The birthday paradox applies if you are asking about the probability that any two shuffled decks throughout history have matched each other, not the probability that a shuffled deck throughout history matches the exact deck you have in your hands today.

The title of this article as posted on hacker news is "A shuffled deck of cards is unique in all human history".

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

I don't think it's that important, really

A quick search turns up this link. Looks like the birthday paradox does not apply in this situation.

http://hatlogic.blogspot.com/2010/04/cards-and-birthdays.htm...

The chances of you sharing a birthday with one of 22 other people is pretty low. the chances of at least two people in a group of 23 is 50%. In the first case, you're fixing the day that some other person must have.

In other words, the odds that a deck you are holding is unique are high; the odds that all decks ever shuffled are unique is low.

Actually, the probability that all decks ever shuffled are unique is also very high. We can approximate the probability that any two of the n decks shuffled in human history were identical as p=1-n^2/52!

Using the same estimate as the OP for n (1.56x10^23) gives p=3.02x10^-22. Still fantastically low.

Nice, I came to the comments to see if someone knew how to calculate this.

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.

The exact answer is (2n choose n) * (1/2)^(2n). This is approximately sqrt(1/Pi n) as n grows large, with error O(n^(-3/2))

That's the same as the probability that 2N flips of a fair coin result in exactly heads: (2N choose N)(.5^N)(.5^N).

Check out: http://en.wikipedia.org/wiki/Binomial_distribution

> We can approximate the probability that any two of the n decks shuffled in human history were identical as p=1-n^2/52!

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)?

Sorry, I was thinking of the complementary probability (that there has been a coincidence).

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},

good for very small `x`, such as ours. Ignoring the cascading errors gives

    \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!.

The error should be roughly of the size

    \frac1 2\sum_{i = 1}^n [(i - 1)/52!]^2
    ≈ n^3/(2(52!)^2),

which is relatively small. (That's stronger, here, than just saying that it is small.) That is to say: I guess I agree with jgershen after all!

Alright, fine, it is comparatively lower. Better? :)