RaTTuS wrote: ↑Tue Feb 13, 2018 8:39 am

I think it is 4824 before you have > 50% chance of having 2 with the same address

if you have n RPis the number of possible pairs is n(n-1)/2.

Each pair has a probability of 1 in 2^24 of having the same MAC.

So for a 50% probability of one pair with both MACs the same you need (2^24)/2 pairs

Putting n(n-1)/2 = (2^24)/2

The 2s cancel leaving

n(n-1) = 2^24

and we see where the square root approximation comes from when n is large (ie n is almost the same as n-1)

Rearranging to the standard quadratic form n^2 - n - 2^24 = 0

Solving exactly gives n = (-(-1) +/- ((-1)^2 - (4 * 1 * -(2^24)))^0.5)/(2 * 1)

For the positive n = (1 + (1 + 2^26)^0.5)/2

=

4096.50003052

(The square root approximation 4096 is quite accurate in this case.)