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New Lincoln is closer to right than you are
The right answer is 52! divided by all of the symmetries. (with some minor corrections for the hands that are fixed under a given symmetry).

What are all of the symmetries?

Well you can choose to swap red and black, and then independently swap the two red and black suits. That's a factor of 8. In the standard configuration of freecell there are 8 piles, of which 4 have 6 cards and 4 have 7. So that is another factor of 24*24.

So the answer for the number of possible freecell games is approximately 52!/8/4!/4! ~ 1.75e64. Which in order of magnitude is a lot closer to 52! (8.07e67) than 13! (6.323e9).

Note that I say approximately because I did not calculate the correction terms for the freecell games which remain the same under some of those symmetries. This will increase the number of possible freecell games somewhat, but the correction will be relatively minor.

Cheers,
Ben
I have come to believe that idealism without discipline is a quick road to disaster, while discipline without idealism is pointless. -- Aaron Ward (my brother)
New That's unexpected (to me)
I figured you'd do the math, if you hadn't already done it for card games in general already. Did you find the answer surprising the first time you did it?
===

Purveyor of Doc Hope's [link|http://DocHope.com|fresh-baked dog biscuits and pet treats].
[link|http://DocHope.com|http://DocHope.com]
New Not really to me
You were talking about a very minor part of the total combinations. It's not like you were taking out 52x51x50x49 by subbing out colours and/or suits, you were taking out the 4x3x2x1 part.

That's why factorial complexity is so nasty; even if you think you've halved or quartered the problem, really you haven't.
--\n-------------------------------------------------------------------\n* Jack Troughton                            jake at consultron.ca *\n* [link|http://consultron.ca|http://consultron.ca]                   [link|irc://irc.ecomstation.ca|irc://irc.ecomstation.ca] *\n* Kingston Ontario Canada               [link|news://news.consultron.ca|news://news.consultron.ca] *\n-------------------------------------------------------------------
New Okay, that helps
Your explanation had the advantage of simplicity and clarity, but something still felt wrong. It took me a few tries thinking through Ben's explanation to finally come to the conclusion you gave.

If I had to pick a nit, it would be that yu have halved or quartered the problem. But looking at orders of magnitude, it didn't really help that much.
===

Purveyor of Doc Hope's [link|http://DocHope.com|fresh-baked dog biscuits and pet treats].
[link|http://DocHope.com|http://DocHope.com]
New Exactly, halving or quartering is not changing by an order o
of magnitude. Not till you get to 1/10th or less are you doing that. One way to think of it is that you have to move that decimal place over by more than one place before you're getting into changes of an order of magnitude, and even then you can still have a long way to go; 1e08 and 1e88 are eighty orders of magnitude apart, moving it one way or another by one isn't going to make that much of a difference... and factorial complexity moves up through those orders pretty damn quick once you get up into anything above about eight or so.
--\n-------------------------------------------------------------------\n* Jack Troughton                            jake at consultron.ca *\n* [link|http://consultron.ca|http://consultron.ca]                   [link|irc://irc.ecomstation.ca|irc://irc.ecomstation.ca] *\n* Kingston Ontario Canada               [link|news://news.consultron.ca|news://news.consultron.ca] *\n-------------------------------------------------------------------
     Looking for the Molson beer commercial - (lincoln) - (13)
         Try this - (imqwerky) - (12)
             THAT'S IT! (as I whack my forhead) - (lincoln) - (11)
                 no prob - (imqwerky) - (10)
                     I used to do that too - (lincoln) - (1)
                         The funny thing is - (imqwerky)
                     Another option - (lincoln) - (7)
                         Not really - (drewk) - (6)
                             Luckily - (imqwerky)
                             Lincoln is closer to right than you are - (ben_tilly) - (4)
                                 That's unexpected (to me) - (drewk) - (3)
                                     Not really to me - (jake123) - (2)
                                         Okay, that helps - (drewk) - (1)
                                             Exactly, halving or quartering is not changing by an order o - (jake123)

Because you should never underestimate the unoriginality of the very stupid.
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