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I play a lot of Solitaire games on my Android phone and love to keep an eye out for the statistics.

Given that the Solitaire version lets you restart the game endlessly, I usually play until I solve it. But I never managed to solve more than 80% of the games played (1000+).

So now I wonder, is every Solitaire game solvable?

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  • 10
    I presume you mean Klondike solitaire?
    – McKay
    Jul 7, 2010 at 20:56
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    I have played thousands of solitaire games both on pc and the old fashioned way (yes with real cards) and have deduced that in order to find a solution to every game you play is to cheat.
    – TheX
    Dec 1, 2011 at 3:37
  • What a solitaire addict! Forever alone :)
    – kokbira
    Jun 25, 2012 at 18:36

7 Answers 7

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No. Example: If all of your cards face up on the board are red, and the cards that come up every third card are also red, and none of them are aces. You lose. Do not pass go, do not collect $200.

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    As a fact, I have come up with almost this exact setup on the computer version of Solitaire (but one card was black, just completely impossible to place anywhere).
    – Grace Note ♦
    Jul 7, 2010 at 21:02
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    Another example that just happened to me: All cards shown are even.
    – McKay
    Jul 14, 2010 at 15:48
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    en.wikipedia.org/wiki/Klondike_%28solitaire%29#Odds_of_winning
    – Margaret
    Aug 11, 2011 at 7:32
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    Even simplier: all aces are on the same column and 2 is above them.
    – SteeveDroz
    Dec 1, 2011 at 10:50
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    @Oltarus Aces in same column and 2 above them is still winnable. Its annoying and probably a loss but doable.
    – MaskedPlant
    Dec 1, 2011 at 14:33
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There is very interesting reading at wikipedia about this topic.

For a "standard" game of Klondike (of the form: Draw 3, Re-Deal Infinite, Win 52) the number of solvable games (assuming all cards are known) is between 82-91.5%.

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  • 16
    Then I was actually doing a great job nearing it to 80%
    – Ivo Flipse
    Jul 21, 2010 at 20:44
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Literally just played a game in which one of the stacks (the one containing 4 cards) was lead by the 9 of diamonds, and the cards inside of it were the King of Spades, the 5 of diamonds, the 10 of spades, and the 10 of clubs (I know this because I had the entire field solved except for this stack and used process of elimination). As far as I can see this makes the game impossible. I have a 9 of diamonds in which can never be moved, as the two 10s that it's eligible to rest upon are trapped underneath it in the stack face down. Attempting to get rid of the 9 by moving it to the diamond stack would also be fruitless, as the 5 of diamonds is stuck underneath it too. Unless someone can tell me some way that this could be solved, I'm pretty darned certain that if a card that is leading a stack is covering a stack that contains the two cards it is capable of resting on, and a lower number of it's own suit, then the game is made impossible right from the get-go.

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Solitaire is a game that precedes its computer version, and that means that all the cards are truly shuffled, without the computer peeking in to verify the game is solvable.

And like McKay mentioned, with a random shuffle you can definitely end up with an unsolvable game.

I'm sure it is possible to design a Solitaire variant in which each game is solvable, though.

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    Would need a LOT of calculation, basically the computer would have to play through an entire game to make sure there's a solution, unless there's some kind of algorithm I'm missing.
    – Arda Xi
    Jul 21, 2010 at 15:42
  • @Arda, there are some conditions that could be easily tested - for example, a card other than a King can be played on only three other cards in the deck (the next-lowest card in its suit, or the foundation for an Ace, and the next-higher cards of the opposite color). If all three of those cards are face down below that card on a pile, the game isn't winnable. Unfortunately I think that's a small percentage, and testing for other conditions might require a ton of recursion.
    – Dave DuPlantis
    May 25, 2011 at 21:05
  • @DaveDuPlantis True, but you will have to test for all of those conditions that exist. I'm not sure whether we even know all of them.
    – Arda Xi
    May 26, 2011 at 5:12
  • @Arda - that's true, that's what I was thinking with respect to recursion. Without some way to demonstrate that any given position is unwinnable, you'd essentially have to play a certain series of cards until you were blocked, back up to the last decision point, and repeat ... it's an intriguing concept, but I've never seen a solitaire program do that.
    – Dave DuPlantis
    May 26, 2011 at 12:31
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    @Arda Could simply work backwards from the solution, randomly moving cards into the deck and onto the board from the four suit piles, always using the reverse of a legal play. Probably won't have the same probability distribution as shuffling and checking for winnability, but I doubt that matters to most players.
    – warbaker
    Jun 5, 2012 at 16:52
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No. Eric Sink decided that he would start a micro-ISV to create a version of solitaire that is always winnable. This was mostly just an experiment to see what it would be like running a software company with one person, but he eventually sold the product which is still available for purchase.

There have been some estimates about the number of Klondike Solitaire games that are unplayable (no moves possible, about 1 in 400), and several guesses about how many games are unwinnable, although this percentage varies wildly from 30%-10%.

The difficulty of this problem stems from the sheer number of initial deals 54! that would need to be evaluated to determine which were winnable and which were not.

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  • would the number of initial deals be 52! ? (unless you expect the jokers to be dealt, too)
    – warren
    Dec 10, 2014 at 18:17
  • Fortunately, one does not need to use the brute force method (look at all possible deals) to compute the odds of winning (since that computation would take longer than the age of the universe - 8x10 to the 68th power decks). An analysis of ways to fail provides an analytical line of attack. As already noted there are clear ways that a single stack may fail. The needed cards might also be unreachable within two stacks, three stacks or four stacks. Once the conformations of cards for locking up the needed cards are known, their individual odds may be computed and combined to get an answer.
    – M Willey
    Dec 9, 2015 at 15:05
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However, if you started a list and enumerated the initial conditions -- I feel like I've seen this on a linux version of Solitare: the numbering of deck order, that is -- and you definitively decide a certain one is un-winnable, you then could compare notes across nodes (share with friends) and VOILA: a list of un-winnable starting deck stacks.

I've been starting to think the Windows 7 version has the un-winnable decks removed, ... I don't know, it's a little heavy-handed and smug about the statistics.

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  • With 52! starting shuffles, you'll need an... inconveniently long... time before you have a good list. Even after you solve the problem of determining unwinnable definitively.
    – Tynam
    May 24, 2012 at 7:59
  • 52 factorial = roughly 8 followed by 67 zeroes. That's a lot of combinations. A 1TB hard drive would store about a trillion of these, and you'd need trillions of terabytes to store even a decent fraction. Not very practical unfortunately, just because of the astronomical number of probabilities involved. Probably easier to just store a certain number of demonstrably winnable games.
    – doppelgreener
    Feb 24, 2014 at 7:54
  • @JonathanHobbs Not all of them have to be stored to make the calculation. for 1 to 52! getdeck, try solving game, add to statistics at each point only one deck need be stored, and the statistics can be quite small.
    – McKay
    Feb 24, 2014 at 13:36
  • @McKay You have to store quite a bit to develop a decent list, though. (I'm not sure which calculation you speak of.) As an aside also regarding the answer: the Windows 7 version actually just stores a few dozen thousand decks, and you're randomly given one each game. It might be they just picked a few dozen thousand decks known to be winnable.
    – doppelgreener
    Feb 24, 2014 at 13:43
  • @JonathanHobbs No, all you have to store is which deck you're looking at (which would need to go up to 52!, which means we'd need about 226 bits), and you'd need to store how many of them were solvable (another 226 bits, or less), and then one game of solitaire (which windows 3.1 was apparently able to store just fine), and the algorithm to actually solve the game. The data storage mechanisms do not need to be very much in order to do a full set of statistics on solvability. We're talking less than 1k of storage. Sure it would take a long time to do all these calculations. But not storage.
    – McKay
    Feb 24, 2014 at 13:57
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To add to the other great answers, this link has a nice explanation of how a deal is un-winnable. Reasons for Getting Stuck in Klondike Solitaire enter image description here The two black 5's are blocking a red 6 and a lower card in their suit. Thus they can not move to a different tableau pile and cannot move to the foundations.

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