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Why can't I win at Solitaire?
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Oct 26, 2006, 01:15 PM
 
I've been playing Solitaire on my iPod for a couple of years, and I've only won once.

Someone told me that it is always possible to complete a game, no matter how the cards are dealt. Is this true? Or are there games that you can't win, no matter how you play?

Is there a trick to winning?

My rules are:

1. Always stack cards on the spaces if you can.

2. Always use cards from the tableaux if you can.

2. Only draw from the stock only if you have to.

Why isn't this working?
     
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Oct 26, 2006, 01:17 PM
 
Cheating always works for me.
     
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Oct 26, 2006, 01:17 PM
 
Beat me to it
     
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Oct 26, 2006, 01:18 PM
 
There's a cheat on the iPod?
     
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Oct 26, 2006, 01:20 PM
 
I dunno, that's just the standard response when someone asks for help with Solitaire
     
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Oct 26, 2006, 01:22 PM
 
I like my water with hops, malt, hops, yeast, and hops.
     
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Oct 26, 2006, 01:34 PM
 
That link is pretty good.

Most people are too set on building up those four ace-piles as fast as possible.

The point is to build them up as EVENLY as possible, because if, say, a three comes up but there's no longer a four to hook onto, because the other three piles are already at five, and you can't put down the three because the two is still missing, then that card is dead. You'll never be able to play it unless the two just *happens* to come up, by sheer luck.

Keep open as many options to stick cards at all time as possible. A card you can't put anywhere is a lost card.
     
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Oct 26, 2006, 01:37 PM
 
I actually hate the game. I play it because I want to learn to win so I can stop playing it.
     
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Oct 26, 2006, 01:38 PM
 
Originally Posted by Tiresias View Post
Someone told me that it is always possible to complete a game, no matter how the cards are dealt.
No, can't be. Except if the dealing mechanism excludes impossible dealings.

-t
     
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Oct 26, 2006, 01:40 PM
 
Originally Posted by Tiresias View Post
Someone told me that it is always possible to complete a game, no matter how the cards are dealt. Is this true?
A statement that contains the word "always" can be disproven by finding one single case in which the statement is false.

A typical deck of cards contains 52 cards. The tableau contains 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 cards, with 7 showing and the other 21 face-down. The stock thus contains 52 - 28 = 24 cards. Since you get to use every third card in the stock, you would see 24 / 3 = 8 usable cards in a single run-through if you didn't actually use any of those cards.

So. Suppose the 7 face-up cards in the tableau are the 2, 3, 4, 5, 6, 7, and 8 of hearts, and the 8 accessible cards in the stock are the 2, 3, 4, 5, 6, 7, 8, and 9 of diamonds. All available cards would be red, and you wouldn't even be able to make a single move. Thus, this game would be unwinnable, and therefore, the statement that it is always possible to complete a game is false.

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Oct 26, 2006, 01:44 PM
 
Originally Posted by CharlesS View Post
A statement that contains the word "always" can be disproven by finding one single case in which the statement is false.

A typical deck of cards contains 52 cards. The tableau contains 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 cards, with 7 showing and the other 21 face-down. The stock thus contains 52 - 28 = 24 cards. Since you get to use every third card in the stock, you would see 24 / 3 = 8 usable cards in a single run-through if you didn't actually use any of those cards.

So. Suppose the 7 face-up cards in the tableau are the 2, 3, 4, 5, 6, 7, and 8 of hearts, and the 8 accessible cards in the stock are the 2, 3, 4, 5, 6, 7, 8, and 9 of diamonds. All available cards would be red, and you wouldn't even be able to make a single move. Thus, this game would be unwinnable, and therefore, the statement that it is always possible to complete a game is false.
That's so...logical!
     
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Oct 26, 2006, 01:45 PM
 
I win at solitaire about half of the time. I use Macsolitaire.

MacSolitaire 1.6 - MacUpdate

Did you google that, CharlesS?
     
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Oct 26, 2006, 01:55 PM
 
Originally Posted by CharlesS View Post
A statement that contains the word "always" can be disproven by finding one single case in which the statement is false.

A typical deck of cards contains 52 cards. The tableau contains 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 cards, with 7 showing and the other 21 face-down. The stock thus contains 52 - 28 = 24 cards. Since you get to use every third card in the stock, you would see 24 / 3 = 8 usable cards in a single run-through if you didn't actually use any of those cards.

So. Suppose the 7 face-up cards in the tableau are the 2, 3, 4, 5, 6, 7, and 8 of hearts, and the 8 accessible cards in the stock are the 2, 3, 4, 5, 6, 7, 8, and 9 of diamonds. All available cards would be red, and you wouldn't even be able to make a single move. Thus, this game would be unwinnable, and therefore, the statement that it is always possible to complete a game is false.


Edit: But if the probability of such a hand is almost zero, the conjecture that "every game is winnable", while false in theory, might still be valid in practice.
( Last edited by Tiresias; Oct 26, 2006 at 02:09 PM. )
     
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Oct 26, 2006, 02:07 PM
 
Deleted by Tiresias.
     
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Oct 26, 2006, 02:11 PM
 
That's not the only combination that won't work, there are quite a few others.
     
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Oct 26, 2006, 06:33 PM
 
Originally Posted by Tuoder View Post
Did you google that, CharlesS?
Nope, although I did start up Solitaire on my iPod just to make sure it used the same rules as the ones I'm familiar with.

Originally Posted by Tiresias View Post
Edit: But if the probability of such a hand is almost zero, the conjecture that "every game is winnable", while false in theory, might still be valid in practice.
I neither know what percentage of possible hands are winnable nor care to spend the time to figure that out. That would be a far, far more difficult problem than simply disproving that simple statement.

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