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:
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...