4. There are five cards each inscribed with the numbers $1,2,3,4,5$, and each time one card is drawn, the number on the card is recorded, and then the card is put back. This process is repeated four times, then the probability that the difference between the maximum and minimum numbers drawn is less than 4 is $\qquad$
4. $\frac{431}{625}$.

First, consider the scenario where the difference between the maximum and minimum numbers drawn is 4, meaning both 5 and 1 are drawn in the four draws. There are $5^{4}$ possible outcomes for the four draws. Among these $5^{4}$ outcomes, there are $4^{4}$ outcomes where 1 is not drawn, and $4^{4}$ outcomes where 5 is not drawn. There are $3^{4}$ outcomes where neither 5 nor 1 is drawn. Therefore, the number of outcomes where both 5 and 1 are drawn is $5^{4}-2 \times 4^{4}+3^{4}=$ 194. The probability of this is $\frac{194}{5^{4}}=\frac{194}{625}$.
Thus, the required probability is $\frac{431}{625}$.