Dealing cards, derangements, and probability: Is the Riddler Express solution incorrect?

Question

Edit. The error was noted in a subsequent Riddler entry with a reference to an earlier MSE post:

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The most recent Riddler Express says:

But suppose you had this same question with only one suit (say, hearts): What is the probability that you get through the 13 hearts without what you say ever matching what you deal?

It seems to me that the method above would suggest computing $(12/13)^{13} = 0.3532\ldots$

This cannot be right, though; the modified question is essentially asking about derangements for a set of $13$ elements. Checking the derangement sequence at OEIS yields as its thirteenth term $2290792932$; among the $13!$ combinations, this would yield a probability of winning the modified game equal to $2290792932/13! = 0.3678\ldots$

The numbers are not very different, but they are certainly different.

Back to the main question:

Is the solution for the "Riddler Express" above incorrect? [I strongly believe it is...]

If so: What is the correct solution, and how does one show this?

I had thought it would require a modification of the derangement formula using the Inclusion-Exclusion Principle or some such thing (cf. this MO answer from Richard Stanley). In fact, I computed it as $0.01623\ldots$ Again, this is "about 1.6 percent" (as in the solution write-up) but the methods are very different and the actual numerical results are not the same.

Am I overthinking? Is the Riddler underthinking? What gives?

Answer 1

The value of $P(win) = \frac{\int_0^{\infty}L_4^{13}(x)e^{-x}dx}{52!/(4!)^{13}} \approx 1.6233\%$ is given here with citations.

Answer 2

Yes, the "Riddler Express" solution is not correct, but it is a reasonably good approximation.

Answer 3

You can see that it is not correct by looking at the simpler case of an eight-card deck with A2 in each suit. To get zero matches, you need precisely 2A2A2A2A; the chance of this is exactly

$$\frac12 \times \frac47 \times \frac12 \times \frac35 \times \frac12 \times \frac23 \times \frac12 =\frac{1}{70}$$

But according to the Riddler's flaky logic, it would be $$\left(\frac12\right)^8=\frac{1}{256}$$