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I found the following example due to J. Jezek and T. Kepka from "Notes on the number of associative triples" Acta Universitatis Carolinae 31 (1990), 15-19 (Example 2.1):

Suppose $Q(+)$ is an abelian group of even order $n\geq 6$. Let $a,b\in Q-\{0\}$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $x\notin \{b,a+b\}$ or $y\notin \{b,a+b\}$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.

 

Then $Q(\cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.

Therefore the probability that three randomly chosen elements associate can be arbitrarily close to 1.

I found the following example due to J. Jezek and T. Kepka from "Notes on the number of associative triples" Acta Universitatis Carolinae 31 (1990), 15-19 (Example 2.1):

Suppose $Q(+)$ is an abelian group of even order $n\geq 6$. Let $a,b\in Q-\{0\}$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $x\notin \{b,a+b\}$ or $y\notin \{b,a+b\}$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.

Then $Q(\cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.

Therefore the probability that three randomly chosen elements associate can be arbitrarily close to 1.

I found the following example due to J. Jezek and T. Kepka from "Notes on the number of associative triples" Acta Universitatis Carolinae 31 (1990), 15-19:

Suppose $Q(+)$ is an abelian group of even order $n\geq 6$. Let $a,b\in Q-\{0\}$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $x\notin \{b,a+b\}$ or $y\notin \{b,a+b\}$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.

Then $Q(\cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.

Therefore the probability that three randomly chosen elements associate can be arbitrarily close to 1.

I found the following example due to J. Jezek and T. Kepka from "Notes on the number of associative triples" Acta Universitatis Carolinae 31 (1990), 15-19 (Example 2.1):

Suppose $Q(+)$ is an abelian group of even order $n\geq 6$. Let $a,b\in Q-\{0\}$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $x\notin \{b,a+b\}$ or $y\notin \{b,a+b\}$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.

Then $Q(\cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.

Therefore the probability that three randomly chosen elements associate can be arbitrarily close to 1.

I found the following example due to J. Jezek and T. Kepka:

Suppose $Q(+)$ is an abelian group of even order $n\geq 6$. Let $a,b\in Q-\{0\}$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $x\notin \{b,a+b\}$ or $y\notin \{b,a+b\}$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.

Then $Q(\cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.

Therefore the probability that three randomly chosen elements associate can approach 1.

I found the following example due to J. Jezek and T. Kepka from "Notes on the number of associative triples" Acta Universitatis Carolinae 31 (1990), 15-19:

Suppose $Q(+)$ is an abelian group of even order $n\geq 6$. Let $a,b\in Q-\{0\}$ be two distinct elements with $2a=0$. Define a new operation on $Q$ by $xy=x+y$ as long as either $x\notin \{b,a+b\}$ or $y\notin \{b,a+b\}$, and $bb=(a+b)(a+b)=2b+a$ together with $b(a+b)=(a+b)b=2b$.

Then $Q(\cdot)$ is a commutative loop with exactly $n^3-16n+64$ associative triples.

Therefore the probability that three randomly chosen elements associate can be arbitrarily close to 1.