In a finite nonabelian group, the probability that two randomly chosen elements commute cannot exceed 5/8. One easy proof also makes it easy to find the smallest groups that attain this bound, namely the two nonabelian groups of order 8:

This makes me wonder: **how large can the probability be that three randomly chosen elements $a,b,c$ of a finite nonassociative loop associate, i.e. obey $(ab)c = a(bc)$?**

You can prove the 5/8 theorem for groups by separately settling two questions:

What is the largest possible fraction of elements of a noncommutative finite group that lie in the center? (Answer: 1/4)

Given a noncentral element of a finite group, what's the largest possible fraction of elements that commute with it? (Answer: 1/2)

The nonabelian groups of order 8 achieve both these upper bounds. We could try a similar strategy for my question, attempting to settle these:

What is the largest possible fraction of elements $a$ in a finite loop such that $(ab)c = a(bc)$ for all elements $b,c$?

If an element $a$ of a finite loop does not have $(ab)c = a(bc)$ for all elements $b,c$, what is the largest possible fraction of elements $b$ such that $(ab)c = a(bc)$ for all $c$?

If a pair $a,b$ does not have $(ab)c = a(bc)$ for all elements $c$, what is the largest possible fraction of elements $c$ such that $(ab)c = a(bc)$?

Unfortunately I don't know how to settle these.

Since the quaternion 8-group $$Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$$ attains the 5/8 bound for commutativity of pairs in a nonabelian group, one might hope that the octonion 16-loop $$O_{16} = \{\pm 1, \pm e_1, \dots, \pm e_7\}$$ attains the maximum probability of associativity for triples in a nonassociative loop. **Does it?**

I'm afraid I haven't even worked out the probability that a triple in $O_{16}$ associates, though it would be easy to do.

nucleus(the associativity analogue of the centre) might work. I have no idea about the second part of the argument. – Robert Furber Sep 25 at 1:31