Break a stick at $n$ random points. What is the probability that the three shortest pieces can form a triangle, as $n\to\infty$?

Question

Question :

On a stick, choose $n$ uniformly random points, and break the stick at those points. What is the limit of the probability that the three shortest pieces can form a triangle, as $n\to\infty$?

Context :

I'm currently interested in questions about broken sticks. They often have nice answers.

My attempt :

I don't know how to find the exact probability, so I ran a simulation on Excel. For each of $n=10,n=100,n=1000$, the probability seems to be roughly $0.45$. So it seems that the probability quickly converges... but to what?

Answer 1

Suppose the stick is of length 1 and let $S_{(1)}< S_{(2)}<\dots<S_{(n)}$ be the lengths of the $n$ pieces ordered by magnitude.

We have (see 4.1)

$$ (S_{(1)}, \dots, S_{(n)})\stackrel{d}{=} (X_{(1)}/T_n, \dots, X_{(n)}/T_n),$$ Where $X_{(i)}$ are the order statistics from i.i.d. Exponential(1) random variables, and $T_n = \sum_{i=1}^n X_i$.

In particular, $$P(S_{(1)}+S_{(2)} > S_{(3)}) = P(X_{(1)}+X_{(2)} > X_{(3)}) =P(X_{(1)} > X_{(3)}-X_{(2)}).$$

However, $nX_{(1)}$ and $(n-2)(X_{(3)} - X_{(2)})$ are i.i.d. Exponential(1), hence $$P(X_{(1)} > X_{(3)}-X_{(2)}) = \frac{\frac{1}{n}}{\frac{1}{n}+\frac{1}{n-2}} = \frac{n-2}{2n-2}\rightarrow \frac{1}{2}.$$

Answer 2

Upon running more simulations, I agree with the comments. The probability tends towards $0.5$, as shown in the result graph below

$n = 100$" />

This simulation consisted of $10000 \cdot 10000$ (nested) iterations for every $n$; $\, 2 \le n \le 100$.

The interesting thing to analyse, is the first $10$ values for $n$; $\, 2 \le n \le 10$.

$n = 10$" />

This is where the most change occurs in probability, and funnily enough can be near perfectly mapped by the density function $$f(x) = \dfrac{1}{\pi}\cdot\tan^{-1}{(n - 1)}$$

Observe, it nicely obeys our convergence of $0.5$, as $n \to \infty$. However, I am unsure what other useful information this function gives us, other than the analytic continuation of the experiment over $\mathbb{R}$.

Answer 3

Earlier , when I saw this Question , I got the Answer $0.5$ but was hesitant to Post it , because the Simulation by OP was getting $0.45$ limit. With user "Jeroen van der Meer" & user "Dstarred" confirming $0.5$ limit , I think I can safely Post my Analysis.

(1) Shortest 3 Consecutive Pieces :

Let the Stick be broken & the Pieces arranged by Size.
Take the Shortest 3 Consecutive Pieces $(X,Y,Z)$ where $X \le Y \le Z$.
We want $X+Y \ge Z$
We can scale these ( by Dividing throughout by $Z$ ) to get $(x=X/Z,y=Y/Z,z=Z/Z=1)$

What we now want is :
$x+y \ge 1 \tag{C1}$
where
$x \le y \le 1 \tag{C2}$

[[
OBSERVATION 1 : In Case , we have the additional Criteria :
$X+Y+Z = Constant \tag{??C3??}$
then it is equivalent to this Case , where Probability is $1/4$ , though that Criteria is not valid here in the limiting Case.
]]

That Shaded Area ( larger triangle ) is satisfying (2) , while that Highlighted Area ( Smaller triangle ) satisfies (1) , hence that will give the necessary Probability $P=0.25/0.5=0.5$ , matching the Simulations.

[[
OBSERVATION 2 : Simulation Probability is moving from the Initial Case where we have the additional Criteria (C3) to the limiting Case where that Criteria (C3) is irrelevant. The Simulation is moving from $0.25$ ( $n$ close to $3$ ) to $0.5$ ( $n$ very large ) gradually.
]]

(2) Some 3 Consecutive Pieces , not necessarily the Shortest , not necessarily the longest :
Let the Pieces be $P_1,\cdots,P_n$ arranged by Size.
When $(P_1,P_2,P_3)$ can give the triangle , Probability is $P=1/2$.
When $(P_1,P_2,P_3)$ can not give the triangle , it is $(1-P)=1/2$ & we can check $(P_2,P_3,P_4)$ , $(P_3,P_4,P_5)$ , $(P_4,P_5,P_6)$ , $(P_5,P_6,P_7)$ , ETC.
Thus total Probability is $P + P \times (1-P) + P \times (1-P)^2+P \times (1-P)^3\cdots=P/(1-(1-P))=1$
Amazingly , we can guarantee a triangle somewhere among 3 Consecutive Pieces in the limiting Case , no matter what $P$ is !
In other words : (A) If we want acute-angled triangle with some non-Zero Probability , then we can guarantee that among 3 Consecutive Pieces when $n$ is very large. (B) If we want obtuse-angled triangle with some non-Zero Probability , then we can guarantee that among 3 Consecutive Pieces when $n$ is very large. (C) If we want all irrational sides of the triangle with some non-Zero Probability , then we can guarantee that among 3 Consecutive Pieces when $n$ is very large. Necessary Criteria is that $P$ must be non-Zero.