Probability and Combinatorial Group Theory.

Solution 1:

Here are some quite similar facts:

The probability that a random pair of elements generates either $S_n$ or $A_n$ is $1 - \frac{1}{n} - \frac{1}{n^2} - \frac{4}{n^3} - \frac{23}{n^4} - \frac{172}{n^5} - \frac{1542}{n^6} - O\left(\frac{1}{n^7}\right)$.

The probability that a random triple of elements generates either $S_n$ or $A_n$ is $1 - \frac{1}{n^2} - \frac{3}{n^4} - \frac{6}{n^5} - O\left(\frac{1}{n^6}\right)$.

The probability that a random pair of elements generates exactly $S_n$ is $\frac{3}{4} + O(\frac{1}{n})$

The probability that a random element generates $\mathbb{Z}_n$ is $\frac{\phi(n)}{n}$, where $\phi$ is Euler's totient function.

The probability that $m$ random elements generate $\mathbb{Z}_{p^n}$, where $p$ is prime, is $1 - \frac{1}{p^m}$.

The probability that $n$ random elements generate $\mathbb{Z}^{n-1}$ is $\prod_{j=2}^n \zeta(j)^{-1}$, where $\zeta$ is Riemann zeta function.

The most general fact of that type is that the probability of $m$ random elements generating an arbitrary finite group $G$ is $\sum_{H \leq G} \mu(G, H) {\left(\frac{|H|}{|G|}\right)}^m$, where $\mu$ is the Moebius function for subgroup lattice of $G$. However, despite being much stronger, than aforementioned facts, this one is very hard to use, because we can only apply it to the groups with known subgroup structure.

Solution 2:

For a finite group you can compute the probability that any two random elements will commute by looking at certain conjugacy classes.