$6$ people each have a hat.  If they shuffle their hats and redistribute them, what is the probability that exactly one person gets their own hat back?
1. **Fix one person with their own hat**: Let's denote the people as \( P_1, P_2, P_3, P_4, P_5, P_6 \) and their corresponding hats as \( H_1, H_2, H_3, H_4, H_5, H_6 \). Assume \( P_1 \) gets \( H_1 \).

2. **Derangement of the remaining hats**: We need to find the number of ways to distribute the remaining 5 hats such that none of the remaining 5 people get their own hat. This is known as the derangement of 5 items, denoted as \( D_5 \).

3. **Calculate the derangement \( D_5 \)**: The formula for the derangement of \( n \) items is given by:
   \[
   D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}
   \]
   For \( n = 5 \):
   \[
   D_5 = 5! \left( \frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} + \frac{(-1)^5}{5!} \right)
   \]
   \[
   D_5 = 120 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right)
   \]
   \[
   D_5 = 120 \left( 1 - 1 + 0.5 - 0.1667 + 0.0417 - 0.0083 \right)
   \]
   \[
   D_5 = 120 \left( 0.3667 \right) = 44
   \]

4. **Total number of ways to distribute hats**: The total number of ways to distribute 6 hats to 6 people is \( 6! \):
   \[
   6! = 720
   \]

5. **Probability calculation**: The probability that exactly one person gets their own hat is the number of favorable outcomes divided by the total number of outcomes. Since any one of the 6 people can be the one who gets their own hat, we multiply the derangement by 6:
   \[
   \text{Probability} = \frac{6 \cdot D_5}{6!} = \frac{6 \cdot 44}{720} = \frac{264}{720} = \frac{11}{30}
   \]

The final answer is \( \boxed{ \frac{11}{30} } \)