Probability of no moves in solitaire I was playing a standard solitaire game on my mobile app and I came across a round where I couldn't perform a single move thus resulting in a loss. I was then thinking as to what the probability of this event to happen in a single game. Would anybody know this probability and show the calculations given a standard deck of 52 cards? 
Here is a link to the rules: https://www.wikihow.com/Play-Solitaire
Note: There are three cards turned at a time but you can only play the second card after you played the top card. Also, the game is klondike.
 A: Assuming Klondike, with 3 cards turned at a time:
http://www.techuser.net/klondikeprob.html

This gives 0.0025 (rounded to two significant figures) as an estimate of the probability of occurrence of unplayable games. This estimate suggests that on average 0.25 percent (one out of 400) of all Klondike games are unplayable.

Unfortunately the best way to get this probability is through statistical estimations rather than exact numerical calculations, but this gives a pretty good idea.
(Note: the way the question was worded, I was assuming you meant what is the probability you start out with an unwinnable hand, not necessarily ended up with an unplayable situation, since that requires assumptions about strategy of play.)
A: Incomplete estimate.
Of the seven cards displayed the first card can not be an Ace.  So there are 48 possible options. 
Other than aces, There are four cards that it can play on or be played on it. Unless the card is a king or a two in which case there are two. 
There will be 6 displayed cards and there are 10 ranks from 3 to Queen and 2 for twos and Kings.
For simplicity (I did say this was an incomplete estimate) let's assume the fourth displayed card is a two or king and the rest are not.
There are 51 cards for the second card.  It can't be an ace, and it can't be one of the 4 fours the first can play or be played on.  So there are 51-4-4=43 options.
For the third there are 50 minus the 4 aces minus the 8 the previous two can be played with.  That's 38.
For the fourth there are 49-4-12=33 options.
For the fifth there are 48-4-14=30.
For the sixth there are 47-4-18=25.
there are 24 well cards and if you flip 3 there are 8 cards that are potential plays.  None of them can be aces, and none of them can be the 22 cards that the six display cards can play with, and they must be different than the six display cards. So there are 20...13 possible options for those 8 playable wellcards.  
The remaining 38 cards can be any remaining cards.
So there are roughly, $48*43*38*33*30*25*\frac {20!}{12!}*38! $ ways to have no play, over $52! $ ways to deal the 52 cars.
So probability is roughly. $\frac {48*43*38*33*30*25*20*19*...*13}{52*....*39}$.  Which is really small.
I calculate it to be roughly $ =0.0000485894$ or $1$ in $20,000$ but the likelihood  I made an error is quite high.
Or so I calculate.  I could be wrong.
