To deal with the $\frac1n$ issue and get the exact result, let's in fact choose $n$ iid uniform postions on a circle circumference $1$ (equivalent in the birthday problem to days in a year), then pick one of them as the starting point and measure the distances clockwise from there: this is equivalent to choosing $n-1$ iid uniform numbers in $(0,1)$ and sorting them into $0 \le x_{(1)} \le x_{(2)}\le \cdots \le x_{(n-1)} \le 1$, and then using $x_{(0)}=0$ and $x_{(n)}=1$ so $n$ gaps $G_i= x_{(i)}-x_{(i-1)}$.
The $n$ gaps $G_i$ are identically distributed and sum to $1$ so they each have expectation $E[G_i]=\frac1n$. We can state their distributions as $P(G_i \le y) = 1-(1-y)^{n-1}$ for $0 < y <1$ so with density $(n-1)(1-y)^{n-2}$ and expectation $\frac 1n$.
The $G_i$ are not independent of each other as they sum to $1$ and so the smallest cannot exceed $\frac1n$, but as $n$ increases the relationship between them becomes weaker. Going back to the circle analogy, the probability the gaps all exceed some $y$ is $(1-ny)^{n-1}$ since you are excluding a combined $ny$ from the circumference of $1$ for the location of the $n-1$ points, so $P(\min(G_i) \le y)=1 - (1-ny)^{n-1}$ with density $n(n-1)(1-ny)^{n-2}$ when $0 < y < \frac1n$ and expectation $E[\min(G_i)]=\frac1{n^2}$, which is what you are looking for.
There is also a convergence property for large $n$:
- If you were to consider $nG_i$ then you would get convergence in distribution to $Exp(1)$ as $n$ increases, because $P(nG_i \le z) = 1-\left(1-\frac zn\right)^{n-1} \to 1-e^{-z}$ for $z>0$. An earlier question was similar.
- Similarly if you were to consider $n^2\min(G_i)$, it also converges in distribution to $Exp(1)$ as $n$ increases because $P(n^2\min(G_i) \le w)= 1-\left(1-\frac wn\right)^{n-1} \to 1-e^{-w}$ for $w>0$ too.
Here is a simulation in R to confirm this expectation, using $n=10$.
mingap <- function(n){
x <- runif(n-1)
gap <- diff(c(0,sort(x),1))
return(min(gap))
}
set.seed(2024)
n <- 10
sims <- replicate(10^5, mingap(n))
mean(sims) # expected 1/n^2
# 0.01000694
The black curve below is the empirical CDF from the simulation, with the red theoretical CDF $1 - (1-ny)^{n-1}$ almost exactly matching it and so obscuring it. The grey exponential CDF for $Exp(n^2)$ gives a fairly close approximation.
plot.ecdf(sims)
curve(1-(1-n*x)^(n-1),from=0, to=1/n, add=TRUE, col="red" )
curve(1-exp(-x*n^2),from=0, to=1/n, add=TRUE, col="grey" )
![CDFs](https://cdn.statically.io/img/i.sstatic.net/QyvsH.png)