When you randomly shuffle a deck of cards, what is the probability that it is a unique permutation never before configured? I just came back from a class on Probability in Game Theory, and was musing over something in my head.

Assuming, for the sake of the question:


Playing cards in their current state have been around for approximately eight centuries
A deck of playing cards is shuffled to a random configuration one billion times per day
Every shuffle ever is completely (theoretically) random and unaffected by biases caused by human shuffling and the games the cards are used for
By "deck of cards", I refer to a stack of unordered $52$ unique cards, with a composition that is identical from deck to deck.


This would, approximately, be on the order of $3 \cdot 10^{14}$ random shuffles in the history of playing cards.

If I were to shuffle a new deck today, completely randomly, what are the probabilistic odds (out of $1$) that you create a new unique permutation of the playing cards that has never before been achieved in the history of $3 \cdot 10^{14}$ similarly random shuffles?

My first thought was to think that it was a simple matter of $\frac{1}{52!} \cdot 3 \cdot 10^{14}$, but then I ran into things like Birthday Paradox.  While it is not analogous (I would have to be asking about the odds that any two shuffled decks in the history of shuffled decks ever matched), it has caused me to question my intuitive notions of Probability.

What is wrong in my initial approach, if it is wrong?

What is the true probability?

And, if the probability is less than $0.5$, if we how many more years (centuries?) must we wait, assuming the current rate of one billion shuffles per day, until we reach a state where the probability is $0.5$+?   $0.9$+?

(Out of curiosity, it would be neat to know the analogous birthday paradox answer, as well)
Your original answer of $\dfrac{3 \times 10^{14}}{52!}$ is not far from being right. That is in fact the expected number of times any ordering of the cards has occurred.

The probability that any particular ordering of the cards has not occurred, given your initial assumptions, is $\left(1-\frac1{52!}\right)^{(3\times10^{14})}$, and the probability that it has occurred is 1 minus this value.  But for small values of $n\epsilon$, $(1+\epsilon)^n$ is nearly $1+n\epsilon$.  In particular, since $52!\approx 8\times 10^{67}$ and so $\dfrac{3\times10^{14}}{52!}\approx 3.75\times 10^{-54}$ is microscopically small, $1-\left(1-\frac1{52!}\right)^{(3\times10^{14})}$ is very nearly $\frac1{52!}\times (3\times10^{14})$.
You would never ever get a deck of cards in the same order.  The odds of doing so would be 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000.  This is done by 52 factorial!