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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 $N=1000$)

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

I would have guessed $p\approx \sqrt{N}$, but numerically I find $p\approx N^\alpha$ with $\alpha\approx 0.57$ distinctly greater than $1/2$.

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 $N=1000$)

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.

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

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 $N=1000$)

Now the desired number $p$ of fortunate guests should follow by solving $s_{p,N}=1/N$ for $p$. I expect an answer $p\approx \sqrt{N}$, but will need to work a bit more for a precise result.

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 $N=1000$)

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

I would have guessed $p\approx \sqrt{N}$, but numerically I find $p\approx N^\alpha$ with $\alpha\approx 0.57$ distinctly greater than $1/2$.

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 $$s_{p,N}=p(n-p+1) e^{1-p} \sqrt{\frac{1}{n+1}}\sqrt{\frac{1}{n-p+2}} (n+1)^{n+1} n^{-p-1} (n-p+2)^{-n+p-1}$$ which agrees very well (compare the dots with the continuous curve in the plot below, for $N=1000$)

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 $N=1000$)

Now the desired number $p$ of fortunate guests should follow by solving $s_{p,N}=1/N$ for $p$. I expect an answer $p\approx \sqrt{N}$, but will need to work a bit more for a precise result.