed by 2 random lower case letters and then lastly a random digit:

"111""2 Random lowercase letters""random digit"

Is my below calculation correct?

fixed value 111 -> log2(1) = 0 bits

2 random lowercase letters -> log2(26)*2 = 9.4 bits

random digit -> log2(10) = 3.3 bits

Total entropy of the password would be 0 + 9.4 + 3.3 bits = 12.7 bits

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1 Answer

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Yes, the computation is correct: the passwords will have an entropy of $log_2(26^2\cdot10)\simeq {12.7}$, which means that they are weaker than a randomly chosen secret key with 13 bits.

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I think you meant 12.7 and not $2^{12.7}$ in your answer :) – Thomas Oct 14 '12 at 9:10
@bob Thanks! :) – null Oct 14 '12 at 9:12
@Thomas: indeed, corrected. – bob Oct 14 '12 at 9:16

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