For any natural number $N$ and $0\le n\le N$ define  

$$ 
f(n) = f(n,N) = \frac{1}{(N+1)!} \sum\_{\substack{{S\subset \{1,\ldots,N\}} \\ {|S|=n}}} \prod\_{s\in S} s. 
$$
(The empty product is interpreted as $1$.) It is easy to see that 
$$
\sum\_{n=0}^{N} f(n) = 1, 
$$ 
so that $f$ may be thought of as a probability distribution. 


As $N$ gets large, what is the maximum value of $f(n)$? What is the value $n\_{max}$ where this peak value is attained, and 
what is the distribution of $f$, especially around $n\_{max}$? 



summation of products of combinatorials
The unnormalized sums are [unsigned Stirling numbers of the first kind](http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind), the coefficient of $x^{N-n}$ in $(x+1)(x+2)\cdots(x+N).$


For $p=1/2$, this is the central Stirling number of the first kind $3,35,735,22449, ..., S\_1(2n-1,n), ...$. See [A129505](http://oeis.org/A129505) which mentions


$$S\_1(2n-1,n) \sim \frac{1+2c}{8c\sqrt{-\pi(1+c)}} \bigg(\frac{-8c^2}{e(1+2c)}\bigg)^n n^{n-3/2}$$


where $c=\operatorname{LambertW}(-1,\frac{-1}{2\sqrt{e}}) \sim -1.75643$ 


in a note updated about a month ago. 




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When you think about this as a probability distribution, it is a sum of independent Bernoulli random variables with probabilities $1/2, 2/3, ... N/(N+1)$. It's easy to read off the mean $N+2-H\_{N+1} \approx N-\log N + c\_1$ and variance $1/4 + 2/9 + ... + N/(N+1)^2 \approx \log N + c\_2$. [Lindeberg's condition](http://en.wikipedia.org/wiki/Lindeberg%27s_condition) is satisfied so the distribution is asymptotically normal. 


Each binomial factor is unimodal, so the product is unimodal, so the asymptotic normality with standard deviation asymptotic to $\sqrt{\log N}$ gives asymptotic lower bounds for the maximum probability asymptotic to $\frac{1}{\sqrt{2 \pi \log N}},$ which agrees with The Masked Avenger's heuristics in the comments. I think there should be asymptotic upper bounds of the same form. 


