Probability of "clock patience" going out Here is a question I've often wondered about, but have never figured out a satisfactory answer for. Here are the rules for the solitaire game "clock patience." Deal out 12 piles of 4 cards each with an extra 4 card draw pile. (From a standard 52 card deck.) Turn over the first card in the draw pile, and place it under the pile corresponding to that card's number 1-12 interpreted as Ace through Queen. Whenever you get a king you place that on the side and draw another card from the pile. The game goes out if you turn over every card in the 12 piles, and the game ends if you get four kings before this happens. My question is what is the probability that this game goes out?
One thought I had is that the answer could be one in thirteen, the chances that the last card of a 52 card sequence is a king. Although this seems plausible, I doubt it's correct, mainly because I've played the game probably dozens of times since I was a kid, and have never gone out!
Any light that people could shed on this problem would be appreciated!

As Ross said, you cannot get out if the bottom card matches it position, because there is no way to access it.  At best, the probability of each bottom card being different from its position is 12/13.  So the chance of all 12 being different (assuming independence which is approximately true) is 12/13 to the power 12, which is 0.38.  So the chance of success is at best 1/13 times 0.38 which is about 1 in 34.  However, it will be slightly less than this, as Ross highlights other problems (such as an 4 being at the bottom of 2 o'clock and 2 being at the bottom of 4 o'clock).  You could go on with circular combinations involving more than 2 that means that certain cards can never be reached.
