The problem seems difficult that has remained unanswered for about 9 years. I could find two solutions to it, while they are both long.
First method:
Let A, B, C, and D denote the event of that the $j\text{th}$ hand, $j=1,2,3,4$, misses at least one suit. Then, using the De Morgan's Law, the probability of event is given by
$$ P(A\cup B\cup C\cup D)=1- P(A'\cap B' \cap C'\cap D')=1-\dfrac{ N }{\binom{52}{13,13,13,13} }$$
where $N$ is the number of cases that no hand misses any suit. $N$ can be obtained by summing all the following terms:
$$\binom{13}{i_1} \binom{13}{i_2}\binom{13}{i_3}\binom{13}{13-i_1-i_2-i_3}\times$$
$$\binom{13-i_1}{j_1}\binom{13-i_2}{j_2}\binom{13-i_3}{j_3}\binom{13-(13-i_1-i_2-i_3)}{13-j_1-j_2-j_3}\times$$
$$\binom{13-i_1-j_1}{l_1}\binom{13-i_2-j_2}{l_1}\binom{13-i_3-j_3}{l_1}$$
$$\binom{13-(13-i_1-i_2-i_3)-(13-j_1-j_2-j_3)}{13-j_1-j_2-j_3}\times$$
$$\binom{13-(13-i_1-i_2-i_3)-(13-j_1-j_2-j_3)-(13-l_1-l_2-l_3)}{13-j_1-j_2-j_3}$$
where the indices $i_1,i_2,_, j_1,j_2, j_3, l_1,l_2,l_3$ vary in the following domain:
$$1\le i_1,i_2,i_3\le10, 13-i_1-i_2-i_3\ge1$$
$$1\le j_1,j_2,j_3\le10, 13-j_1-j_2-j_3\ge1$$
$$1\le l_1,l_2,l_3\le10, 13-l_1-l_2-l_3\ge1$$
$$13-i_1-j_1-l_1\ge1, 13-i_2-j_2-l_2\ge1, 13-i_3-j_3-l_3\ge1, 13-(13-i_1-i_2-i_3)-(13-j_1-j_2-j_3)-(13-l_1-l_2-l_3)\ge1.$$
Using a Python code, I could compute
$$N=4.378664133339871e+28.$$
Hence, the probability is
$$1- \frac{4.378664133339871e+28}{5.36447377654888e+28}=0.18376632718730682$$
This probability is very close to one obtained by assuming that all the four events A, B, C, and D are independent, that is, $$1-(1-0.051065521)^4=0.189141811,$$ which shows that the events A, B, C, and D are very weakly dependent (note that the void probability given in Wikipedia 1, i.e. 0.0512, is not accurate; see 2 for more details and other related problems).
Second method:
For each $i=1,2,3,4$, let us define
$A_i$,$B_i$,$C_i$, and $D_i$ as the event of that the $j\text{th}$, $j=1,2,3,4$,hand has no card of the $i\text{th}$ suit, respectively.
Then, the probability of interest is
$$ P\{(\cup_{i=1}^4A_i) \cup (\cup_{i=1}^4B_i) \cup (\cup_{i=1}^4C_i) \cup (\cup_{i=1}^4D_i)\}. $$
By the inclusion-exclusion principle, the above probability is equal to
$$4\times4P(A_1)$$
$$-4{4 \choose 2}P(A_1\cap A_2) -{4 \choose 2}4 \times 3 P(A_1\cap B_2) -
{4 \choose 2}4 P(A_1\cap B_1)$$
$$4{4\choose 3} P(A_1\cap A_2 \cap A_3) + {4\choose 2} {4\choose 2} 2 P(A_1\cap A_2 \cap B_3)+ {4\choose 2} {4\choose 2} 2 P(A_1\cap A_2 \cap B_1)+ {4\choose 3}4\times3\times2 P(A_1\cap B_2 \cap C_3)+{4\choose 3}4\times3 P(A_1\cap B_1 \cap C_2)+{4\choose 3}4 P(A_1\cap B_1 \cap C_1) $$
$$-{4\choose 2}{4\choose 3}P(A_1\cap A_2 \cap A_3\cap B_4)-{4\choose 2}{4\choose 3}3P(A_1\cap A_2 \cap A_3\cap B_1)-
{4\choose 2}{4\choose 2}P(A_1\cap A_2 \cap B_3\cap B_4)-{4\choose 2}{4\choose 2}P(A_1\cap A_2 \cap B_1\cap B_2)-{4\choose 2}{4\choose 2}2\times 2P(A_1\cap A_2 \cap B_1\cap B_3)-{4\choose 3}{4\choose 2}2P(A_1\cap A_2 \cap B_1\cap C_1)-{4\choose 3}{4\choose 2}2P(A_1\cap A_2 \cap B_3\cap C_4)-{4\choose 3}{4\choose 2}2P(A_1\cap A_2 \cap B_3\cap C_3)-4!P(A_1\cap B_2 \cap C_3\cap D_4)-{4\choose 2}4\times 3 \times 2 P(A_1\cap B_2 \cap C_2\cap D_4)-4\times 4 \times 3P(A_1\cap B_2 \cap C_2\cap D_2)$$
$$+\dots - 4! P(A_1\cap A_2 \cap A_3\cap B_1\cap B_2 \cap B_4 \cap C_1\cap C_3 \cap C_4 \cap D_2\cap D_3 \cap D_4).$$
The terms including more than 4 events can be generated similarly. Combinations related to different situations need to be separated, while they may finally have the same probability (c.f. $P(A_1\cap B_1)$ and $P(A_1\cap B_2)$ below). Many combinations of 4 or more events can be ignored as their intersections are empty. A useful rule is that in a reasoanable combination each hand or each suit cannot appear more than three times, e.g.,
$$P(A_1\cap A_2 \cap A_3\cap A_4)=P(A_1\cap B_1 \cap C_1\cap D_1)=0.$$
Using this rule, one can see that the maximum number of events whose intersection is not empty is 12.
Each element of the above expansion can be computed considering how the corresponding subset of 16 events defines a subset of suits for each hand ($A_1A_2B_1$ means that the first hand includes cards only from the 3rd and 4th suits, the second includes cards only from the 2nd, 3rd, and 4th suits, and the third and forth hands can have cards from all suits). For example,
$$ P(A_1)= \dfrac{ \binom{39}{13}\binom{39}{13,13,13} }{ \binom{52}{13,13,13,13} } $$
$$P(A_1\cap A_2)=\dfrac{ \binom{26}{13}\binom{39}{13,13,13} }{ \binom{52}{13,13,13,13} }$$
$$P(A_1\cap B_1)=\dfrac{ \binom{39}{13}\binom{26}{13}\binom{26}{13,13} }{ \binom{52}{13,13,13,13} }$$
$$P(A_1\cap B_2)=\dfrac{ \binom{39}{13}\binom{26}{13}\binom{26}{13,13} }{ \binom{52}{13,13,13,13} }$$
$$P(A_1\cap A_2 \cap A_3)=\dfrac{ \binom{13}{13}\binom{39}{13,13,13} }{ \binom{52}{13,13,13,13} }$$
$$P(A_1\cap A_2 \cap B_3)=\dfrac{ \sum_{i=1}^{13}\binom{13}{i}\binom{13}{13-i}\binom{26+i}{13}\binom{26}{13,13} }{ \binom{52}{13,13,13,13} }$$
$$ \dots$$
$$P(A_1\cap A_2 \cap A_3\cap B_1\cap B_2 \cap B_4 \cap C_1\cap C_3 \cap C_4 \cap D_2\cap D_3 \cap D_4)=\dfrac{ 1 }{ \binom{52}{13,13,13,13} }.$$
$1- \dfrac{{13 \choose 1,1,1,1,9}^4 {36\choose 9,9,9,9}}{52 \choose 13,13,13,13}$
To get this "solution," find the probability that no player is missing a suit. The total number of ways to deal 13 cards each to 4 players is ${52 \choose 13,13,13,13}$. To have no player missing a suit, choose four of the 13 cards from each suit to give to each of the players, then deal 9 cards each to each of the four players. Why precisely is this wrong?
– Bridge counter Sep 05 '14 at 00:32You are trying to divide 4 suits of 13 objects each into 4 subsets such that each subset gets atleast one. So ${4!^4 \cdot S(13,4)}^4$ should be the number of ways to do it. I don't understand combinatorics much but sometime back I had asked a question which was similar. The answer was given by @André Nicolas. http://math.stackexchange.com/questions/804088/probability-of-getting-k-different-coupons-in-a-set-of-s-coupons-from-an-infini So I am just trying to reapply the same principle here.
– Durin Sep 05 '14 at 11:09