Two Marks in the Same Region
A stick of length 1 is divided into $n$ equal regions, each of length $1/n$. Two marks are placed independently and uniformly at random along the stick.
Find the probability that both marks fall in the same region when $n = 5$.
Hints
- Once the first mark is placed, think about what probability the second mark lands in the same region -- which region the first landed in does not matter by symmetry.
- Each region has length $1/n$, so the second mark (uniform on $[0,1]$) lands in any specific region with probability $1/n$, regardless of where the first mark is.
- For $n = 5$: the probability is simply $1/5 = 0.2$. The general formula for $n$ equal regions is $P = 1/n$.
Worked Solution
How to Think About It: The first mark lands somewhere. Since all regions have equal length, the region it lands in does not matter -- by symmetry, it is equivalent to asking: given the first mark is in some region, what is the probability the second mark also falls there? The second mark is uniform on $[0, 1]$, and each region has length $1/n$, so this is just $1/n$.
Quick Estimate: With $n = 5$ equal regions of length $1/5$ each, the second mark has a $1/5 = 0.2$ chance of landing in the same region as the first. Sanity check: if $n = 1$ (one region covering the whole stick), probability is 1. If $n = 100$, probability is $1/100$. Both make sense.
Approach: Use symmetry to reduce to a single unconditional probability.
Formal Solution:
Let $X$ and $Y$ be the positions of the two marks, independently uniform on $[0, 1]$. The stick is divided into $n = 5$ regions: $[0, 1/5)$, $[1/5, 2/5)$, $[2/5, 3/5)$, $[3/5, 4/5)$, $[4/5, 1]$.
Method 1 (symmetry): The first mark $X$ falls in exactly one region. Since all regions have equal length $1/n$, by symmetry we can condition on $X$ being in, say, region $[0, 1/5)$. The probability that $Y$ also falls in this region is: $$P(Y \in [0, 1/5)) = \frac{1/5}{1} = \frac{1}{5}$$
By symmetry of the regions, this probability is the same regardless of which region $X$ landed in. Therefore: $$P(X, Y \text{ in same region}) = \frac{1}{5}$$
Method 2 (direct computation): Each mark independently falls in each of the 5 regions with probability $1/5$. The probability both fall in region $k$ is $(1/5)^2 = 1/25$. Summing over all 5 regions: $$P(\text{same region}) = 5 \times \frac{1}{25} = \frac{5}{25} = \frac{1}{5}$$
General formula: For $n$ equal regions, $P(\text{same region}) = 1/n$.
Answer: $P = \dfrac{1}{5} = 0.2$.
Intuition
This problem is a clean illustration of symmetry reducing a two-dimensional calculation to a trivial one-dimensional one. You do not need to integrate over the joint distribution of $(X, Y)$ -- you just observe that all regions are interchangeable, so conditioning on the first mark's region costs you nothing, and the answer is just the probability a uniform random variable lands in a fixed interval of length $1/n$.
The generalization is $P = 1/n$ for $n$ equal regions, which is also the expected fraction of collisions in a hash table with $n$ buckets and 2 keys. This birthday-problem style reasoning -- what is the probability two randomly chosen items hash to the same bucket -- underlies hash collision analysis, multi-dimensional histogram binning, and discretization effects in simulation. In trading, the analog is the probability two random trade arrivals fall in the same time bucket when you discretize a continuous-time model.