There are $n$ girls and $n$ boys. Any boy and girl know each other with probability $p$. In random order, a boy marries a girl to whom he is connected to. If is the turn of a guy who knows no girl or all the girls he knows are taken, he remains alone.
What is the probability that a guy will actually stay single?
For constant $p$, the probability is $O(\frac1n)$ and is basically determined by the order in which the men get married.
A lower bound on the probability is $\frac{1-p}{n}$. The probability that no man is single is at most $p$: the probability that if the first $n-1$ men get married, the last unmarried man happens to know the last unmarried woman. So at least $1-p$ of the time, some man is single.
If $q$ is the probability of a man staying single (which is the same for all men, by symmetry) then the probability that any man is single is at most $qn$. We have $qn \ge 1-p$, so $q \ge \frac{1-p}{n}$.
On the other hand, an upper bound is $\frac{1-p}{pn}$. The man that is $k^{\text{th}}$ from last has at least $k$ unmarried women to choose from, so his probability of staying single is at most $(1-p)^k$; any man is equally likely to end up in any position, so the probability of a man staying single is at most $$ \sum_{k=1}^n \frac1n \cdot (1-p)^k \le \frac1n \sum_{k=1}^\infty (1-p)^k = \frac1n \cdot \frac{1-p}{1-(1-p)} = \frac{1-p}{pn}. $$
For probabilities $p$ that are functions of $n$, the bounds above still apply, but they mean less. Still, if $np \to \infty$ as $n \to \infty$, the probability that a man is single is $\frac{1-p}{pn} \le \frac1{pn} \to 0$ by the above.
When $np \to C$ as $n \to \infty$, the upper bound above is basically $\frac{1}{C}$; we can give a better lower bound of $(1 - \frac Cn)^n \sim e^{-C}$, the probability that a man knows no women. So in this range, we know that a man's probability of staying single is constant. (By the way, by symmetry, a woman's probability of staying single is exactly the same.)
