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Re: Billiards Puzzle
Posted:
Oct 1, 2004 4:47 AM


poopdeville@gmail.com (Acid Pooh) wrote in message news:<4765002.0409301719.70ddf1ba@posting.google.com>... > 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?
I don't understand your description of "standard configuration." What do you mean by "every S is mapped to a T"? There are seven of each so it is always possible to think of each S being paired with a T. Also, the two back corners cannot be the same (by the rules of the 8 ball) and hence, the rack is never symmetric across the vertical axis.



