How does a darlington pair work?
If the first transistor is over-saturated with high base current, then \$I_{b2} = (\beta_1+1) I_{b1}\$ doesn't hold anymore, it will be: \$I_{b2} &gt; (\beta_1+1) I_{b1}\$.

The second transistor will decide its base current \$I_{b2}\$ in darlington. Say, if the max current passing through the emitter side of the second transistor is \$I_{c2max}\$, the max base current to the second transistor will be \$I_{b2max}=I_{c2max}/\beta_2\$, as it will be in active mode.