# Marking And Breaking Sticks

### Problem

A person makes two marks - randomly and independently - on a stick, after which the stick is broken into $n$ pieces. What is the probability that the two marks are found on the same piece?

Compare two cases: when the pieces are equal and when the division is random.

As usual, when no distribution is specified the word "random" refers to the uniform distribution. "Independently" means *independent* of any previous action. This is especially important in the the second part of the problem. To avoid ambiguity, assume that, prior to breaking the stick, the $n-1$ marks are made *randomly* and *independently* of all the marks already made.

### Hint

For the first part, think of the probability of the second mark falling onto the piece which contains the first mark. The second part is rather combinatorial. In all there are $n+1$ marks; of interest are those markings in which the first two are located successively, with no "break" marks between them.

### Solution

For the case of equal pieces, one of the marks ought to be located on one of the pieces. The second mark is located on the same piece with the probability of $1/n.$

For the random lengths, imagine that first $n-1$ marks are placed at the break points, making the total number of marks $n+1.$ There are ${n+1\choose 2}$ ways to pick $2$ marks out of $n+1.$ Of interest are those in which two original marks follow each other, with no "break" marks in-between. In other words, if the marks are numbered from $1$ through $n+1,$ we are interested in the distributions where the original marks bear successive numbers: $1$ and $2,$ or $2$ and $3,\ldots,$ or $n$ and $n+1.$ There are $n$ such cases (out of ${n+1\choose 2}),$ implying that the thought probability is

$\displaystyle\frac{n}{{n+1\choose 2}}=\frac{n\cdot 2!(n-1)!}{(n+1)!}=\frac{2}{n+1},$

showing a rather remarkable increase compared to the case of equal pieces.

### Question to ponder

We actually had two problems, each with its own solution. It is obvious that the solution of the first problem could not apply to resolve the second one. However, it is worth giving a thought to the question whether the solution to the second problem could not be used to solve the first one. If it could, then the two answers conflict with each other. If it could not, then it is reasonable to inquire, *Why?*

### References

- Paul Nahin,
*Will You Be Alive In 10 Years From Now?*, Princeton University Press, 2013 (36-41)

### Geometric Probability

- Geometric Probabilities
- Are Most Triangles Obtuse?
- Barycentric Coordinates and Geometric Probability
- Bertrand's Paradox
- Birds On a Wire (Problem and Interactive Simulation)
- Buffon's Noodle Simulation
- Averaging Raindrops - an exercise in geometric probability
- Rectangle on a Chessboard: an Introduction
- Marking And Breaking Sticks
- Random Points on a Segment
- Semicircle Coverage
- Hemisphere Coverage
- Overlapping Random Intervals
- Random Intervals with One Dominant
- Points on a Square Grid
- Flat Probabilities on a Sphere
- Probability in Triangle

|Contact| |Front page| |Contents| |Algebra| |Probability|

Copyright © 1996-2018 Alexander Bogomolny71741227