Break a stick at two random points. The probability that the longest piece is at least twice as long as each of the other pieces is $1/2$. Why?

Question

Choose two independent uniformly random points on a stick, and break the stick at that those points. The probability that the longest piece is at least twice as long as each of the other pieces is $1/2$.

My proof below is fairly technical. Given the simplicity of the final answer, I am looking for a more intuitive explanation that does not involve so much calculation, possibly based on symmetry.

(Here is another question of mine about a breaking a stick at random points, that has an intuitive explanation. And here is one with a simple answer but perhaps no intuitive explanation.)

---

My non-intuitive proof:

Assume the stick has length $1$. Hold the stick horizontally.

$x=$ first chosen point's distance from left end.
$y=$ second chosen point's distance from left end.

In the graph, the blue regions contain combinations that meet the condition in the question. The red region contains combinations that do not meet the condition. The regions can be reflected across the diagonal, by symmetry.

I worked out the lines by considering $x$-values one by one, starting from $x=0$ and going up in small increments. Critical points emerged: $(0,\frac13),(0,\frac23),(\frac14,\frac12),(\frac14,\frac34),(\frac13,\frac13),(\frac13,1),(\frac12,\frac34),(\frac23,\frac23),(\frac23,1)$.

The areas of the blue and red regions are equal, so the probability that the condition is met is $1/2$.

---

Related fact: The probability that the longest piece is at least twice as long as the shortest piece is exactly $90\text{%}$ (an amusing geometrical probability).

Answer 1

Here’s another attempt at a derivation from the ternary graph with as little calculation as possible:

The dark region in the middle shows where the triangle is acute, i.e. contains its circumcentre. This is well-known to be $\frac14$ of the area, which is also straightforward to see in the graph, since this region has the same area as the three lighter regions. In that region, the largest side is less than $\frac12$ and thus not twice as large as the others. It remains to show that the condition also fails in $\frac13$ of the remaining, lighter areas. They decompose into six triangles, for instance the one at the lower right bounded by the black line and divided by the red line. The red line is the boundary for our condition, where one side is exactly twice another, so it begins $\frac23$ along the base line and divides the base of the small triangle, and thus also its area, in the ratio $1:2$, as required.

Answer 2

Self-answering. The following answer seems long, but it make intuitive sense, and the only calculation required is $\frac23\times\frac34=\frac12$.

As suggested in the comments by @joriki, I made a ternary plot. Each point represents a set of two randomly chosen points on the stick: the point's perpendicular distances to the edges of the graph are the fragment lengths. Once again, the blue regions contain combinations that meet the condition in the question, and the red region contains combinations that do not meet the condition.

Here is a close-up of the bottom-right region, whose upper-left vertex is the centre of the ternary graph.

In the close-up diagram, we could have worked out intuitively:

• the map of colors;
• the fact that the blue region's base is $\frac23$ of the combined red-blue region's base; and
• the fact that the blue region's height is $\frac34$ of the combined red-blue region's height

(which implies that the blue and red regions have equal area, so the probability in the OP is $\frac12$) by the following reasoning.

In the close-up diagram, the points on the bottom edge represent breakages in which one fragment has $0$ length. So on this line, the points that satisfy the condition in the question are such that the distance to the left edge of the entire ternary graph is at least twice the distance to the right edge. This implies that, in the close-up diagram, the blue region's base is $\frac23$ of the combined red-blue region's base.

Now consider a horizontal line slightly above the bottom edge. As we move this horizontal line up, the boundary point between red and blue has a straight line locus, by the previous reasoning.

Keep moving this horizontal line up, until we reach the horizontal line $0.25$ from the bottom edge (assuming the original stick had length $1$). This is a critical height. Consider a horizontal line $0.26$ above the bottom edge. It is no longer that case that, to meet the condition in the question, points only need their distance to the left edge be at least twice the distance to the right edge. So the blue region must have a vertex here.

Someone arguing based on the lower half of the right edge (instead of the right half of the bottom edge) would have arrived at this same vertex, so this vertex is at the top edge of the close-up diagram. In the close-up diagram, the combined red-blue region has height $\frac13$, so the height of the blue region is $\frac34$ the height of the combined red-blue region.

All of this implies that the probability in the OP is $\frac12$.

Answer 3

The probability of the occurence $$x_1+x_2+x_3=1, x_1\geq2x_2, x_1\geq 2x_3 $$ $(x_1,x_2,x_3\geq0)$ is the area of the region $$x_1\geq2x_2, 3x_2+2x_2\geq2$$ on the unit square $[0,1]\times[0,1]$ which is $\frac16$. Three times $\frac16$ is $\frac12.$

Answer 4

Nice question!

Here is quite a short method, making use of some basic properties of exponential random variables. It generalises nicely to other related questions (including the one mentioned at the bottom of the post).

The following fact is very useful. Let $S_1, S_2, \dots, S_n$ be the lengths (from left to right) when a unit stick is broken at $n-1$ independent and uniform points. That is, $(S_1, S_2, \dots ,S_n)$ is uniformly distributed on the simplex $\{(s_1, s_2, \dots, s_n)\in[0,1]^n: s_1+\dots+s_n=1\}$.

Then $S_1, S_2, \dots, S_n$ has the same distribution as $$ \frac{X_1}{X_1+X_2+\dots+X_n}, \frac{X_2}{X_1+X_2+\dots+X_n}, \dots, \frac{X_n}{X_1+X_2+\dots+X_n}, $$ where $X_1, X_2, \dots, X_n$ are i.i.d. uniform Exp($1$) random variables. See for example Chapter V.2 of Luc Devroye's book. $\newcommand{\Exp}{{\operatorname{Exp}}}$

Since we only care about ratios, we can ignore the denominators above. Let's take the original case $n=3$ for the moment. The question of whether one of $S_1, S_2, S_3$ is more than twice as big as each of the others becomes:

What is the probability that one of $X_1, X_2, X_3$ is more than twice as big as each of the others, where $X_1, X_2, X_3$ are i.i.d. $\Exp(1)$ random variables.

Rewrite $X_1, X_2, X_3$ in increasing order as $X_{(1)}\leq X_{(2)}\leq X_{(3)}$. These order statistics for an i.i.d. exponential sample have a very nice structure. The minimum of $n$ i.i.d. $\Exp(1)$ random variables has $\Exp(n)$ distribution. Applying this repeatedly, along with the memoryless property, we get that $(X_{(1)}, X_{(2)}-X_{(1)}, X_{(3)}-X_{(2)})=(E_3, E_2, E_1)$, where $E_3\sim\Exp(3)$, $E_2\sim\Exp(2)$, $E_1\sim\Exp(1)$, and $E_3, E_2, E_1$ are independent.

Rewriting, $(X_{(1)}, X_{(2)}, X_{(3)})=(E_3, E_2+E_3, E_1+E_2+E_3)$.

We want the probability that $X_{(3)}> 2 X_{(2)}$. This is the event $E_1>E_2+E_3$.

So finally we can rewrite the question as:

What is the probability that $E_1>E_2+E_3$, where $E_1, E_2, E_3$ are independent with $E_i\sim\Exp(i)$?

Now we get to calculate, applying the memoryless property yet again, along with the fact that if $Y\sim\Exp(\lambda)$ and $Z\sim\Exp(\mu)$, then $P(Y<Z)=\lambda/(\lambda+\mu)$: $$\begin{align} P(X_{(3)}>2X_{(2)})&= P(E_1>E_2+E_3) \\ &=P(E_1>E_3)P(E_1>E_2+E_3|E_1>E_3)\\ &=P(E_1>E_3)P(E_1>E_2)\\ &=\frac{3}{4}\times\frac{2}{3}\\ &=\frac{1}{2}, \end{align}$$ giving the desired answer!

---

Now for some generalisations. Using the same method, the probability that the largest of $n$ pieces is more than twice as large as each of the others is $$ \begin{align*} P(X_{(n)}>2X_{(n-1)})&= P(E_1>E_2+\dots+E_n) \\ &= \dots\\ & = \frac{n}{n+1}\times\dots\times\frac{3}{4}\times \frac{2}{3} \\ &= \frac{2}{n+1}. \end{align*} $$ (Again we had $E_i\sim \Exp(i)$ independently for different $i$.) For any $r>0$, we can also ask for the probability that the largest piece is more than $(1+r)$ times as large as each of the others. This comes out as $$ \begin{align*} P(\frac{1}{r}E_1>E_2+\dots+E_n) &= \frac{n}{n+r}\times\dots\times\frac{3}{3+r}\times\frac{2}{2+r}. \end{align*} $$ If $r$ is an integer, you get some nice cancellations. For example, for the probability that one piece is more than 3 times each of the others is $6/(n+1)(n+2)$. For general integer $r$ this seems to give $r!n!/(n+r-1)!$. (Are there nice geometric explanations of these facts?!)

More generally, what's the probability that the $k$th largest of $n$ is at least $(1+r)$ times as long as the $(k+1)$st largest? This seems to come out as $$ P(\frac{1}{r}E_k > E_{k+1} +\dots + E_n) = \prod_{i=k+1}^n \frac{i}{i+rk}. $$

At the other end we can also look at the nice $9/10$ fact mentioned in passing in the question! What's the probability that the $k$th largest is at least $(1+r)$ times the smallest? Looks like one gets $$ \begin{align*} P(E_n+\dots+E_k > (1+r)E_n) &=1-P(E_n > \frac{1}{r}(E_{n-1}+\dots+E_k))\\ &=1-\prod_{i=k}^{n-1}\frac{ri}{n+ri}. \end{align*} $$ For example putting in $r=1$, $n=3$, $k=1$, you get $9/10$ as expected.