I'm sure you're looking for something deeper, but one example I like where using a rough probabilistic argument actually gives the right answer is when considering the number
of derangements (fixed-point free) permutations of $\{1,2,\ldots, n\}.$ Putting the uniform 
probablity distribution on $S_n$, the probablity that a given symbol $i$ is fixed by 
a random permutation $\tau$ is $1 - \frac{1}{n}$. Therefore, you might expect the
probability that a given permutation should fix no symbol would be around
$(1 - \frac{1}{n})^{n}$, so around $e^{-1}$. But all sorts of unjustified independence
assumptions have been made here. However, if we do a more precise analysis, an inclusion-exclusion
argument shows that the number of permutations in $S_n$ fixing no symbol is exactly $\sum_{j=0}^{n} (-1)^{j}$ " n choose j" ` $|S_{n-j}|$. Dividing by $n!$, we are left with the 
$n$-th degree Taylor polynomial for $e^{-x}$ at $x =1$, which soon gets very close to $e^{-1}$.
The unjustified independence assumptions seem to be compensated for by the fact that 
$(1- \frac{1}{n})^{n}$ is only an approximation to $e^{-1}$. In fact, the proportion of derangments
clearly approaches $e^{-1}$ much faster than $(1- \frac{1}{n})^{n}$ does.

In another direction, there are situations in finite group theory where it is possible to show
that the probability that a given pair of elements with given properties generate the group is close to $1$, yet it may be very difficult to exhibit an explicit pair of such generators. See
for example the work of Martin Liebeck and various co-authors on this topic.