How likely is it not to be anyone's best friend? A teenage acquaintance of mine lamented:

Every one of my friends is better friends with somebody else.

Thanks to my knowledge of mathematics I could inform her that she's not alone and $e^{-1}\approx 37\%$ of all people could be expected to be in the same situation, which I'm sure cheered her up immensely.
This number assumes that friendships are distributed randomly, such that each person in a population of $n$ chooses a best friend at random. Then the probability that any given person is not anyone's best friend is $(1-\frac{1}{n-1})^{n-1}$, which tends to $e^{-1}$ for large $n$.
Afterwards I'm not sure this is actually the best way to analyze the claim. Perhaps instead we should imagine assigning a random "friendship strength" to each edge in the complete graph on $n$ vertices, in which case my friend's lament would be "every vertex I'm connected to has an edge with higher weight than my edge to it". This is not the same as "everyone choses a best friend at random", because it guarantees that there's at least one pair of people who're mutually best friends, namely the two ends of the edge with the highest weight.
(Of course, some people are not friends at all; we can handle that by assigning low weights to their mutual edges. As long as everyone has at least one actual friend, this won't change who are whose best friends).
(It doesn't matter which distribution the friendship weights are chosen by, as long as it's continuous -- because all that matters is the relative order between the weights. Equivalently, one may simply choose a random total order on the $n(n-1)/2$ edges in the complete graph).
In this model, what is the probability that a given person is not anyone's best friend?
By linearity of expectations, the probability of being mutually best friends with anyone is $\frac{n-1}{2n-3}\approx\frac 12$ (much better than in the earlier model), but that doesn't take into account the possibility that some poor soul has me as their best friend whereas I myself has other better friends. Linearity of expectation doesn't seem to help here -- it tells me that the expected number of people whose best friend I am is $1$, but not the probability of this number being $0$.

(Edit: Several paragraphs of numerical results now moved to a significantly expanded answer)
 A: Here's another take at this interesting problem. 
Consider a group of $n+1$ persons $x_0,\dots,x_n$ with $x_0$ being myself.
Define the probability
$$
P_{n+1}(i)=P(\textrm{each of $x_1,\dots,x_i$ has me as his best friend}).
$$
Then we can compute the wanted probability using the inclusion-exclusion formula as follows:
\begin{eqnarray}
 P_{n+1} &=& P(\textrm{I am nobody's best friend}) \\
 &=& 1-P(\textrm{I am somebody's best friend}) \\
 &=& \sum_{i=0}^n (-1)^i\binom{n}{i}P_{n+1}(i). \tag{$*$}
\end{eqnarray}
To compute $P_{n+1}(i)$, note that for the condition to be true,
it is necessary that of all friendships between one of $x_1,\dots,x_i$ and anybody,
the one with the highest weight is a friendship with me.
The probability of that being the case is
$$
\frac{i}{in-i(i-1)/2}=\frac{2}{2n-i+1}.
$$
Suppose, then, that that is the case and let this friendship be $(x_0, x_i)$.
Then I am certainly the best friend of $x_i$.
The probability that I am also the best friend of each of $x_1,\dots,x_{i-1}$ is unchanged.
So we can repeat the argument and get
$$
P_{n+1}(i)
=\frac{2}{2n}\cdot\frac{2}{2n-1}\cdots\frac{2}{2n-i+1}
=\frac{2^i(2n-i)!}{(2n)!}.
$$
Plugging this into $(*)$ gives a formula for $P_{n+1}$ that agrees with Henning's results.
To prove $P_{n+1}\to e^{-1}$ for $n\to\infty$, use that the $i$-th term of  $(*)$ converges to, but is numerically smaller than, the $i$-th term of
$$
e^{-1}=\sum_{i=0}^\infty\frac{(-1)^i}{i!}.
$$
By the way, in the alternative model where each person chooses a best friend at random, we would instead have $P_{n+1}(i)=1/n^i$ and $P_{n+1}=(1-1/n)^n$.
A: This answer is quite similar to Lancel's answer, but I think it deals successfully with the subtle issue of dependence between friendships (if I am the best friend of $x_1$, then I am slightly more likely to also be the best friend of $x_2$, because my being the best friend of $x_1$ suggests that the friendship strength between $x_1$ and $x_2$ is "low", making it more likely that my strength to $x_2$ beats all of the other friendship strengths for $x_2$). This answer also avoids needing to worry about convergence of the power series.
We use Brun's Sieve, which is essentially a repackaging of the inclusion-exclusion method. Suppose that there are $n$ people aside from myself in the graph, and let $x_1, \ldots, x_r$ be any set of $r$ people. Let $B_i$ be the event that I am the best friend of person $i$. We need to show that, for each fixed $r$, as $n \to \infty$ we have ${n \choose r}P(B_1 \wedge \cdots \wedge B_r) \to \frac{1}{r!}$. To do this, we will show that $P(B_1 \wedge \cdots \wedge B_r) \to \frac{1}{n^r}$.
Now $P(B_1) = 1/n$: of the $n$ people that $x_1$ sees, I must have the highest strength. For higher $i$ we need to obtain conditional probabilities, and we only obtain upper and lower estimates, but these estimates will asymptotically approach each other as $n \to \infty$. For $P(B_2 | B_1)$, we have the lower bound $P(B_2 | B_1) \geq P(B_2) \geq 1/n$, since conditioning on $B_1$ only "pushes down" the value of the edge $x_1x_2$ (it is somewhat easier to see this by flipping the situation around: if we consider controlling the value of $x_1x_2$ and picking all other edge weights randomly, the probability of $B_1$ and the probability of $B_2$ each go down as the weight of $x_1x_2$ goes up). On the other hand, $P(B_2 | B_1) \leq 1/(n-1)$, since my weight to $x_2$ needs to beat each of $x_2$'s weights to vertices other than $x_1$, and conditioning on $B_1$ gives no information about those weights.
Continuing in this manner, we see that $1/n \leq P(B_i | B_1, \ldots, B_{i-1}) \leq 1/(n-i+1)$ for each $i$. When $r$ is fixed and $n \to \infty$, this gives $P(B_i | B_1, \ldots, b_{i-1}) \to 1/n$, so that $P(B_1 \wedge \cdots \wedge B_r) \to 1/n^r$.
Thus, by Brun's Sieve, we have $P\left(\bigwedge\overline{B_i}\right) \to e^{-1}$, and, what's more, the number of people whose best friend you are is (asymptotically) Poisson-distributed with mean $1$.
