Re: [B.A.T.M.A.N.] IP from MAC

2009/9/16 Lick A Prize <lick.prize(a)gmail.com> What is the occurrence probability of "r" MAC-Adress twon agree on their last two bytes? This is similar to the birthday paradox: At a meeting of r> 2 people which is the number of people needed for the probability that two men have same birthday sould be 99%? The solution is in many books and that can view at: http://en.wikipedia.org/wiki/Birthday_problem Then, At a meeting of 60 people there is the 99% probability of the least two men have a birthday the same day. So,, Similarly to the case of MAC-Adress is: P(r)=1-65536!/(65536^r * (65536-r)!) where *! is factorial operator, and "r" is the Mac-Adress. Well if the calculations are not mistake, we obtain r = 13,107 MAC-Adress minimun in the same site then 99% probability two matches in his last two bytes. Now a new MAC entering, which is the probability that given "n" MAC-Adress,,, all with two last byte differents this new MAC matches with any MAC-Adress in its last two bytes? Well this is: P(n)=1-(65535/65536)^n then when n=13107 is P(13107)=0.18 or 18%. That is MAC-paradox...