3 ms·

It would be interesting to have an exact number of how probable is to have a MAC collision in the wild.

bigiain 8y ago
It's 48bits, so a space or around 2.8 million billion possibilities, so you'd need a over a million billion devices to have a 50% chance of collision. _If_ everything was statistically randomly distributed across the entire address space. But they aren't. There's a few kinds of structure in MAC address formats that reduce that but potentially a _lot_... Theoretical worst care, a MAC address has 24bit of organisation identifier and 24 bits of device identifier. So If an organisation/manufacturer only makes one model of device, they'd "only" need to build ~16.7 million (24bits) of them before they repeated a device identifier (if they chose not to use up any of their organisation bits to reflect that rollover). Again, maybe half that if they just randomly choose a device ID each time instead of enumerate the space. Practically? Manufacturers screw up... https://www.raspberrypi.org/forums/viewtopic.php?p=902294 https://www.raspberrypi.org/forums/viewtopic.php?p=902294 Oooops... (Also, many Wi-Fi adaptors have easily changeable MAC addresses. Back in the day when cafes used to charge for Wi-Fi access, it wasn't uncommon to sniff the network for a "paid up" MAC address, and either wait til they left and use it, or de-auth them and do a hostile takeover of their paid-for internet access. Apologies to anyone who used to pay for "unreliable" Wi-Fi at Atlas Cafe on Alabama St back in the late 90s/early 2000s...)
- bboreham 8y ago
  The Birthday Problem, as it is known, has sqrt(N) as an approximate answer to get a 50% chance of a clash. So 16 million devices.
  - bigiain 8y ago
    Yep - the chance of any collisions at all, vs the chance of a collision with _your_ specific device/MAC. That does though drop the chance of any collision at all (aka the birthday paradox) of devices discriminated solely by the 24 bit device identifier down to sort(2^24) which is only 4096. A significantly smaller number than I expected...

It's 48bits, so a space or around 2.8 million billion possibilities, so you'd need a over a million billion devices to have a 50% chance of collision. _If_ everything was statistically randomly distributed across the entire address space. But they aren't. There's a few kinds of structure in MAC address formats that reduce that but potentially a _lot_... Theoretical worst care, a MAC address has 24bit of organisation identifier and 24 bits of device identifier. So If an organisation/manufacturer only makes one model of device, they'd "only" need to build ~16.7 million (24bits) of them before they repeated a device identifier (if they chose not to use up any of their organisation bits to reflect that rollover). Again, maybe half that if they just randomly choose a device ID each time instead of enumerate the space. Practically? Manufacturers screw up... https://www.raspberrypi.org/forums/viewtopic.php?p=902294 https://www.raspberrypi.org/forums/viewtopic.php?p=902294 Oooops... (Also, many Wi-Fi adaptors have easily changeable MAC addresses. Back in the day when cafes used to charge for Wi-Fi access, it wasn't uncommon to sniff the network for a "paid up" MAC address, and either wait til they left and use it, or de-auth them and do a hostile takeover of their paid-for internet access. Apologies to anyone who used to pay for "unreliable" Wi-Fi at Atlas Cafe on Alabama St back in the late 90s/early 2000s...)

bboreham 8y ago
The Birthday Problem, as it is known, has sqrt(N) as an approximate answer to get a 50% chance of a clash. So 16 million devices.
- bigiain 8y ago
  Yep - the chance of any collisions at all, vs the chance of a collision with _your_ specific device/MAC. That does though drop the chance of any collision at all (aka the birthday paradox) of devices discriminated solely by the 24 bit device identifier down to sort(2^24) which is only 4096. A significantly smaller number than I expected...