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(4) $\left(\frac{1}{3}\right)^{5} \frac{\left(\mathrm{C}_{4}^{1}\right)^{12}}{\mathrm{C}_{48}^{2}}$ Hint: First consider that one blank card must be placed in one box, the probability that the second card is also placed in that box is $\frac{1}{3},$ and the probability that the 4 ones are placed in that box is $\left(\frac{1}{3}\right)^{4}$. The subsequent process involves selecting 12 cards from the remaining 48 cards, with each number being chosen exactly once. Therefore, the probability of a super box appearing is $\left(\frac{1}{3}\right)^{5} \frac{\left(\mathrm{C}_{4}^{1}\right)^{12}}{\mathrm{C}_{48}^{2}}$.