What is the probability that two hyperspheres of radii $x$ overlap within a bigger hypersphere? Having a sphere of dimension $N = 5$, its volume can be calculated as $$\frac{8}{15}\pi^2R^5.$$ 
If two smaller spheres of radii $x$ are dropped randomly inside of the bigger sphere, how can we find the probability of both spheres overlapping? Can someone point me towards a direction?  
I am trying to understand the probability of different sizes of smaller spheres.
 A: Possible approach / too long for a comment.
If the two smaller spheres must be entirely inside the big sphere, then the center $C_1, C_2$ of each smaller sphere is uniform within a sphere $S$ of radius $R-x$, concentric with the big sphere.
So we're looking at the prob that $C_1, C_2$ both uniform in $S$ are within $2x$ of each other.  This can be calculated, at least in theory (with quite a lot of calculus?), by conditioning on $D=$ distance of $C_1$ from the center.
Let $V=$ the intersection between $S$ and a sphere $S'$ centered at $C_1$ of radius $2x$.  Then the two small spheres intersect iff $C_2 \in V$, i.e. the prob is $volume(V)/volume(S)$
Conditioned on $D=d$, it should be possible to calculate $volume(V)$, right?  Since $D$ controls how much $S'$ "sticks out" of $S$ so to speak.  (I personally don't know how... lost my calculus skills years ago.)
It should be easier to calculate the density $f_D(d)$.
A: I assume that the radii $x$ of the smaller spheres satisfy $x<r/2$ (otherwise they intersect with probability $1$), and that their centers are chosen uniformly from the volume of the larger sphere. Notice that this would imply that the smaller spheres might not actually be fully contained inside of the larger sphere. If this is not what you meant, you should be able to tweak my solution to get what you want.
Let $C_1$ and $C_2$ be the randomly chosen centers of the small spheres. Let’s place $C_1$ first. Notice that its distance from the center from the large sphere has the following probability density function for $d\in [0,r]$:
$$f(d)=\frac{5d^4}{r^5}$$
This is because the probability of $C_1$ falling on one of the infinitesimal 4-spherical “shells” in the volume of the larger 5-sphere is proportional to its “surface area,” which is in turn proportional to $r^4$.
Now notice that the small spheres will intersect if and only if $C_2$ falls inside of the 5-sphere of radius $2x$ surrounding $C_1$. The probability of this occurring can be stated as follows. Define $g_5(r_1,r_2,d)$ to be the volume of the intersection of two 5-spheres of radii $r_1,r_2$ whose centers are separated by a distance of $d$. Then the probability of $C_2$ falling inside of the 5-sphere of radius $2x$ centered at $C_1$ is equal to
$$1-\frac{g_5(2x,r,d)}{\frac{8\pi^2}{15}r^5}$$
Then we have that the total probability of intersection is given by
$$\int_0^r \bigg(1-\frac{g_5(2x,r,d)}{\frac{8\pi^2}{15}r^5}\bigg) \cdot \frac{5d^4}{r^5}\space dd$$
Notice that $g_5(2x,r,d)$ simply equals $\frac{8\pi^2}{15}(2x)^5$ when $2x+d<r$, or when $d<r-2x$, so we can split this integral up as
$$\int_0^{r-2x} \bigg(1-\frac{\frac{8\pi^2}{15}(2x)^5}{\frac{8\pi^2}{15}r^5}\bigg) \cdot \frac{5d^4}{r^5}\space dd+\int_{r-2x}^r \bigg(1-\frac{g_5(2x,r,d)}{\frac{8\pi^2}{15}r^5}\bigg) \cdot \frac{5d^4}{r^5}\space dd$$
The leftmost integral is straightforward:
$$\bigg(1-\frac{\frac{8\pi^2}{15}(2x)^5}{\frac{8\pi^2}{15}r^5}\bigg) \cdot \frac{(r-2x)^5}{r^5}+\int_{r-2x}^r \bigg(1-\frac{g_5(2x,r,d)}{\frac{8\pi^2}{15}r^5}\bigg) \cdot \frac{5d^4}{r^5}\space dd$$
To solve the other integral, you’ll need to find a closed form expression for $g_5(2x,r,d)$, the hypervolume of the intersection of a 5-sphere of radius $2x$ and a 5-sphere of radius $r$ whose centers are separated by a distance of $d$. This is definitely possible, but it requires a lot of messy hand-calculation that I’m not willing to do. If you want, try to find a closed form for $g_5$, and then you can evaluate the integral in the above expression.
