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?

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 $\left(\frac {12}{13}\right)^{52}\approx 0.0157$
