How can I construct the following distribution?

A French-suited cards pack consist of $52$ cards where $13$ are clovers. $4$ players play a game where every player has $13$ cards in his/her hands (, in other words the full pack has been shared). Let $X$ denotes a random variable, which shows the number of players who's got at least $1$ clover. After shuffling, every outcome has the same probability. How can I calculate $\mathbf{P}\left(X=1\right)$, $\mathbf{P}\left(X=2\right)$, $\mathbf{P}\left(X=3\right)$ and $\mathbf{P}\left(X=4\right)$?

My idea was the following...

I listed all the possible outcomes:

When we calculate $\mathbf{P}\left(X=1\right)$, then one player has all the clovers, so there are $13-0-0-0$ clovers in the different packs.

When I want to calculate $\mathbf{P}\left(X=2\right)$, then there are the following cases: $12-1-0-0$, $11-2-0-0$, $10-3-0-0$, $9-4-0-0$, $8-5-0-0$, $7-6-0-0$.

In the case of $\mathbf{P}\left(X=3\right)$, the following cases can occur: $11-1-1-0$, $10-2-1-0$, $9-3-1-0$, $9-2-2-0$, $8-4-1-0$, $8-3-2-0$, $7-5-1-0$, $7-4-2-0$, $7-3-3-0$, $6-6-1-0$, $6-5-2-0$, $6-4-3-0$, $5-5-3-0$, $5-4-4-0$.

I know the previous $3$ probabilities determinates $\mathbf{P}\left(X=4\right)$, but for the sake of completeness: $10-1-1-1$, $9-2-1-1$, $8-3-1-1$, $8-2-2-1$, $7-4-1-1$, $7-3-2-1$, $7-2-2-2$, $6-5-1-1$, $6-4-2-1$, $6-3-3-1$, $6-3-2-2$, $5-5-2-1$, $5-4-3-1$, $5-4-2-2$, $5-3-3-2$, $4-4-4-1$, $4-4-3-2$, $4-3-3-3$.

I tried to calculate the probabilities of the previous events. I demonstrate how I did it via an example.

Let's check the case when there are $10-2-1-0$ clovers in the packs. All deals have the same probabilities, so it is enough to check that case when the 1st player gets 10 clovers, the second player gets 2 clovers, the 3rd player gets 1 clover and the last palyer gets no clovers. Because all deals have the same probability there should be a $\frac{4!}{1!2!1!}$ factor in the following solution. In one deal there are $39!$ ways to share the non-clovers, and $13!$ ways to share the clovers. We can choose $10$ clovers out of $13$ cards $\binom{13}{10}$ ways in the case of the 1st player, $1$ clover out of $13$ clovers can be choosed $\binom{13}{1}$ ways in the case of the 2nd player, $\binom{13}{1}$ ways in the case of the 3rd player. All cases are $52!$, so the probability of that event when {there are $10$ clovers in one pack, $1$ clover in one pack, $1$ clover in another pack, $0$ clover in the last pack} is $$\frac{39!\cdot13!\cdot\binom{13}{10}\cdot\binom{13}{1}\cdot\binom{13}{1}}{52!}\cdot\frac{4!}{1!\cdot2!\cdot1!}.$$ Calculating all cases listed above we should get the proper distribution.

I would appreciate much easier solution for the problem, for example somehow calculating the $\mathbf{P}\left(X\leq i\right)$ distribution function, and in this case it would be easy to calculate with $$\mathbf{P}\left(X=i\right)=\mathbf{P}\left(X\leq i\right)-\mathbf{P}\left(X\leq i-1\right)=\mathbf{P}\left(X\leq i\right)-\left(\mathbf{P}\left(X=1\right)+...+\mathbf{P}\left(X=i-1\right)\right).$$

Or somehow calculating the probabilities with the complementer events.

This looks like a good place to apply the Generalized Principle of Inclusion and Exclusion. You can find a statement of the principle, references, and proofs on this site in Generalized inclusion-exclusion principle.

Let's say a deal of the deck has "Property $i$" if player $i$ has no clovers, for $1 \le i \le 4$, and let $S_j$ be the total of the probabilities of all the deals with $j$ of the properties, for $1 \le j \le 3$. Then $$\begin{align} S_1 &= \frac{\binom{4}{1} \binom{39}{13}}{\binom{52}{13}} \\ S_2 &= \frac{\binom{4}{2} \binom{39}{26}}{\binom{52}{26}} \\ S_3 &= \frac{\binom{4}{3} \binom{39}{39}}{\binom{52}{39}} \\ \end{align}$$ The probability that a deal has exactly $m$ of the properties is $X_m$. By the Generalized PIE, $$\begin{align} P(X=1) &= S_1 - \binom{2}{1} S_2 + \binom{3}{1} S_3 \\ P(X=2) &= S_2 - \binom{3}{2} S_3 \\ P(X=3) &= S_3 \end{align}$$

For $X=1$, the calculation is relatively simple, as there are only $4$ essentially different cases (there are four choices as to which players has all the clovers). For $X=2$, there are six choices as to which players have all the clovers. These two players have $26$ cards between them, and they each get $13$ cards, so we have $26C13$ ways of splitting cards between them. However, this doesn't actually get you $P(X=2)$, it gets you $P(X\leq2)$, so you have to subtract out the double-counted cases from $X=1$. And this double counting is more complicated than it may appear at first; when you find the probability that Player 1 and Player 2 are the ones with clovers, you've double counting both the case where Player 1 is the only one and the case where Player 2 is the only one. So the math on that can be tricky. With problems like these, it can be helpful to consider simpler problems (for instance, four clovers and four non-clovers distributed among four players, or four clovers and eight non-clovers) to see whether the formulas that you've come up with are giving results that make sense (with twelve cards, it's easy to come up with a computer program that gets the answer by brute force).

For $X=3$, there are four choices for who doesn't get any clovers, and there are $39C13$ choices for what non-clovers to give them. For $X=4$, if you calculate the previous three probabilities correctly, and you subtract the double-counted cases correctly, you can just do $P(X=4) = 1-P(X=3)-P(X=2)-P(X=1)$.

You also need to make sure that you're using the same denominator for each calculation. For instance, you can consider a hand to be different if the cards are dealt in a different order, or you can consider that the same hand, but you have to be consistent. So, for instance, when I said that there are $39C13$ choices for $X=3$, you may need to also multiply by the number of ways of distributing the remaining cards among the other three players, depending on how you're counting "different" configurations.