Let \(\alpha \) be a type, let \(k\) be a natural number, and let \(x, y \in \alpha \). Writing a hash function as a map \(h : \alpha \to \mathrm{Fin}\, k\), the collision event on which \(x\) and \(y\) receive the same hash value decomposes as the union of collision pieces over all possible common values:

Fix a type \(\alpha \), a natural number \(k\), and two elements \(x, y : \alpha \). For each value \(a \in \{ 0, \dots , k-1\} \), let \(C_a\) denote the set of hash functions \(h : \alpha \to \{ 0, \dots , k-1\} \) with \(h(x) = a\) and \(h(y) = a\). Then the family \((C_a)_a\) is pairwise disjoint: whenever \(a \ne b\), the sets \(C_a\) and \(C_b\) have empty intersection.

Let \(k \ge 1\) be a natural number and let \(a\) be an element of the finite type \(\mathrm{Fin}\, k = \{ 0, 1, \dots , k-1\} \), equipped with its canonical (uniform) volume measure. Then the measure of the singleton \(\{ a\} \) equals \(1/k\), as an element of the extended nonnegative reals \([0, \infty ]\).

Let \(k \ge 1\), and equip the finite set \(\mathrm{Fin}\, k = \{ 0, 1, \dots , k-1\} \) with its uniform probability measure. Then for every element \(a \in \mathrm{Fin}\, k\), the real-valued measure of the singleton \(\{ a\} \) equals \(1/k\).

Let \(\alpha \) be a finite type and let \(k \ge 1\). Fix two distinct elements \(x \ne y\) of \(\alpha \) and a value \(a \in \mathrm{Fin}\, k\), and equip the space of hash functions \(\alpha \to \mathrm{Fin}\, k\) with the product of the uniform probability measures on the coordinates. Then the collision piece consisting of those hash functions \(h\) with \(h(x) = a\) and \(h(y) = a\) has measure