Yes, but conversely if there's 10 simultaneous orders the probability is 1 - (4095/4096) * (4094/4096) * ... * (4087/4096) = 1.093% to have a collision wi

Yes, but conversely if there's 10 simultaneous orders the probability is 1 - (4095/4096) * (4094/4096) * ... * (4087/4096) = 1.093% to have a collision with a code consisting of 3 hexadecimal characters.