Probability of card decks being in the same order for N shufflers over X amount of time? So I wanted to figure out a meaningful probability for the card deck order problem.  I chose arbitrary values in an attempt to show the magnitude of permutations:

If you and everyone on the Earth right
  now shuffled a million randomized
  decks every millisecond that universe
  has existed, the probability that any
  two decks at one moment were the same would be:

What is the probability?
 A: Let's make some assumptions for the purposes of calculation. Suppose 6.7 billion people have been alive for 13.5 billion years each shuffling one deck a thousand times a second.
The total number of rounds of shuffles $N=1000 \times 3600 \times 24 \times 365 \times1.35 \times 10^{10}.$
So $N = 4.25736 \times 10^{20}.$
Let $n=52! \approx 8.0658 \times 10^{67}$ and let $m=6.7 \times 10^9.$
When everyone shuffles the cards once in the first millisecond the probability that they are all different is
$$ \prod_{i=1}^{m-1} \left( 1- \frac{i}{n} \right) \approx 1 - \frac{m(m-1)}{2n},$$
ignoring the cross products since $n$ is much bigger than $m$.
Each shuffle is independent, so after $N$ steps the probabilty that we have had no match is
$$\left( 1 - \frac{m(m-1)}{2n} \right)^N \approx 1 - \frac{m(m-1)N}{2n}, \qquad (1)$$
since $m(m-1)/2n$ is very small.
Hence the probability that there has been a match is 1 minus RHS of (1), that is
$$\frac{m(m-1)N}{2n}.$$
I make this about $1.18 \times 10^{-28}$ with the figures I've used.
REMARK: Note that if you drop the "at one moment" requirement. That is, if you now ask if two shuffles have ever been the same, the problem is slightly quicker. The total number of shuffles that have ever taken place is $mN$ and so the probability that they were all different (as in the birthday problem) is $\prod_{i=0}^{mN-1} \left( 1- \frac{i}{n} \right),$ where as before $n=52!,$ So the probability that we've had a repeat is  $1- \prod_{i=0}^{mN-1} \left( 1- \frac{i}{n} \right),$ which, similar to before, is approximately $mN(mN-1)/2n,$ or $5 \times 10^{-8}.$
Note that as the index $i$ in the product gets larger, better approximations are obtained by passing to logarithms (as on the wikipedia page).
A: Here is what I came up with and my assumptions.  Please point out any errors. 
First, I attempted to find the probability of 2 people having the same deck order of N people shuffling decks in one trial.  I could not figure out how to model it in a more standard way, so my leap of faith was that for N people, there were (N+N-1+N-2+...)=Nt.  For 6.7e9 people on Earth, that's 2.2445*^19 pairs of people.  The probability that one pair had the same deck order was (1/52!) = 1/8e67 = p1.  Since the events are not independent, I need to multiply by (52!)!/(52!-N).
Next, I wanted the probability that any of them had a match.  So that would be 1 minus the probability that none of them had any matches:
(1 - (52!)!/(52!-N)*(p1)^0*(1 - p1)^(Nt)) = (1 - (52!)!/(52!-N)*(1 - p1)^(Nt)) = p2.
Now, I wanted to do a million randomized trials every millisecond that the universe existed.  So that's 1.4e10 years * 365 days/year * 24 hours/day * 3600e3 ms / hour * 1e9 trials/ms = T = 4.5e29.  Again, I wanted at least one match, that would bring it to:
(1 - (1 - p2)^(T)) = 
(1 - (1 - (1 - (52!)!/(52!-N)*(1 - p1)^(Nt)))^(T)) =
(1-(1-(1-(52!)!/(52!-6.7e9)*(1-1/8e67)^(2.2445*^19)))^(4.5e29)) = 6.31 × 10^-20
