24 decks of 52 standard playing cards are each shuffled independently.

What is the probability that at least one of them has the Ace through 4 of hearts appearing in it in the right order?

For example if you turned them over one at a time, the Ace would appear first and so on...

**
Assumption
**
: The cards can be anywhere in the deck, and don't necessarily need to be consecutive.

The answer is 0.64.

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First, let us consider one deck. We don't care where the Ace through 4 of hearts appear in the deck, so long as they appear in the right order. There are $4! = 24$ possible orders for these four cards, and only one of them is acceptable. So the probability of

onedeck having those four cards in order is $\frac{1}{24}$ and the probability of those cards being out of order is $1-\frac{1}{24}=\frac{23}{24}$ .Thus, the probability of

noneof our 24 decks having the desired order is $\left(\frac{23}{24}\right)^{24}$ . Finally, the probability that at least one deck has this order is $1-\left(\frac{23}{24}\right)^{24}\approx \boxed{0.64}$