The limiting probability that three randomly chosen sticks can form a triangle is $1-1/\binom{2k}{k}$. Is there a combinatorial proof?

Question

Consider sticks of lengths $1^{1/k},\space 2^{1/k},\space 3^{1/k},\space\dots,\space n^{1/k}$ where $k,n\in\mathbb{Z^+}$.

Here is an example with $k=4$ and $n=50$.

I have discovered that, as $n\to\infty$, the probability that three randomly chosen sticks can form a triangle approaches $1-\dfrac{1}{\binom{2k}{k}}$.

Is there a combinatorial proof for this fact?

Non-combinatorial proof

Suppose, for a given $n$, we have already chosen $a^{1/k}$ and $b^{1/k}$, and we are about to choose $c^{1/k}$.

If $a^{1/k},b^{1/k},c^{1/k}$ can be the side lengths of a triangle, then $\left|a^{1/k}-b^{1/k}\right|<c^{1/k}<a^{1/k}+b^{1/k}$, that is, $\left|a^{1/k}-b^{1/k}\right|^k<c<\left(a^{1/k}+b^{1/k}\right)^k$.

So the probability that $a^{1/k},b^{1/k},c^{1/k}$ can be the side lengths of a triangle is

$$\frac{\min\left(\left(a^{1/k}+b^{1/k}\right)^k,n\right)-\left|a^{1/k}-b^{1/k}\right|^k}{n}$$

(I'm ignoring the fact that the numerator should be an integer, and the fact that we cannot choose the same stick twice; these facts will be irrelevant when we take the limit as $n\to\infty$.)

Then we take the average as $a$ and $b$ run from $1$ to $n$.

$$\frac{1}{n^2}\sum_{a=1}^n\sum_{b=1}^n\frac{\min\left(\left(a^{1/k}+b^{1/k}\right)^k,n\right)-\left|a^{1/k}-b^{1/k}\right|^k}{n}$$

As $n\to\infty$, the limiting probability is the integral

$$\int_0^1\int_0^1\left(\min\left(\left(x^{1/k}+y^{1/k}\right)^k,1\right)-\left|x^{1/k}-y^{1/k}\right|^k\right)\mathrm dx\mathrm dy$$

$$=\int_0^1\left(\int_0^{(1-y^{1/k})^k}\left(x^{1/k}+y^{1/k}\right)^k\mathrm dx+\int_{(1-y^{1/k})^k}^1 1\mathrm dx-\int_0^y\left(y^{1/k}-x^{1/k}\right)^k\mathrm dx-\int_y^1\left(x^{1/k}-y^{1/k}\right)^k\mathrm dx\right)\mathrm dy$$

which can be evaluated using hypergeometric functions, and simplifies to

$$1-\frac{1}{\binom{2k}{k}}$$

Desmos agrees.

In fact, $k$ does not have to be a positive integer; it can be any positive real number. (Fun fact: As $k\to 0$, the probability that the three sticks can form a triangle approaches $\dfrac{\pi^2}{6}k^2$.) But since I am looking for a combinatorial proof, I am restricting $k$ to positive integers.

Generalization

Numerical investigation suggests the following generalization (which I am not asking to prove):

As $n\to\infty$, the probability that $m$ randomly chosen sticks can form an $m$-gon approaches $1-\dfrac{k!^{m-1}}{((m-1)k)!}$.

Context

This is another question of mine about the probability that three sticks can form a triangle. My previous question was here.

Answer 1

Divide all the stick lengths by $n^{1/k}$. Then as $n → ∞$, the distribution of each stick choice approaches the same distribution as that of the largest of $k$ independent samples from the uniform distribution $U(0, 1)$. That is, if we select $3k$ independent $U(0, 1)$ samples $a_1, …, a_k, b_1, …, b_k, c_1, …, c_k$, then we seek the probability that $a = \max_i a_i$, $b = \max_i b_i$, and $c = \max_i c_i$ form a triangle. To see that the distributions have the same CDF:

$$\Pr\left[a ≤ \frac{m^{1/k}}{n^{1/k}}\right] = \Pr\left[a_1 ≤ \frac{m^{1/k}}{n^{1/k}} ∧ ⋯ ∧ a_k ≤ \frac{m^{1/k}}{n^{1/k}}\right] = \left(\frac{m^{1/k}}{n^{1/k}}\right)^k = \frac mn.$$

Now, with probability $\frac13$, $c$ is the largest of $a, b, c$ (the other cases are symmetric). Under this condition, let $x_i = \frac{a_i}{c}$, $y_i = 1 - \frac{b_i}{c}$, and observe that $x_1, …, x_k, y_1, …, y_k$ are themselves $2k$ independent $U(0, 1)$ samples. The only way for $a, b, c$ to fail to form a triangle is if $a + b < c \iff \frac ac < 1 - \frac bc$, or equivalently, all the $x_i$ are less than all the $y_i$. So we get a triangle with probability $1 - \frac{1}{\binom{2k}{k}}$.