If a unit cylinder is dropped on floor, is there equal chances to be horizontal or vertical? For example, if a cylinder has radius 1m and height 1m, it drops to floor randomly, is it has equal chance to stop and stay either horizontally or vertically? If so, how to explain or prove it?
 A: I interpret the question this way: the cylinder's staying vertical requires it to hit the floor so that its centre of mass (CoM) is "inside" the vertical line passing through the hitting point. If, instead, the CoM is "outside" this line then the cylinder will stay horizontal. In the picture, the CoM is inside, so the cylinder would be vertical.

Now, imagine the cylinder fixed inside a minimal sphere so that they meet in two circles, these being the ends of the cylinder. These two circles cut the sphere surface into $3$ parts: $2$ caps and a central area between them.
Dropping the cylinder randomly means that when the sphere hits the floor, any point on it is equally likely to touch the floor. Call this point $A$. If $A$ lies in one of the caps this equates to the CoM being inside the vertical line and thus the cylinder staying vertical. Therefore,
$$P(\text{Cylinder vertical}) = \dfrac{\text{Area of $2$ spherical caps}}{\text{Area of sphere}}.$$
Area of a sphere is $4\pi r^2$. Area of a spherical cap is $2\pi rh$, where $h$ is the cap height.
The dimensions of our cylinder mean that $r=\dfrac{\sqrt{5}}{2}$ and $h=\dfrac{\sqrt{5}-1}{2}$, which are found by simple geometry. Therefore,
$$P(\text{Cylinder vertical}) = \dfrac{2\times 2\pi rh}{4\pi r^2} = \dfrac{h}{r} = \dfrac{\sqrt{5}-1}{\sqrt{5}} \approx 0.55.$$
A: If you compare the surface areas, The side is $2\pi$ and the top+bottom is also $2\pi$ so yes it has an equal chance.
