Here's a neat little puzzle I thought of, though I still haven't figured out its answer.
Suppose you're racking up 15 billiard balls in one of the standard configurations (I'm not going to try to typeset these, so just picture an equilateral triangle instead of a right one):
S T S S E T T S T S S T S T T
where S is a "solid," T is a stripe, and E is the eight ball. A configuration is also standard if every S is mapped to a T, or if the triangle is reflected across its verticle axis of symmetry. Anyway, so you're racking up and you dump 15 balls into the rack randomly. Assuming you don't make any mistakes, what's the maximum number of two ball permuations necessary to get to any of the 4 standard configurations?