Solitaire probability

I would like to know the exact probability of the following game. I start counting from one to 13 and do this totally four times. On each turn I say a number and turn a card from a deck. Ace is considered to be as 1. If on one turn I say the same number as it is on the card, I lose. What is my winning probability?

Solution 1:

The exact answer is

$$P(win) = \frac{\int_0^{\infty}L_4^{13}(x)e^{-x}dx}{52!/(4!)^{13}} \approx 1.6233\%$$

Where $L_4(x)$ is the 4th Laguerre polynomial. The numerator is the number of derangements of a multiset with 13 different elements each repeated 4 times, and it is from Derangements and Laguerre Polynomials by S. Even and J. Gillis and described in this MSE answer. Also a short derivation using rook polynomials can be found in Laguerre polynomials and derrangements by D.M. Jackson. The denominator of course is total permutations of the deck where we ignore suit. This exact game was also analyzed in On the Asymtotic Behavior of a Card Matching Problem by Knudsen and Skau.