Hi, On Sep 14, 2011, at 2:34 AM, T wrote:
> Thanks for the info. > > On Mon, Sep 12, 2011 at 11:01 PM, Karsten Nohl <n...@virginia.edu> wrote: >> On Sep 12, 2011, at 11:21 PM, T wrote: >>> Why are only some bursts able to be cracked? >> The coverage of all tables combined is 0.04%. >>> What percentage of bursts is expected to be cracked? >> Each burst gives you 114-63 tries with 0.04% success probability each. >>> Should at least one of the four bursts in a message always be cracked? >> Four burst are cracked with 92% success. > > How do you get to the 92% success rate for four bursts? > > If I assume a 0.04% success rate for each try, I get: > p=0.04e-2; > Probability of success = 1 - (1-p)^(4*(114-63)) = 7.8% The calculation is not quiet as simple as the A5/1 state space is not uniformly distributed. But you are right, there was a little inconsistency in my calculation: I was assuming 8 bursts, ie. 2 full messages. > Also, how does that compare to the success rate quoted here: > https://media.blackhat.com/bh-ad-10/Nohl/BlackHat-AD-2010-Nohl-Attacking-Phone-Privacy-wp.pdf > "Given two encrypted known plaintext messages (ie. Cipher mode complete and a > System Information message), the table set finds a secret key with > almost 90% probability." 2 messages => 92% success. Cheers, -Karsten _______________________________________________ A51 mailing list A51@lists.reflextor.com http://lists.lists.reflextor.com/cgi-bin/mailman/listinfo/a51
