Take two sticks (not necessarily of the same length). Break one of them at a uniformly random point, support the other one at a uniformly random point, and place the pieces of the former on the ends of the latter. What's the probability that the shorter end goes down?
P.S.: As Rahul rightly pointed out, I was considering only the weight of the first stick and regarding the second one as massless. We could consider the same question with two identical sticks (same length and mass density).
On
another (and longer) approach: consider 2 random variables:
$X$ - the length of the first piece. uniformly random between 0 and 1(proportion).
$Y$ - the distance between the support and the first piece.uniformly random between 0 and 1(proportion)
the first piece goes down when $XY > (1-X)(1-Y)$, could also be written as $(X+Y) > 1$
Now, back to the question, if we force $X$ to be uniformly random between $0$ and $\frac{1}{2}$; than, we can consider $(X+Y)$, and by simple convolution, we can conclude that the area of the density function between $1$ and $\frac{3}{2}$ is equal to the area between $0$ and $\frac{1}{2}$, which is exactly half of the area between $\frac{1}{2}$ and $1$. so the chance for the piece to fall is $\dfrac{1}{4}$
The stick is balanced if the weights are in inverse proportion to the lever lengths. Thus if the shorter end has length $x$, the probability that it goes down is $x$, so the overall probability is
$$ 2\int_0^\frac12x\mathrm dx=\frac14\;. $$