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% 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." -Todd _______________________________________________ A51 mailing list A51@lists.reflextor.com http://lists.lists.reflextor.com/cgi-bin/mailman/listinfo/a51
