I have $m=10^6$ balls and $n=13\cdot 10^3$ bins. I need to know how I can calculate maximum load for each bin. I'm aware of [1], but it's to loose (e.g. max_load: $e\cdot m/n$. And in practice using the formula overflow does not happen as I have implemented it (in c++).
I need a formula to get a bin max-overflow and it does not use any big $O$ notation, so I can replace my parameters and get the result.
[1]. http://www.ic.unicamp.br/~celio/peer2peer/math/czumaj-balls-into-bins.pdf
You can use the normal approximation. The expected number of balls in a bin is $1000000/13000 \approx 77$. The standard deviation is $\sqrt {10^6 \cdot \frac 1{13000} \cdot (1-\frac 1{13000})}\approx 8.77$ You expect one bin in $13000$ to be about $3.8 \sigma$ high from a z-score table, so your max bin would be around $77+3.8\cdot 8.8 \approx 111$