Probability Two Friends Sit Adjacent in a Row
Ten people are seated randomly in a row of 10 chairs. Two of them are friends. What is the probability that the two friends end up seated next to each other?
Hints
- Fix one of the friends in a seat. How many of the remaining 9 seats are adjacent?
- Count the total number of adjacent seat pairs in a row of 10 chairs -- there are 9 such pairs. Then count the total number of ways to choose 2 seats out of 10.
- Alternatively, treat the two friends as a single block. The block can be placed in 9 positions along the row, and the friends can swap within the block in $2!$ ways.
Worked Solution
How to Think About It: Fix one friend's position and ask: what fraction of the remaining seats are adjacent? That gives you the answer immediately, no combinatorics needed. Alternatively, you can count favorable arrangements over total arrangements -- but the "fix one person" approach is cleaner and less error-prone.
Quick Estimate: There are 10 seats in a row, so 9 adjacent pairs. There are $\binom{10}{2} = 45$ total pairs of seats. So the probability two specific people are adjacent is $9/45 = 1/5$. Done.
Approach: Let us verify this with two different methods.
Formal Solution:
Method 1 -- Fix one person:
Fix person A in some seat. Person B is equally likely to be in any of the 9 remaining seats. How many of those are adjacent to A?
- If A is in seats 2 through 9 (8 positions): 2 adjacent seats each.
- If A is in seat 1 or seat 10 (2 positions): 1 adjacent seat each.
Average number of adjacent seats, averaged over A's position:
$$\frac{8 \times 2 + 2 \times 1}{10} = \frac{18}{10} = 1.8$$
$$P(\text{adjacent}) = \frac{1.8}{9} = \frac{1}{5}$$
Method 2 -- Direct counting:
Total arrangements: $10!$
Favorable: Treat the two friends as a glued block. There are 9 positions for this block in a row of 10 chairs (positions $(1,2), (2,3), \ldots, (9,10)$). Within the block, the friends can swap, giving $2!$ orderings. The remaining 8 people fill the other 8 seats in $8!$ ways.
$$P = \frac{9 \times 2 \times 8!}{10!} = \frac{9 \times 2}{10 \times 9} = \frac{2}{10} = \frac{1}{5}$$
Answer: The probability is $\dfrac{1}{5} = 0.2$.
Intuition
The cleanest way to think about adjacent-pair probabilities is the "fix one, count neighbors" trick. Once you fix one person's seat, the other person is uniformly distributed over the remaining seats, so you just need the fraction that are adjacent. This avoids the factorial counting that trips people up.
The deeper pattern here is that for any adjacency question in a row of $n$ seats, the probability two specific people are adjacent is $2(n-1)/(n(n-1)) = 2/n$. For $n = 10$, that is $2/10 = 1/5$. For a circular table, every seat has exactly 2 neighbors, so the probability becomes $2/(n-1)$ -- slightly higher because end effects disappear. These formulas are worth memorizing; they come up often in interview warm-ups and serve as quick sanity checks for more complex seating problems.