Probability of antisymmetric nor symmetric Let $R$ be the set of all possible relation on set $A$. If you pick an element from $R$ uniformly at random, what is the probability that the picked one is neither antisymmetric nor symmetric?
 A: Let $A=\{1,2,3,\cdots,n\}$ WLOG.
Consider $(1,1), (2,2), (3,3), \cdots, (n,n)$. The relation can include any of these without making it antisymmetric or symmetric. There are $n$ of these pairs.
Consider $(a,b)$ and $(b,a)$ for $a \ne b$. If a relation is symmetric, then either both of them are included, or none of them is included, which gives $2$ possibilities. If a relation is anti-symmetric, then either none of them is included, or one of them is included, which gives $3$ possibilities. There are $\dfrac12(n^2-n)$ pairs of pairs of this kind.
The number of symmetric relations is $2^n 2^{\frac 1 2 (n^2-n)}$, while the number of anti-symmetric relations is $2^n 3^{\frac 1 2 (n^2-n)}$, and the number of symmetric and anti-symmetric relations is $2^{n^2}$.
Hence, the number of relations that are neither symmetric nor anti-symmetric is $2^{n^2} - 2^n \left[ 2^{\frac12(n^2-n)} + 3^{\frac12(n^2-n)} - 1 \right]$.
So, the required probability is $\dfrac{2^{n^2} - 2^n \left[ 2^{\frac12(n^2-n)} + 3^{\frac12(n^2-n)} - 1 \right]}{2^{n^2}}$.
Note that the probability $\to 1$ as $n \to \infty$, so if $A$ is an infinite set, you could say that the probability is $1$.
