Let $p_n(x)$ be the probability that $n$ points in the unit square are at distances at least $x$ from each other. The expected minimum distance $E_n$ is $\int_0^\sqrt{2} p_n(x)dx$. If we can estimate $p_n(x)$, we get estimates on the integral.

Here are two crude estimates for $p_n(x)$. If we have drawn $k$ points, the probability that the next is within $x$ of one of the points is at most $k \pi x^2$, and at least $k \frac{\pi}4 (\frac{x}2)^2 = k \pi x^2/16$. The latter comes from putting a disk of radius $x/2$ at each of the previous points. These disks can't overlap, and at least a quarter of each disk is inside the square. We could improve this significantly on a torus.

$$\begin{eqnarray} p_n(x) & \ge & \prod_{k=0}^{n-1} \max(0,1-k\pi x^2) \newline &\ge &\prod_{k=0}^{n-1} \max(0,1-n\pi x^2)^{k/n} \newline & = & \max(0,1-n\pi x^2)^{(n-1)/2} \newline \int_0^\sqrt2 p_n(x)dx & \ge & \int_0^{1/\sqrt{n\pi}} (1-n\pi x^2)^{(n-1)/2 } dx. \end{eqnarray}$$

A trigonometric substitution works. Mathematica evaluates that integral as $$E_n \ge \frac{\Gamma((n+1)/2)}{n^{3/2} \Gamma(n/2)} = \frac{n\choose n/2} {2^n\sqrt{\pi}}.$$

$$\begin{eqnarray}p_n(x) & \le & \prod_{k=0}^{n-1} \max(0,1-k\pi x^2/16) \newline & \le & \prod_{k=0}^{n-1} \max(0,1-\pi x^2/16)^k \newline & = & \max(0,1-\pi x^2/16)^{n(n-1)/2}\newline \int_0^\sqrt{2} p_n(x)dx &\le & \int_0^{4/\sqrt{\pi}}(1-\pi x^2/16)^{n(n-1)/2}dx. \end{eqnarray}$$

Mathematica evaluates that integral as $$E_n \le \frac{2 \Gamma(1 + n(n-1)/2)}{\Gamma(3/2 + n(n-1)/2)} = \frac{2 \sqrt{t(n)}{t(n) \choose t(n)/2}}{2^{t(n)}\sqrt \pi}.$$

where $t(n) = {n \choose 2}$.

For $n=100$, the lower bound on the expected value of the minimum distance is $0.00705 \sim 1/142$. The upper bound on the expected value of the minimum distance is $0.0284 \sim 1/35$. For $n=1000$ we can sandwich the expected value $1/1415 \le E_{1000} \le 1/353$. So, these crude estimates get you within about a factor of $4$ of the expected value, and for large $n$, $1/n$ might be about right.

Let $p_n(x)$ be the probability that $n$ points in the unit square are at distances at least $x$ from each other. The expected minimum distance $E_n$ is $\int_0^\sqrt{2} p_n(x)dx$. If we can estimate $p_n(x)$, we get estimates on the integral.

Here are two crude estimates for $p_n(x)$. If we have drawn $k$ points, the probability that the next is within $x$ of one of the points is at most $k \pi x^2$, and at least $k \frac{\pi}4 (\frac{x}2)^2 = k \pi x^2/16$. The latter comes from putting a disk of radius $x/2$ at each of the previous points. These disks can't overlap, and at least a quarter of each disk is inside the square. We could improve this significantly on a torus.

Lower bound:

$$\begin{eqnarray} p_n(x) & \ge & \prod_{k=0}^{n-1} \max(0,1-k\pi x^2) \newline &\ge &\prod_{k=0}^{n-1} \max(0,1-n\pi x^2)^{k/n} \newline & = & \max(0,1-n\pi x^2)^{(n-1)/2} \newline \int_0^\sqrt2 p_n(x)dx & \ge & \int_0^{1/\sqrt{n\pi}} (1-n\pi x^2)^{(n-1)/2 } dx. \end{eqnarray}$$

A trigonometric substitution works. Mathematica evaluates that integral as $$E_n \ge \frac{\Gamma((n+1)/2)}{n^{3/2} \Gamma(n/2)} = \frac{{n\choose n/2}\sqrt{\pi}} {2^n 2\sqrt{n}} \sim \frac{1}{n\sqrt{2}}.$$

Upper bound:

$$\begin{eqnarray}p_n(x) & \le & \prod_{k=0}^{n-1} \max(0,1-k\pi x^2/16) \newline & \le & \prod_{k=0}^{n-1} \max(0,1-\pi x^2/16)^k \newline & = & \max(0,1-\pi x^2/16)^{n(n-1)/2}\newline \int_0^\sqrt{2} p_n(x)dx &\le & \int_0^{4/\sqrt{\pi}}(1-\pi x^2/16)^{n(n-1)/2}dx. \end{eqnarray}$$

Mathematica evaluates that integral as $$E_n \le \frac{2 \Gamma(1 + n(n-1)/2)}{\Gamma(3/2 + n(n-1)/2)} = \frac{2 \sqrt{\pi}{t(n) \choose t(n)/2}}{2^{t(n)}} \sim \frac{2\sqrt{2}}{n}.$$

where $t(n) = n^2 -n + 1$.

So, these crude upper and lower bounds only differ by a factor of $4$, and for large $n$, $1/n$ might be about right.

Let $p_n(x)$ be the probability that $n$ points in the unit square are at distances at least $x$ from each other. The expected minimum distance is $\int_0^\sqrt{2} p_n(x)dx$. If we can estimate $p_n(x)$, we get estimates on the integral.

Here are two crude estimates for $p_n(x)$. If we have drawn $k$ points, the probability that the next is within $x$ of one of the points is at most $k \pi x^2$, and at least $k \frac{\pi}4 (\frac{x}2)^2 = k \pi x^2/16$. The latter comes from putting a disk of radius $x/2$ at each of the previous points. These disks can't overlap, and at least a quarter of each disk is inside the square. We could improve this significantly on a torus.

$$\begin{eqnarray} p_n(x) & \ge & \prod_{k=0}^{n-1} \max(0,1-k\pi x^2) \newline &\ge &\prod_{k=0}^{n-1} \max(0,1-n\pi x^2)^{k/n} \newline & = & \max(0,1-n\pi x^2)^{(n-1)/2} \newline \int_0^\sqrt2 p_n(x)dx & \ge & \int_0^{1/\sqrt{n\pi}} (1-n\pi x^2)^{(n-1)/2 } dx. \end{eqnarray}$$

Mathematica evaluates that integral as $$\frac{\Gamma((n+1)/2)}{n^{3/2} \Gamma(n/2)}.$$

$$\begin{eqnarray}p_n(x) & \le & \prod_{k=0}^{n-1} \max(0,1-k\pi x^2/16) \newline & \le & \prod_{k=0}^{n-1} \max(0,1-\pi x^2/16)^k \newline & = & \max(0,1-\pi x^2/16)^{n(n-1)/2}\newline \int_0^\sqrt{2} p_n(x)dx &\le & \int_0^{4/\sqrt{\pi}}(1-\pi x^2/16)^{n(n-1)/2}dx. \end{eqnarray}$$

Mathematica evaluates that integral as $$\frac{2 \Gamma(1 + n(n-1)/2)}{\Gamma(3/2 + n(n-1)/2)}.$$

I might simplify those gamma expressions later.

For $n=100$, the lower bound on the expected value of the minimum distance is $0.00705 \sim 1/142$. The upper bound on the expected value of the minimum distance is $0.0284 \sim 1/35$. For $n=1000$ we can sandwich the expected value between $1/1415$ and $1/353$. So, these crude estimates get you within about a factor of $4$ of the expected value, and $1/n$ might be about right.

Let $p_n(x)$ be the probability that $n$ points in the unit square are at distances at least $x$ from each other. The expected minimum distance $E_n$ is $\int_0^\sqrt{2} p_n(x)dx$. If we can estimate $p_n(x)$, we get estimates on the integral.

Here are two crude estimates for $p_n(x)$. If we have drawn $k$ points, the probability that the next is within $x$ of one of the points is at most $k \pi x^2$, and at least $k \frac{\pi}4 (\frac{x}2)^2 = k \pi x^2/16$. The latter comes from putting a disk of radius $x/2$ at each of the previous points. These disks can't overlap, and at least a quarter of each disk is inside the square. We could improve this significantly on a torus.

$$\begin{eqnarray} p_n(x) & \ge & \prod_{k=0}^{n-1} \max(0,1-k\pi x^2) \newline &\ge &\prod_{k=0}^{n-1} \max(0,1-n\pi x^2)^{k/n} \newline & = & \max(0,1-n\pi x^2)^{(n-1)/2} \newline \int_0^\sqrt2 p_n(x)dx & \ge & \int_0^{1/\sqrt{n\pi}} (1-n\pi x^2)^{(n-1)/2 } dx. \end{eqnarray}$$

A trigonometric substitution works. Mathematica evaluates that integral as $$E_n \ge \frac{\Gamma((n+1)/2)}{n^{3/2} \Gamma(n/2)} = \frac{n\choose n/2} {2^n\sqrt{\pi}}.$$

$$\begin{eqnarray}p_n(x) & \le & \prod_{k=0}^{n-1} \max(0,1-k\pi x^2/16) \newline & \le & \prod_{k=0}^{n-1} \max(0,1-\pi x^2/16)^k \newline & = & \max(0,1-\pi x^2/16)^{n(n-1)/2}\newline \int_0^\sqrt{2} p_n(x)dx &\le & \int_0^{4/\sqrt{\pi}}(1-\pi x^2/16)^{n(n-1)/2}dx. \end{eqnarray}$$

Mathematica evaluates that integral as $$E_n \le \frac{2 \Gamma(1 + n(n-1)/2)}{\Gamma(3/2 + n(n-1)/2)} = \frac{2 \sqrt{t(n)}{t(n) \choose t(n)/2}}{2^{t(n)}\sqrt \pi}.$$

where $t(n) = {n \choose 2}$.

For $n=100$, the lower bound on the expected value of the minimum distance is $0.00705 \sim 1/142$. The upper bound on the expected value of the minimum distance is $0.0284 \sim 1/35$. For $n=1000$ we can sandwich the expected value $1/1415 \le E_{1000} \le 1/353$. So, these crude estimates get you within about a factor of $4$ of the expected value, and for large $n$, $1/n$ might be about right.