Five People Around a Table

Think about the problem this way. Imagine that the 5 seats are positioned around the table, and they sit down one by one, starting with whoever has the middle (3rd oldest) age. This person can sit in any of the seats and it doesn't change anything as we can rotate the table around. Now there are 4 seats remaining. Now let's bring in somebody to sit to the left of the 3rd oldest person, and then somebody to sit on the right of the 3rd oldest person. We have 4 choices and 3 choices respectively, meaning that there are 12 possibilities. However, there are 2 possibilities which suit our needs: 2nd oldest sits to the right and 4th oldest sits to the left, or 2nd oldest sits to the left and 4th oldest sits to the right (as age order can be clockwise or anticlockwise). This gives us a probability of $\frac{2}{12} = \frac{1}{6}$ . Now, assuming that one of these scenarios has occurred, there are 2 people (oldest and youngest) left to sit in the 2 remaining seats. By the seating of the 2nd and 4th oldest, we have already determined the clockwise/anticlockwise way, meaning there is only one possibility that we are interested in. However, there are 2 ways for 2 people to sit in 2 seats (AB, BA), so this gives us a probability of $\frac{1}{2}$ . Therefore, the total probability is $\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$ .