Here is a contrived game, that there could be some benefit for solving. I hope it will turn out somehow like the Monty Hall problem.
A gameshow. First here is the simpler game. A contestant spins a spinner on a target.
Then the gameshow host spins the target, without looking at where the contestant's spinner wound up.
Then the host measures how far right and left the spinner is from the center, and draws a circle that width.
Finally, a blindfolded monkey throws darts in the general direction of the target until one of them does hit the target. If it hits the green circle, the contestant wins.
I think we can agree that this is basicly a fair game where the contestant has a one-half chance to win. The circle spinning gives them a random angle which gives a circle of random size, on average one half. The monkey is no more likely to hit the center of the target than an edge, so the chance to win is just the area of the circle divided by the area of the target.
Now here is the more interesting game. There are two contestants. They each spin an arrow randomly, at the same time so they don't have any idea where the other arrow will land. Then the host spins the target to randomize things more. Then they draw two circles. Then the blindfolded monkey throws the dart.
Both contestants win if the dart either lands inside both circles, or outside both circles.
A famous statistician argues that their chance to win ought to be one half, on average. The chance for each of them is one half. Their spins are independent.
But experience shows that they in fact win around 59% or 60% of the time.
How can that be?
Edit: In case it isn't obvious that the single-player game has probability 1/2:
The player has some angle after the host's spin, each angle is equally probable, so for the whole probability we would find the probability for a game with some angle $\theta$, and then integrate over $\theta$ from 0 to $\pi$. (Cosines from $\pi$ to $2\pi$ should be the same.)
The probability of winning given some $\theta$ is $cos^2\theta$, because the circle with radius $\theta$ has area $\pi \cos^2\theta$ while the whole target has area $\pi$.
$$ \sideset{}{_0^{\pi}}\int \cos^2(\theta) d\theta = \frac{\cos(\theta)\sin(\theta) + \theta}{2} = \frac{\cos(\pi)\sin(\pi) + \pi}{2}-\frac{\cos(0)\sin(0) + 0}{2}= \pi/2 $$
Since the domain is $[0,\pi]$ the average result is 1/2.