Hoop Game Throw Ring Toss I have a problem figuring out what seems a paradoxe.
This is a Hoop game in which we throw a skittle (or pole) on a square and flat surface, and then we have to throw (at random) successively $n$ rings of same radius at this pole. The goal is to catch the pole with one of the rings.
When we simulate this game, the centers of the rings are uniformly distributed on all the possible position of the square surface. For the sake of concreteness, we can assume the square surface to be the continuous domain $[0, 1] \times [0,1]$.
What we would like is to figure out whether the events $E_i$ := "catch the pole with the $i$-th ring" are independent or not.
We can propose another game which seems (but maybe wrongly) equivalent to the first one: In this alternative game, we first throw at random the $n$ rings, and then we throw the pole. Intuitively, the events $E_i$ seem to be independent. Thinking the game in this sense gave me great confidence in their independence.
A simulation reveals that the events are not independent.
I am looking for hints to help me better understand this paradoxal problem.
 A: I think that the $E_i$ should be independent given the point of the pole, $Z = (X, Y)$. If the $i$-th ring has radius $R_i$ then $P(E_i | Z)$ is the probability of picking a point in the region of the circle with radius $R_i$  and center $Z$ that is in the unit square that is in the unit square, which is just the area of this circle intersected the unit square. I do not think the $E_i$ will be independent, however, if not conditioned on $Z$. As the probability $P(E_j | Z)$ is maximized if the circle $R_i$ centered at $Z$ is contained in the unit square, I think $P(E_i | E_j) > P(E_i)$.
I know this is a pretty informal argument, but I'm not too sure what you're looking for (or if I have the interpretation right), but I can try to spell this out more later (or look again if I'm wrong).
A: I can give an intuitive explanation of why the events are dependent. First, consider this toy example:

Imagine choosing a random real number $p$ uniformly between $0$ and $1$. This $p$ becomes probability of heads for a biased coin. This coin is flipped twice. Let $A$ a be the event that the first flip is heads, and let $B$ be the event the second coin is heads. Are $A$ and $B$ independent?

At a first glance, it might seem that $A$ and $B$ are independent, since they are distinct flips of a coin. However, we find that
$$
P(A)=\int_0^1 P(A\mid p)\,dp=\int_0^1 p\,dp=\frac12,\qquad P(A)P(B)=\frac12\times \frac 12=\frac14,\\
P(A\cap B)=\int_0^1 P(A\cap B\mid p)\,dp=\int_0^1 p^2\,dp = \frac13 \tag{$*$}
$$
Huh, turns out $P(A\cap B)\neq P(A)P(B)$, so the events are dependent. What gives? The idea is that $A$ occurring gives information about $p$; if $A$ occurs, it is more likely that $p$ was large, which in turn affects the probability of $B$.
Now, back to your problem. Suppose there are just two rings, for simplicity. There is a similar phenomenon going on in the ring problem. If the peg happens to land near the edge of the square, then each ring it less likely to hit it. Conversely, if the first ring happens to hit the peg, then that makes it less likely that the peg is on the edge, which makes it more likely that the second ring hits the peg. Therefore, the first ring actually affects the probability of the second ring indirectly, so they are dependent.
It is possible to make all this rigorous by doing a calculation similar to $(*)$, but it would not be especially illuminating, and the geometry makes the math too tricky for what its worth. Therefore, I leave that as an exercise to the reader.
