Example 1 Remove the big and small jokers from a deck of cards, and randomly draw five cards from the remaining 52 cards. The probability that at least two of the cards have the same number (or letter $K, Q, J, A$) is $\qquad$ (require calculating the numerical value of this probability, accurate to 0.01).
【Analysis】Let event $A$ be “drawing five cards randomly from 52 cards, where at least two cards have the same number (or letter K, Q, J, A)”. Then event $\bar{A}$ is “drawing five cards randomly from 52 cards, where each card has a different number (or letter K, Q, J, A)”.

The total number of scenarios is $\mathrm{C}_{52}^{5}$; the number of scenarios where event $\bar{A}$ occurs: choosing five numbers (or letters) from 13 numbers (or letters), there are $\mathrm{C}_{13}^{5}$ ways to choose, and each number (or letter) has four suits, giving $4^{5}$ ways, so the number of scenarios where event $\bar{A}$ occurs is $\mathrm{C}_{13}^{5} \times 4^{5}$.
$$
\text{Thus, } P(\bar{A})=\frac{\mathrm{C}_{13}^{5} \times 4^{5}}{\mathrm{C}_{52}^{5}} \approx 0.5071 \text{. }
$$

Therefore, the probability of event $A$ occurring is approximately
$$
P(A)=1-P(\bar{A}) \approx 0.49 \text{. }
$$

This problem uses the basic method of combinatorial counting. When dealing with “at least” or “at most” problems, one can think in reverse, starting from the easier-to-calculate complementary event, and then using the fact that the sum of probabilities is 1 to find the probability of the original event. This demonstrates the mathematical idea of “if the direct approach is difficult, then think in reverse”.