(This question is a follow-up on an [older one](https://mathoverflow.net/questions/272679/limit-of-biggest-share-of-the-pie).)


A huge pie is divided among $N$ guests. The first guest gets $\frac{1}{N}$ of the pie. Guest number $k$ guest gets $\frac{k}{N}$ of what's left, for all $1\leq k\leq N$. (In particular, the last guest gets all of what is left.) 


A guest is said to be **fortunate** if his share of pie is strictly greater than the average share (which is $1/N$ of the original pie). Let $f(N)$ denote the number of fortunate guests out of the total of $N$ guests. What is the value of $$\lim\sup\_{N\to\infty}\frac{f(N)}{N}$$
?



Share of fortunate people in some pie splitting setting
The $p$-th guest receives a share of 
$$s\_{p,N}=p N^{-p-1} (N-p+1) \frac{\Gamma(N+1)}{\Gamma(N-p+2)}$$
For large $N$ I substitute the asymptotic expansion of the Gamma function,
$$\Gamma(z)\approx e^{-z}z^z(2\pi/z)^{1/2},$$ 
to arrive at the approximation
$$s\_{p,N}\approx p(N-p+1) e^{1-p} \sqrt{\frac{1}{N+1}}\sqrt{\frac{1}{N-p+2}} (N+1)^{N+1} N^{-p-1} (N-p+2)^{-N+p-1}$$
which agrees very well with the exact value (compare the dots with the continuous curve in the plot below, for $N=1000$)


![](https://ilorentz.org/beenakker/MO/pochhammer.png)


Now the desired number $f(N)$ of fortunate guests should follow by solving $s\_{p,N}=1/N$ for $p\equiv f$. As a check, for $N=1000$ the exact integer result is $f=94$, while the large-$N$ asymptotics gives $f=94.334$.


I would have guessed $f\approx \sqrt{N}$, but numerically I find a slightly more rapid increase, the plot below is a comparison with $f=\sqrt{2N}\log\log N$. In any case, the large-$N$ limit of $f/N$ seems to be zero.


![](https://ilorentz.org/beenakker/MO/fortunate.png)


*Gold curve: Number $f$ of fortunate guests versus $N$, calculated form the asymptotics of $s\_{p,N}$. The blue curve is $f=\sqrt{2N}\log\log N$.*


