2 circles

Call the other intersection point of the two circles $\gamma'$ . Now move $\gamma$ (and $\gamma'$ ) close enough to $\beta$ so that the circle with center $\beta$ goes through $\alpha$ . At that position, both circles have equal radii, and $\alpha \beta = \alpha \gamma = \beta \gamma$ so $\triangle \alpha \beta \gamma$ is equilateral making arc $\gamma \beta \gamma' = \frac{2\pi}{3}$ , i.e one third of the circumference of the circle with center $\alpha$ . Clearly for the circle with center $\beta$ to intersect $\alpha \beta, \gamma$ has to be on that third of the circumference, so the probability is $\boxed{\frac{1}{3}}$