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?
 A: It turns out there is an easier solution to this, following Byron's approach in this answer. The rook polynomial for each $4\times4$ block of excluded matches is $R(x)=24x^4+96x^3+72x^2+16x+1$, and the desired probability is
$$
\frac1{52!}\int_0^\infty x^{52}R(-1/x)^{13}\mathrm e^{-x}\mathrm dx\;.
$$
Wolfram|Alpha refuses to evaluate this, but in Sage
y=(-1/x)
a=(x^52 * (24*y^4 + 96*y^3 + 72*y^2 + 16*y + 1)^13).full_simplify ()
integral (a * exp (-x),0,infinity)/factorial (52)

yields
$$
\frac{4610507544750288132457667562311567997623087869}{284025438982318025793544200005777916187500000000}
$$
in agreement with the result quoted by Barry.
A: 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
$$\begin{align}
{R_{13} \over 52!} &= {4610507544750288132457667562311567997623087869 \over 284025438982318025793544200005777916187500000000}\cr
\cr
&= 0.01623272746719463674 \ldots
\end{align}$$
A: The probability that a card causes you to fail is $\frac 1{13}$  An approximation, which shouldn't be too far off, is that each card is independent.  This is not true, as the fact that the first Ace didn't make you fail increases (slightly) the chance that the next two makes you fail.  If we make that approximation, the chance of winning is the chance that no card makes you fail, which is $\left(\frac {12}{13}\right)^{52}\approx 0.0157$
A: After our first draw the probability of winning will be $\frac {12}{13}$. For the second card we have to split the problem into two cases. The first one is we draw two in the first draw and the second is we didn't draw two in the first draw. So the probability will be: $\frac{12}{13} \cdot \frac{47}{51} + \frac{1}{13} \cdot \frac{48}{51}$. So the probability of not losing after two draws will be:
$$\frac {12}{13} \cdot \left(\frac{12}{13} \cdot \frac{47}{51} + \frac{1}{13} \cdot \frac{48}{51}\right) \approx 85,2 \% $$
For the third draw the calulcation are even more complicated. $\frac{12}{13} \cdot \frac{47}{51} \cdot \frac{46}{50} + \frac{1}{13} \cdot \frac{48}{51} \cdot \frac{47}{50} + \frac{12}{13} \cdot \frac{4}{51} \cdot \frac{47}{50} + \frac{1}{13} \cdot \frac{3}{51} \cdot \frac{48}{50}$
So after the third draw our chance of not losing will be $\approx 78.65 \%$, but note that:
$$78.65 \% \approx \left(\frac{12}{13}\right)^3$$
I don't know whether this will continue, but maybe it's good try:
I tried to make a program that will compute the chance of winning after 52 draws(although I must admit I didn't have very good skills in programing), and it computed $\approx 0.0162$ which is very close to the result Ross Millikan's formula gave$
