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Freecell (Posted on 2006-09-06) Difficulty: 2 of 5
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.

No Solution Yet Submitted by Cory Taylor    
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re(2): Solution complications | Comment 5 of 6 |
(In reply to re: Solution by Tristan)

I agree with the basic solution of 52!/(4!^2).

However, complications ensue if we accept Tristan's alternative interpretation, whereby Red and Black can be switched without change, so can hearts and diamonds, clubs and spades.  One cannot simply divide by 2^3, because this undercounts the total number of different deals.  

Consider, for instance, if all the black suits are in sequence within suit.  Column 1 = A-7 of spades, column 2 = A-7 of clubs, column 5 = 8-K of spades, column 6 = 8-K of clubs.  We can already swap suits by swapping columns, and we have already accounted for all column swaps.  So we do not need to reduce the total number of deals further to allow for black suit symmetry in this case.

I'm not sure how much 52!/(4!^2*2^3) undercounts by, but I am sure that accounting for suit symmetries, the total number of essentially different deals is something more than 52!/(4!^2*2^3).
  Posted by Steve Herman on 2006-09-11 12:09:36

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