Probability of winning the game 1-2-3-4-5-6-7-8-9-10-J-Q-K A similar question to mine was answered here on stackexchange:
Probability of winning the game "1-2-3"
However, I am unable to follow the formulas so perhaps someone could show the calculation and the way they arrived at it to answer this question.
My card game is very similar, except the counting of cards goes all the way from 1 (Ace) through to King and then restarts.  If you can make it through an entire deck of cards without hitting the same ranked card you win.  So, for example, if you call out Ace - 2 -3 - 4 - 5 - 6, etc. and hit "7" when you've just called out "7", you lose.
What is the chance of winning this card game?

I think you'll find what you want in the paper Frustration solitaire by Doyle, Grinstead, and Snell at http://arxiv.org/pdf/math/0703900.pdf -- it looks like they get the answer
$$\begin{align}
{R_{13} \over 52!} &= {4610507544750288132457667562311567997623087869 \over 284025438982318025793544200005777916187500000000}\cr
\cr
&= 0.01623272746719463674 \ldots
\end{align}$$
