To convince myself of the solution, I compared angular speeds.
When the line joining the centres, CO, has turned through an angle [MATH]\theta[/MATH] radians, the radius line of the 'small' circle, OA, has turned through [MATH]\boxed{\; \phi = \phi' + \theta \;}[/MATH] radians (from the original downward vertical direction), and the arc lengths AX on the two circles are equal (because rolling).
This last fact means that:
[MATH]\begin{align*} \hspace25ex r\theta &= \alpha r \phi'\\
\phi'&=\tfrac{1}{\alpha}\theta\\
\therefore \phi&=\left(1+\tfrac{1}{\alpha}\right)\theta \end{align*}[/MATH]
The ratio of the angular speeds is [MATH]1+\tfrac{1}{\alpha}[/MATH]
[MATH]\begin{align*} \text{In the example originally posted, } \;\alpha&=\tfrac{1}{3}\\
\therefore \phi&=4\theta \end{align*}[/MATH]The 'small' circle makes 4 complete turns as [MATH]\theta[/MATH] maps out 1 complete turn.
Another way of looking at the motion is to keep A in contact with the big circle and
slide the small circle (don't roll it) down the big circle, so that A lands at X. By this sliding OA is already [MATH]\theta[/MATH] radians from the original downward vertical direction. Now
rotate the small circle (without rolling) to give the current picture. OA turns a further [MATH]\phi'[/MATH] radians.