Expectation of random callers distribution In a country with $N$ people, if everyone calls one random person in the country, what is the expected number of people who dont receive a call?
If $f(m,n,a,b)=$Expected fraction of people who receive $t$ calls, for $a<t<b$, in a country with $n$ people if everyone calls $m$ random persons, what distribution arise as $n$ goes to infinity?
 A: This, and Francesco's answer, are based on the question as originally posed. Set $X=1$ or $X=m$ and $N=n$ to fit the current statement.
Assuming that no one calls themself, nor calls anyone else more than once, the probability of receiving exactly $k$ calls is
$$
\binom{N-1}{k}\left(\frac{X}{N-1}\right)^k\left(1-\frac{X}{N-1}\right)^{N-k-1}\tag{1}
$$
Thus, the probability of receiving no calls is
$$
\left(1-\frac{X}{N-1}\right)^{N-1}\tag{2}
$$
By linearity of expectation, the expected number of people who don't get a call is
$$
N\left(1-\frac{X}{N-1}\right)^{N-1}\tag{3}
$$
As $N\to\infty$, $(1)$ becomes
$$
\frac{X^k}{k!}e^{-X}\tag{4}
$$
which says that the probability of getting $k$ calls follows a Poisson distribution with mean $X$.
By linearity of expectation, the expected number of people who get exactly $k$ calls is $N$ times $(1)$:
$$
N\binom{N-1}{k}\left(\frac{X}{N-1}\right)^k\left(1-\frac{X}{N-1}\right)^{N-k-1}\tag{5}
$$
Summing over $a<k<b$, yields
$$
f(X,N,a,b)=\sum_{k=a+1}^{b-1}N\binom{N-1}{k}\left(\frac{X}{N-1}\right)^k\left(1-\frac{X}{N-1}\right)^{N-k-1}\tag{6}
$$
which as $N\to\infty$, becomes
$$
\lim_{N\to\infty}\frac{f(X,N,a,b)}{N}=\sum_{k=a+1}^{b-1}\frac{X^k}{k!}e^{-X}\tag{7}
$$
A: The probability of not receiving a call is $(1 - \frac{X}{N-1})^{N-1}$ (assuming uniform distribution of the calls, that everybody has a phone number and that one doesn't call himself). The expected number of people which will not receive a call is thus obtained by summing over all the N people in the country.
Do you recognize a well known limit?
ps: This seems like a homework question so I tried to give only a hint.
