Suppose you have $n$ triangles whose corners are random points on a sphere $S$
in $\mathbb{R}^3$.
Viewing the triangles as built from rigid bars as edges,
two triangles are *linked* if they cannot be separated without two
edges passing through one another.
A triangle that is not topologically linked with any other is *loose*.
In the example below of $n=15$ triangles, $11$ are linked to at least
one other triangle, and $4$ are loose.

^{ $n=15$. Magenta triangles $\{1,5,10,11\}$ are loose. }

It is easy to surmise that the proportion of linked triangles approaches $1$ as $n \to \infty$:

^{ Fraction of triangles linked to at least one other triangle. }

But I wonder about the largest

*linked component*, the collection of all triangles linked into one "giant" component, in the sense that if you picked up one triangle all the others would follow. I wonder that when $n \to \infty$, what is the probability that this linked component includes

*all*the triangles.

. As $n \to \infty$, what is the probability that every triangle is linked into one giant component?Q

My sense is that this probability is zero: Even though the probability that each triangle is linked to another approaches $1$, the probability that all triangles are linked to one another approaches $0$. This would contrast with an earlier related question, Random rings linked into one component?, whose answer was the opposite: The rings form one component as $n \to \infty$.

it could be removed without disturbing the others." That would be true if the bars were stretchable and bendable. With rigid bars you may easily have metric linkages that are not topological ones (link 2 large equilateral triangles near vertices and surround the linkage by a small equilateral triangle). $\endgroup$ – fedja Apr 28 '18 at 3:19Proc. London Math. Soc.(3), 81(2):485-512, 2000. "Under what conditions does a set of arcs in $\mathbb{R}^3$ have a large entangled subset?" $\endgroup$ – Joseph O'Rourke Apr 28 '18 at 15:38