Finitely generated group unlikely to be generated by randomly chosen elements. A well known interesting fact is that if two integers are picked "at random" (in an appropriate asymptotic sense), the chances they generate the integers is $6/\pi^2$. So, the integers can be generated by two randomly selected elements with non-vanishing probability.
I am wondering if there exists a (finitely generated) group that is almost certainly not generated by any finite number of "randomly chosen" elements. That is, is there a group $G$ with generators $\{g_1, ... , g_n\}$ such that for every $k$, our if we pick $k$ elements of $G$ at random, there is a vanishingly small probability that the selected elements generate $G$. 
To clarify, select elements randomly from balls of a given radius in the group, using the word metric with given generators, and see if the probability $k$ elements chosen from the ball generate $G$ tends to $0$ as the radius of the balls grows. This definition is compatible with the above result about the integers. 
 A: All non-abelian free groups have this property, although it really depends how we flesh out the phrase "randomly chosen".
The most obvious way of addressing this problem is to do as suggested in the question. That is, consider balls of increasing radius and go as follows:


*

*fix a basis $X$ of your free group $F$,

*find a function $f: \mathbb{N}\rightarrow\mathbb{R}$ such that
$$
\frac{\#\{\text{non-generating sets of size $n$ consisting of words of length $\leq k$}\}}
{\#\{\text{sets of size $n$ consisting of words of length $\leq k$}\}}\geq f(k)
$$

*Prove that $\lim_{k\rightarrow\infty}f(k)=1$.


You can prove that such a function exists*, and is in fact of the form $1-ae^{bk}$ (so the number of generating sets decreases exponentially with $k$). Therefore, all non-abelian free groups have this "random non-generation" property.
In fact, if you are comfortable with presentations, instead of taking the ball $\{\text{non-generating sets of size $n$ consisting of words of length $\leq k$}\}$ we could have taken the ball $$\{\text{sets of size $n$ consisting of words of length $\leq k$ and with $\langle X\mid S\rangle$ non-trivial}\}.$$ Refining this ball further gives us the notion of a "random group"; this is called the "few relator" or "$0$-density" model of random groups, and such groups are known to be infinite (indeed metric small cancellation and so hyperbolic), as well as lots of other things. What is interesting is that the theory has been developed to be much more general than the "growing balls" idea, and there is a theorem which says that a "random group of density $>\frac12$ is either trivial or cyclic of order $2$". Therefore, changing the model changes the result (I do not know though how this impacts on the question at hand, about generation). For more details on random groups and their various models see the book Yann Ollivier, A January 2005 invitationto random groups which you can find on Ollivier's webpage here (Section I.3.c gives further properties of the random groups I describe above).
*standard arguments use small cancellation theory though, and so are non-trivial but prove much more.
A: I think that a free group (not free abelian) on two generators $a$ and $b$ has this property --- the odds of two longish words "cancelling" so that their product is one of the four generators $a$, $a^{-1}$, $b$, $b^{-1}$ is pretty small. (OK, other generators are possible, too, but...) There's probably a nice topological proof of this having to do with the universal cover of the figure-8 curve, but I don't see it offhand. 
