2. Given that the four surfaces of a regular tetrahedron are labeled with the numbers $1, 2, 3, 4$, if four such uniformly dense regular tetrahedrons are simultaneously thrown onto a table, the probability that the sum of the four numbers on the surfaces in contact with the table is divisible by 4 is $\qquad$
2. $\frac{1}{4}$.

The sum that can be divisible by 4 can be divided into the following cases:
(1) The numbers on the four faces are the same, with a total of 4 kinds;
(2) The numbers on the four faces are $1,3,2,2$, with a total of $\mathrm{A}_{4}^{2}$ kinds;
(3) The numbers on the four faces are $1,3, 1, 3$, with a total of $\mathrm{C}_{4}^{2}$ kinds;
(4) The numbers on the four faces are $1,3,4,4$, with a total of $\mathrm{A}_{4}^{2}$ kinds;
(5) The numbers on the four faces are $2, 2,4,4$, with a total of $\mathrm{C}_{4}^{2}$ kinds;
(6) The numbers on the four faces are $1, 1, 2, 4$, with a total of $\mathrm{A}_{4}^{2}$ kinds;
(7) The numbers on the four faces are $3, 3, 2, 4$, with a total of $\mathrm{A}_{4}^{2}$ kinds;
In summary, there are $4+4 \mathrm{~A}_{4}^{2}+2 \mathrm{C}_{4}^{2}=64$ kinds.
Therefore, the required probability is $\frac{64}{4^{4}}=\frac{1}{4}$.