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

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.

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

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 $$\left(\frac12\right)^8=\frac{1}{256}$$