I have an addiction of sorts - I can't keep from playing Freecell. (Most "Windows" users have access to this game from their "Start" menu.) There was once a theory that every possible deal is winnable in this game (this has apparently been disproven).
How many essentially different deals are there in Freecell?
Freecell setup: deal a standard 52 card deck out into eight columns: four with seven cards and four with six. Two deals with only column order changed (i.e., that can be made identical by only switching the locations of particular columns) are not considered different in this context.
(In reply to Solution
by Old Original Oskar!)
This doesn't count for symmetries between the suits. Red and Black can be switched without change, so can hearts and diamonds, clubs and spades. So the modified solution is 52!/(4!^2*2^3). This, of course, is a matter of interpretation.
Posted by Tristan
on 2006-09-06 14:04:26