TITLE: Tangled random triangles: One giant component?
QUESTION [10 upvotes]: 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.

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

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 $n \to \infty$.

REPLY [6 votes]: Here is @fedja's clever example: The magenta triangle is topologically loose
but metrically "stuck":

          


          

fedja: "(link two large ... triangles [blue & green] near vertices


          

and surround the linkage by a small ... triangle [magenta])"