On a disk, choose $n$ uniformly random points. Then draw the smallest circle enclosing those points. (Here are some algorithms for doing so.)
The circle may or may not lie completely on the disk. For example, with $n=7$, here are examples of both cases.
What is $\lim\limits_{n\to\infty}\{\text{Probability that the circle lies completely on the disk}\}$?
Is the limiting probability $0$? Or $1$? Or something in between? My geometrical intuition fails to tell me anything.
The case $n=2$
I have only been able to find that, when $n=2$, the probability that the smallest enclosing circle lies completely on the disk, is $2/3$.
Without loss of generality, assume that the perimeter of the disk is $x^2+y^2=1$, and the two points are $(x,y)$ and $(0,\sqrt t)$ where $t$ is uniformly distributed in $[0,1]$.
The smallest enclosing circle has centre $C\left(\frac{x}{2}, \frac{y+\sqrt t}{2}\right)$ and radius $r=\frac12\sqrt{x^2+(y-\sqrt t)^2}$. If the smallest enclosing circle lies completely on the disk, then $C$ lies within $1-r$ of the origin. That is,
$$\sqrt{\left(\frac{x}{2}\right)^2+\left(\frac{y+\sqrt t}{2}\right)^2}\le 1-\frac12\sqrt{x^2+(y-\sqrt t)^2}$$
which is equivalent to
$$\frac{x^2}{1-t}+y^2\le1$$
The area of this region is $\pi\sqrt{1-t}$, and the area of the disk is $\pi$, so the probability that the smallest enclosing circle lies completely on the disk is $\sqrt{1-t}$.
Integrating from $t=0$ to $t=1$, the probability is $\int_0^1 \sqrt{1-t}dt=2/3$.
Edit
From the comments, @Varun Vejalla has run trials that suggest that, for small values of $n$, the probability (that the enclosing circle lies completely on the disk) is $\frac{n}{2n-1}$, and that the limiting probability is $\frac12$. There should be a way to prove these results.
Edit2
I seek to generalize this question here.
First, let me state two lemmas that demand tedious computations. Let $B(x, r)$ denote the circle centered at $x$ with radius $r$.
Lemma 1: Let $B(x,r)$ be a circle contained in $B(0, 1)$. Suppose we sample two points $p_1, p_2$ inside $B(0, 1)$, and $B(x', r')$ is the circle with $p_1p_2$ as diameter. Then we have $$\mathbb{P}(x' \in x + dx, r' \in r + dr) = \frac{8}{\pi} r dxdr.$$
Lemma 2: Let $B(x,r)$ be a circle contained in $B(0, 1)$. Suppose we sample three points $p_1, p_2, p_3$ inside $B(0, 1)$, and $B(x', r')$ is the circumcircle of $p_1p_2p_3$. Then we have $$\mathbb{P}(x' \in x + dx, r' \in r + dr) = \frac{24}{\pi} r^3 dxdr.$$ Furthermore, conditioned on this happening, the probability that $p_1p_2p_3$ is acute is exactly $1/2$.
Given these two lemmas, let's see how to compute the probability in question. Let $p_1, \cdots, p_n$ be the $n$ points we selected, and $C$ is the smallest circle containing them. For each $i < j < k$, let $C_{ijk}$ denote the circumcircle of three points $p_i, p_j, p_k$. For each $i < j$, let $D_{ij}$ denote the circle with diameter $p_i, p_j$. Let $E$ denote the event that $C$ is contained in $B(0, 1)$.
First, a geometric statement.
Claim: Suppose no four of $p_i$ are concyclic, which happens with probability $1$. Then exactly one of the following scenarios happen.
There exists unique $1 \leq i < j < k \leq n$ such that $p_i, p_j, p_k$ form an acute triangle and $C_{ijk}$ contains all the points $p_1, \cdots, p_n$. In this case, $C = C_{ijk}$.
There exists unique $1 \leq i < j \leq n$ such that $D_{ij}$ contains all the points $p_1, \cdots, p_n$. In this case, $C = D_{ij}$.
Proof: This is not hard to show, and is listed on wikipedia.
Let $E_1$ be the event that $E$ happens and we are in scenario $1$. Let $E_2$ be the event that $E$ happens and we are in scenario $2$.
We first compute the probability that $E_1$ happens. It is $$\mathbb{P}(E_1) = \sum_{1 \leq i < j < k \leq n} \mathbb{P}(\forall \ell \neq i,j,k, p_\ell \in C_{ijk} , C_{ijk} \subset B(0, 1), p_ip_jp_k \text{acute}).$$ Conditioned on $C_{ijk} = B(x, r)$, Lemma 2 shows that this happens with probability $\frac{1}{2}r^{2(n - 3)} \mathbb{1}_{|x| + r \leq 1}$. Lemma 2 also tells us the distribution of $(x, r)$. Integrating over $(x, r)$, we conclude that $$\mathbb{P}(\forall \ell \neq i,j,k, p_\ell \in C_{ijk} , C_{ijk} \subset B(0, 1), p_ip_jp_k \text{acute}) = \int_{B(0, 1)}\int_0^{1 - |x|} r^{2(n - 3)} \cdot \frac{12}{\pi} r^3 dr dx.$$ Thus we have $$\mathbb{P}(E_1) = \binom{n}{3}\int_{B(0, 1)}\int_0^{1 - |x|} r^{2(n - 3)} \cdot \frac{12}{\pi} r^3 dr dx.$$ We can first integrate the $x$-variable to get $$\int_{B(0, 1)}\int_0^{1 - |x|} r^{2(n - 3)} \cdot \frac{12}{\pi} r^3dr dx = 12 \int_0^1 r^{2n - 3}(1 - r)^2 dr.$$ Note that $$\int_0^1 r^{2n - 3}(1 - r)^2 dr = \frac{(2n - 3)! * 2!}{(2n)!} = \frac{2}{2n * (2n - 1) * (2n - 2)}.$$ So we conclude that $$\mathbb{P}(E_1) = \frac{n - 2}{2n - 1}.$$ We next compute the probability that $E_2$ happens. It is $$\mathbb{P}(E_2) = \sum_{1 \leq i < j \leq n} \mathbb{P}(\forall \ell \neq i,j, p_\ell \in D_{ij} , D_{ij} \subset B(0, 1)).$$ Conditioned on $D_{ij} = B(x, r)$, this happens with probability $r^{2(n - 2)} \mathbb{1}_{|x| + r \leq 1}$. Lemma 1 tells us the distribution of $(x, r)$. So we conclude that the probability that $$\mathbb{P}(\forall \ell \neq i,j, p_\ell \in D_{ij} , D_{ij} \subset B(0, 1)) = \int_{B(0, 1)}\int_0^{1 - |x|} r^{2(n - 2)} \cdot \frac{8}{\pi} r dr dx.$$ So $$\mathbb{P}(E_2) = \binom{n}{2}\int_{B(0, 1)}\int_0^{1 - |x|} r^{2(n - 2)} \cdot \frac{8}{\pi} r dr dx.$$ We compute that $$\int_{B(0, 1)}\int_0^{1 - |x|} r^{2(n - 2)} \cdot \frac{8}{\pi} r dr dx = 8\int_0^1 r^{2n - 3} (1 - r)^2 dr = 8 \frac{(2n - 3)! 2!}{(2n)!}.$$ So we conclude that $$\mathbb{P}(E_2) = 8 \binom{n}{2} \frac{(2n - 3)! 2!}{(2n)!} = \frac{2}{2n - 1}.$$ Finally, we get $$\mathbb{P}(E) = \mathbb{P}(E_1) + \mathbb{P}(E_2) = \boxed{\frac{n}{2n - 1}}.$$ The proofs of the two lemmas are really not very interesting. The main tricks are some coordinate changes. Let's look at Lemma 1 for example. The trick is to make the coordinate change $p_1 = x + r (\cos \theta, \sin \theta), p_2 = x + r (-\cos \theta, -\sin \theta)$. One can compute the Jacobian of the coordinate change as something like $$J = \begin{bmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ \cos \theta & \sin \theta & -\cos \theta & -\sin \theta \\ -r\sin \theta & r\cos \theta & r\sin \theta & -r\cos \theta \\ \end{bmatrix}.$$ And we can compute that $|\det J| = 4r$. As $p_1, p_2$ has density function $\frac{1}{\pi^2} \mathbf{1}_{p_1, p_2 \in B(0, 1)}$, the new coordinate system $(x, r, \theta)$ has density function $$\frac{4r}{\pi^2} \mathbf{1}_{p_1, p_2 \in B(0, 1)}$$ The second term can be dropped as it is always $1$ in the neighbor of $(x, r)$. To get the density of $(x, r)$ you can integrate in the $\theta$ variable to conclude that the density of $(x, r)$ is $$\frac{8r}{\pi}$$ as desired.
The proof of Lemma 2 is analogous, except you can use the more complicated coordinate change from $(p_1, p_2, p_3)$ to $(x, r, \theta_1, \theta_2, \theta_3)$ $$p_1 = x + r (\cos \theta_1, \sin \theta_1), p_2 = x + r (\cos \theta_2, \sin \theta_2), p_3 = x + r (\cos \theta_3, \sin \theta_3).$$ The Jacobian $J$ is now $6$ dimensional, and Mathematica tells me that its determinant is $$|\det J| = r^3|\sin(\theta_1 - \theta_2) + \sin(\theta_2 - \theta_3) + \sin(\theta_3 - \theta_1)|.$$ So we just need to integrate this in $\theta_{1,2,3}$! Unfortunately, Mathematica failed to do this integration, but I imagine you can do this by hand and get the desired Lemma.