Probability of no friends meeting at a coffee house

Question

I came across what seems to be in my opinion a very challenging problem:

A number n of friends each visit Old Slaughter’s coffee house independently and uniformly at random during their lunch break from noon to 1pm. Each leaves after δ hours (or at 1pm if that is sooner), where δ < 1/(n−1). Show that the probability that none of them meet inside is $(1−(n−1)\delta)^n$.

I've tried to consider the events $A_{i,j}$ that correspond to "$i$ and $j$ meet inside the coffee house", and by calling the arrival times of the $k$-th person $U_k \sim Uniform([0,1])$, I can indeed compute the $\mathbb{P}(A_{i,j}) = \mathbb{P}(|U_i - U_j| < \delta)$ by using the continuous version of the total probability formula and chopping up my integral in several pieces : $\mathbb{P}(A_{i,j}) = \int_0^1 \mathbb{P}(U_j - \delta < U_i < U_j + \delta | U_j = x)dx = \int_0^1\int_{max(0,x-\delta)}^{min(1,x+\delta)}dydx$

etc...

But I don't think I'm on the right track because I've got no plan to go from $\mathbb{P}(A_{i,j})$ to $\mathbb{P}(\cap_{i<j}A_{i,j}^C)$. By peeking at the Grimmet & Stirzaker solution manual I get the impression that it might have something to do with the "order statistic" but it all seems pretty cryptic to me, humble probability novice.

Answer 1

First approach:

Selecting $n$ arrival points $x_1,...,x_n$ from the interval of $(0,1)$ such that the distance between each pair of consecutive points is at least $\delta$ is equivalent to select $n$ points $y_1,...,y_n$ from the interval $(0,1−(n−1)\delta)$; which means that all the $n$ persons arrive before $1−(n−1)\delta$. The one-to-one relation is given by

$$x_{(1)}=y_{(1)},x_{(2)}=y_{(2)}+\delta,...,x_{(n)}=y_{(n)}+\delta.$$

The probability of that the $n$ points $y_1,...,y_n$ are selected from the interval $(0,1−(n−1)\delta)$ is $$(1−(n−1)\delta)^n.$$

Second approach:

Another method is to directly compute the probability

$$\mathbb P \left ( X_{(2)} \ge X_{(1)}+\delta, \dots ,X_{(n)} \ge X_{(n-1)} +\delta\right ).$$

For $0 \le x_1 \le x_2\dots\le x_n \le 1$, we have

$$f_{X_{(1)},\dots,X_{(n)}}(x_1,...,x_n)= n!f_{X_1}(x_1)f_{X_2}(x_2)\dots f_{X_n}(x_n)=n! .$$

Hence, the probability is

$$\int_0^{1−(n−1)\delta} \int_{x_1+\delta}^{1−(x_1+\delta)-(n−2)\delta} \dots \int_{x_{n-1}+\delta}^{1} n! \text{d}x_n \text{d}x_{n-1} \dots \text{d}x_1. $$

For $n=2$, we have

$$\int_0^{1−\delta} \int_{x_1+\delta}^{1} 2! \; \text{d}x_2 \text{d}x_1 =\int_0^{1−\delta} 2!(1-x_1-\delta) \text{d}x_1= - 2\frac{(1-\delta-x_1)^2}{2} \Big |_0^{1-\delta}=(1-\delta)^2.$$

Solving the general case needs more work, which can be done by induction. Now assume that the probability is $(1-(n−2)\delta)^{n-1}$ for $n-1$. This implies that when the interval $(a,1)$ is considered instead of $(0,1)$, the probability is $\left (\frac{1-a-(n−2)\delta}{1-a} \right )^{n-1}$ , which yields

$$\int_a^{1-a−(n−2)\delta} \int_{x_1+\delta}^{1−(x_1+\delta)-(n−3)\delta} \dots \int_{x_{n-2}+\delta}^{1} \frac{(n-1)!}{(1-a)^{n-1}} \text{d}x_{n-1} \text{d}x_{n-2} \dots \text{d}x_1=\left (\frac{1-a-(n−2)\delta}{1-a} \right )^{n-1}. $$

Thus, using the above for $x_2,...,x_n$ with $a=x_1+\delta$, the statement for $n$ can be proven as follows:

$$\int_0^{1−(n−1)\delta} \int_{x_1+\delta}^{1−(x_1+\delta)-(n−2)\delta} \dots \int_{x_{n-1}+\delta}^{1} n! \text{d}x_n \text{d}x_{n-1} \dots \text{d}x_1 = \int_0^{1−(n−1)\delta} \frac{n!(1-(x_1+\delta))^{n-1}}{(n-1)!} \left (\frac{1-(x_1+\delta)-(n−2)\delta}{1-(x_1+\delta)} \right )^{n-1} \text{d}x_1 =- n\frac{(1-(x_1+\delta)-(n−2)\delta)^n}{n} \Big |_0^{1-(n−1)\delta}=(1-(n−1)\delta)^n.$$