To add some context to Anthony Quas's answer:

The two problems are two special cases of a general question. Suppose you have a map $T: X \to
X$ which preserves a measure $\mu$. For simplicity assume that $\mu$ is ergodic, so any invariant set has measure $0$ or $1$.
A map $\alpha: \mathbf{Z} \times X \to GL(d,\mathbf{R})$ is called a (dynamical) cocycle if
$$\alpha(n+m,x) = \alpha(n,T^m x) \alpha(m,x).$$

The Osceledec multiplicative ergodic theorem says that (in particular) there exists a number $\lambda_1$ called the top Lyapunov exponent for $\alpha$ such that for $\mu$-almost all $x \in X$,
$$ \lim_{n \to \infty} \frac{1}{n} \log \|\alpha(n,x)\| = \lambda_1.$$

(The theorem also guarantees the existence of other Lyapunov exponents $\lambda_2, \dots, \lambda_d$. One quick way to define them is to say that $\lambda_1 + \dots + \lambda_k$ measures the exponent of the growth of the norm of the cocycle acting on the $k$'th exterior power of $\mathbf{R}^d$.)

Now both Visvanath's number and the Khinchine-Levy constant are the top Lyapunov exponent for certain cocycles. In Visvanath's case, the space $X$ is the infinite product of $GL(2,\mathbf{R})$'s and $T$ is the shift map. In the Khinchine-Levy case, $T$ is the Gauss map of the interval to itself, and $\alpha$ is closely related to the derivative cocycle, (so $\alpha(n,x)$ is given in terms of the derivative of $T^n$).

In general Lyapunov exponents are virtually impossible to compute, except in the situation where you are dealing with the derivative cocycle of a flow on a homogeneous space. (One exception is derivative cocycles for flows on the moduli space of curves where at least certain combinations of Lyapunov exponents have interpretations in terms of algebraic geometry). But the Gauss map is essentially equivalent to a flow on a homogeneous space, since it is (very closely related to) the coding of the geodesic
flow on the modular surface. Thus, it is not surprising that the Khinchine-Levy constant can be evaluated explicitly.

Visvanath's calculation is different: he guesses the stationary measure and then uses the Furstenberg formula for the (top) Lyapunov exponent in terms of the stationary measure. It seems very hard to guess the stationary measure in most cases.