**Definitions:**

Lagrange's theorem implies that for each prime $p$, the factors of $(p − 1)!$ can be arranged in unequal pairs, with the exception of $±1$, where the product of each pair $≡ 1 \pmod p$. *See Wiki article on Wilson's theorem.*

From the example in the link above, for $p=11$ we have

$$(11-1)!=[(1\cdot10)]\cdot[(2\cdot6)(3\cdot4)(5\cdot9)(7\cdot8)] \equiv [-1]\cdot[1\cdot1\cdot1\cdot1] \equiv -1 \pmod{11}$$

Let the products of the pairs that $≡ 1 \pmod p$ be the multiset $A_p$, and $A_{p_n}$ the multiset for the $n$th prime.

For the above example then, $A_{p_5}=\{(2\cdot6),(3\cdot4),(5\cdot9),(7\cdot8)\}=\{12,12,45,56\}$.

**Conjecture:**

$$\lim\limits_{n\rightarrow\infty}\dfrac{\sum\limits_{k \in A_{p_n}}(k-1)}{(p_n)^3}\approx\frac18$$

where $p_n$ is the $n$th prime.

**Examples:**

For $p=11$ we have

$$\dfrac{11+11+44+55}{11^3}=\dfrac{1}{11}$$

For $p=997$ we have

$$\dfrac{123218233}{997^3}=\dfrac{123218233}{991026973}$$

**Comments:**

As @YCor noted below, the $-1$ in the $k-1$ can be removed, since its contribution tends to $0$. The conjecture can therefore be simplified to

$$\lim\limits_{n\rightarrow\infty}\dfrac{\sum\limits_{k \in A_{p_n}}k}{(p_n)^3}\approx\frac18$$

I have no idea whether the above statement is correct, or how to go about trying to find a proof. Any comments on the any of the above are most welcome.