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Re: MD5 math question
From: Charles Miller <cmiller () pastiche org>
Date: Wed, 4 Jan 2006 14:54:14 +1100
On 04/01/2006, at 12:18 PM, Jeff Robertson wrote:
Assume that a password between 1 and 24 ASCII characters was stored as
an MD5 hash. No salt. What is the probability that someone cracking
the
password will find not the password that the user originally chose,
but
a different password that happens to collide with it? Intuitively it
seems so unlikely that you wouldn't ever expect to see it. But what is
the probability really?
From my back-of-the-envelope calculation, your intuition is
misplaced. :)
Even if you assume only 6 bits of variance per password character
(which is just a-zA-Z0-9 plus two punctuation chars), that's 2^144
possible 24-character passwords. MD5 is a 128 bit hash, so that's
2^16 passwords for every hash value, or only a 1 in 65,000 chance
that the first matching hash you come across in the password space
is, in fact, the correct password.
And that's only if you assume the original password lives inside [a-
zA-Z0-9.!]{24}, not the "1-24 ASCII characters" of the original
question.
Charles
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