$1/2 = 1 - e^{-n(n-1)/2d}$ General Birthday Formula?

Question

When solving $\frac{1}{2} = 1 - e^{-n(n-1)/2d}$ for $n$, the solution should be $n \approx \sqrt{\ln(4)d}$. But I can't get the calculations done.

Any advice would be highly appreciated.

Thank you!

PS: The formula has to do with the General Birthday Formula and it simply states the 50% chance that at least 2 people (among $n$ people) share the same birthday when there are $d$ days.

Answer 1

We have $$ \frac{1}{2} = 1 - e^{ - n(n - 1)/2d} \Leftrightarrow \frac{1}{2} = e^{ - n(n - 1)/2d} \Leftrightarrow \log 2 = \frac{{n(n - 1)}}{{2d}} $$ $$ \Rightarrow 2d\log 2 = n(n - 1) \approx n^2 \Rightarrow n \approx \sqrt {2d\log 2} = \sqrt {d\log 4} . $$ Note that since $$ n(n - 1) = \left( {n - \tfrac{1}{2}} \right)^2 - \tfrac{1}{4} $$ we have in fact $$ n = \tfrac{1}{2} + \sqrt {\tfrac{1}{4}+d\log 4}. $$