Derangements of a deck of cards, where ranks are equal Two players have randomly shuffled decks. They compare the cards in order one at a time, checking if there is a match in rank (i.e.  suit doesn't matter). What's the probability that there is no match? 
If suit matters, this problem boils down to the number of derangements of 52. However, since suit doesn't matter the problem is a bit different and seems to be harder from what I can tell. 
 A: This answer by joriki to an earlier question is easily modified to provide an answer to this one. This is a generalized derangement problem, and in the notation of that answer we have $r=13$ and $n_i=4$ for $i=1,\ldots,13$. The $4$th Laguerre polynomial is 
$$L_4(x)=\frac1{24}\left(x^4-16x^3+72x^2-96x+24\right)\;,$$
and the $4$ cards of each rank are distinguishable, so there are
$$(4!)^{13}\int_0^\infty\big(L_4(x)\big)^{13}e^{-x}\,dx=\int_0^\infty\left(x^4-16x^3+72x^2-96x+24\right)^{13}e^{-x}\,dx$$
permutations that do not yield a match. According to WolframAlpha this yields a probability of
$$\frac{4,610,507,544,750,288,132,457,667,562,311,567,997,623,087,869}{52!}\approx0.016233\;.$$
A: I started with solving it for a simple case, when there are 1 cards of each value, i.e. 13 cards in total in each deck. Then you get derangements: 
$$
\frac{13!^2}{13!^2} - \frac{\binom{13}{1}\binom{13}{1}12!^2}{13!^2} + \frac{\binom{13}{2}\binom{13}{2}2!11!^2}{13!^2} - \frac{\binom{13}{3}\binom{13}{3}3!10!^2}{13!^2} + \ldots \\
= 1-1+\frac{1}{2!} - \frac{1}{3!}+\ldots = \sum_{k=2}^{13}\frac{(-1)^k}{k!}
$$
I'm not sure I was able to extend to the normal case with 4 cards each value. For example, the second term (first after $\frac{52!}{52!}$) should be 
$$
\frac{\binom{13}{1} \times \binom{52}{4}4!^2 48!^2}{52!^2}
$$
is just roughly $\frac{1}{20000}$, which is really small, so I probably made a mistake somewhere.
