Odd bean among twelve with three weighings

Twelve beans look identical, but exactly one has a different weight, either heavier or lighter than the rest (you are not told which). Using only a two-pan balance, what is the least number of weighings that always suffices to single out the odd bean?

$3$. Weigh four against four. If balanced, the odd bean is among the remaining four and two more weighings against known-good beans find it. If unbalanced, the tilt direction tells you each of the eight weighed beans is either 'possibly heavy' or 'possibly light'; a clever second weighing that swaps and rotates beans between pans splits these into groups of at most three, and a third weighing settles it. Two weighings give only $9$ outcomes, fewer than the $24$ cases, so $3$ is optimal.