A team of four students goes to LMT, and each student brings a lunch. However, on the bus, the students’ lunches get mixed up, and during lunch time, each student chooses a random lunch to eat (no two students may eat the same lunch). What is the probability that each student chooses his or her own lunch correctly?
1. **Determine the total number of ways to distribute the lunches:**
   Each of the four students can receive any of the four lunches. The total number of permutations of four items is given by \(4!\) (4 factorial).
   \[
   4! = 4 \times 3 \times 2 \times 1 = 24
   \]

2. **Identify the number of favorable outcomes:**
   There is only one favorable outcome where each student receives their own lunch. This is the identity permutation where each item is in its original position.

3. **Calculate the probability:**
   The probability \(P\) that each student gets their own lunch is the ratio of the number of favorable outcomes to the total number of possible outcomes.
   \[
   P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{1}{24}
   \]

\[
\boxed{\frac{1}{24}}
\]