Doesn't the birthday paradox also come into play here, significantly increasing the chances of a duplicated shuffle?
http://hatlogic.blogspot.com/2010/04/cards-and-birthdays.htm...
Note that it does not say "If you shuffle a deck of cards, it will be unique in all of human history," which is the argument that the actual article is making.
The article title here is logically equivalent to "Any shuffled deck of cards is unique in all of human history" which is logically equivalent to "No two shuffled decks of cards in human history are equivalent". Therefore, the title of the article as it appears on hacker news needs to be changed
Using the same estimate as the OP for n (1.56x10^23) gives p=3.02x10^-22. Still fantastically low.
Related question. If there are exactly 2N people who vote in a binary election (ie: for presidential candidates) and they have an even 50/50% chance of voting either way, how do I compute the odds that they will have a even split? This is a generous estimate for the probability my vote will matter.
Check out: http://en.wikipedia.org/wiki/Binomial_distribution
I think that this is way too low. Shouldn't it be the quite large number
1 - \prod_{i = 1}^n (1 - (i - 1)/52!)
(a la the birthday paradox)?
Mathematica overflowed when I tried to compute this by brute force. The next best thing I can think of is to use the exponential approximation
1 - x ≈ e^{-x},
\prod_{i = 1}^n (1 - (i - 1)/52!)
≈ \prod_{i = 1}^n e^{-(i - 1)/52!}
= e^{-n(n - 1)/52!}
≈ 1 - n(n - 1)/52!.
\frac1 2\sum_{i = 1}^n [(i - 1)/52!]^2
≈ n^3/(2(52!)^2),