Another proof of Fibonacci sequence in classic probability puzzle

Recently Scientific American reported that Fibonacci sequence appears as an answer to the following problem:

If I have n sticks with random lengths between 0 and 1, what is the probability that no three of those sticks can form a triangle?

The answer to this puzzle is $\Pi_{i=1}^n \frac{1}{F_i} = \frac{1}{1} \frac{1}{1} \frac{1}{2} \dots \frac{1}{F_n}$ where $F_i$ is the i-th Fibonacci number. This is the Arxiv link for the paper which contains a proof of it. Someone posted a novel proof of this puzzle which is more linear algebraic in nature. I made a small modification on top of this proof which is what I present here. Most of the credit to this proof goes to the Reddit user.

0. Outline of the proof

Here is the overall idea - sort the lengths $l_i$ from shortest to longest. The vector of sorted lengths can be shown to be a linear combination of a certain basis. Once we get this basis, we use it express the longest length $l_n$ in terms of it. Finally we use the condition that $\text{longest length} \le 1$ to get a simplex.

These steps are implemented below:

1. Steps 1 and 2 - Expressing sorted stick lengths as a matrix-vector product

We do this for two cases - the case where the lengths can not form a triangle and the general case (the lengths may or may not form a triangle).

1.1 Lengths which cannot form a triangle

Given the shortest length is $l_1 = \varepsilon_1$, we have that $l_2 = l_1 + \varepsilon_2$. Note that $l_3$ has to be bigger than $l_1 + l_2 = 2l_1 + \varepsilon_2$. We can assume that $l_3 = 2l_1 + \varepsilon_2 + \varepsilon_3$. Using the same argument, the next three lengths can be written down as:

Thus the vector of sorted lengths can be written down as:

1.2 Any arbitrary vector of lengths

This case is much simpler. As before, start with $l_1 = \varepsilon_1$. Then since $l_2 \ge l_1$, we have that $l_2 = l_1 + \varepsilon_2$ and so on. For sake of completion, the remaining lengths look something like:

In this case, the vector of sorted lengths can be written down as:

2. Step 3 - largest length in each case

2.1 First case

In this case, the largest length is $\sum_{i=1}^n \textcolor{#b45309}{F_i}y_i$ for positive coefficients $y_i$.

2.2 Second case

In this case, the largest length is $\sum_{i=1}^n x_i$ for positive coefficients $x_i$.

3. Step 4 - obtain simplexes for each case

Since these lengths are sampled from the interval $[0,1]$, they have to be less than or equal to one. Applying this condition to each of the largest lengths above we get:

i) $\sum_{i=1}^n \textcolor{#b45309}{F_i} y_i \le 1$: the points that satisfy this condition form a volume $V_1$. This is the volume of the simplex formed by origin and vectors $\underbrace{\frac{1}{F_i} \textcolor{#b45309}{e_i}}_{\text{standard basis vector scaled by} \ 1/F_i}$ for $i$ from 1 to n.

ii) $\sum_{i=1}^n x_i \le 1$: this is the standard simplex and has a volume $V_2$ of $\frac{1}{n!}$. The exact volume is not needed for the proof though.

4. Final step

The probability is given by $\frac{V_1}{V_2}$. How do we find $V_1$ in terms of $V_2$? By defining a map $y_i = x_i / F_i$. It is easy to see that the determinant of Jacobian of this map is $|J| = \Pi_i \frac{1}{F_i}$. Thus the volume $V_1 = |J|V_2$. This gives us the probability to be: