Expected Adjacent Couples at a Round Table
$n$ married couples are seated uniformly at random around a circular table with $2n$ seats. What is the expected number of couples who end up sitting next to each other?
For concreteness, "sitting next to each other" means a husband and wife occupy two consecutive seats in the circle. A couple counts as adjacent regardless of who else sits nearby.
Hints
- Think about defining an indicator variable for each couple and using linearity of expectation -- you don't need to worry about dependence between couples.
- Fix one spouse at a seat. How many of the remaining $2n - 1$ seats are adjacent to that person in a circular arrangement?
- The probability a specific couple is adjacent is $\frac{2}{2n-1}$. Multiply by $n$ and simplify.
Worked Solution
How to Think About It: Whenever you see "expected number of [something] in a random arrangement," your first reflex should be linearity of expectation with indicator variables. Define one indicator per couple, compute a single probability, and multiply. No need to think about joint distributions or complicated combinatorics -- linearity does not care about dependence.
The key probability question is: given a random circular seating of $2n$ people, what is the chance that two specific people end up next to each other?
Quick Estimate: For $n = 4$ (8 people around a table), fix one spouse. She has 2 neighbors out of $2n - 1 = 7$ remaining people, so the probability her partner is adjacent is $2/7$. Expected adjacent couples: $4 \times 2/7 = 8/7 \approx 1.14$. That feels right -- with 4 couples, you expect a bit more than one lucky pair. For large $n$, the answer should approach 1 since $2n/(2n-1) \to 1$.
Approach: Indicator variables plus a straightforward counting argument for circular adjacency.
Formal Solution:
Define $X_i = \mathbf{1}\{\text{couple } i \text{ sits adjacent}\}$ for $i = 1, 2, \ldots, n$. We want $E\left[\sum_{i=1}^{n} X_i\right]$.
By linearity of expectation:
$$E\left[\sum_{i=1}^{n} X_i\right] = \sum_{i=1}^{n} E[X_i] = n \cdot P(\text{a specific couple is adjacent})$$
To compute $P(\text{adjacent})$: fix one member of the couple (say the wife) at an arbitrary seat. In a circular arrangement of $2n$ people, fixing one person eliminates the rotational symmetry, and the remaining $2n - 1$ people are equally likely to sit in any of the remaining $2n - 1$ seats. The wife has exactly 2 neighboring seats (one on her left, one on her right). So the probability her husband occupies one of those 2 seats is:
$$P(\text{adjacent}) = \frac{2}{2n - 1}$$
Therefore:
$$E[\text{adjacent couples}] = n \cdot \frac{2}{2n - 1} = \frac{2n}{2n - 1}$$
Sanity checks:
- $n = 2$ (4 people): $E = 4/3 \approx 1.33$. Two couples around 4 seats -- more often than not, exactly one couple ends up adjacent, occasionally both do, giving an average above 1.
- $n = 4$ (8 people): $E = 8/7 \approx 1.14$, matching our quick estimate.
- $n \to \infty$: $E \to 1$. As the table gets huge, each couple has a vanishingly small chance of sitting together, but there are many couples, and these effects balance to give an expected count of about 1.
Note: for $n = 1$ (2 people, 2 seats), the couple is always adjacent, so $E = 1$. The formula gives $2/1 = 2$, which is off because with only 2 seats in a circle each person has just 1 distinct neighbor, not 2. The formula $2/(2n-1)$ assumes $n \geq 2$ so that each person genuinely has 2 distinct neighbors.
Answer: $E[\text{adjacent couples}] = \dfrac{2n}{2n - 1}$ for $n \geq 2$.
Intuition
This problem is a textbook showcase for linearity of expectation. The beautiful thing about linearity is that it works regardless of dependence between the indicators -- whether couple 1 sits together can absolutely affect the probability that couple 2 does, but we never need to deal with that. We just compute one marginal probability, multiply by the number of couples, and we are done. This is one of the most powerful tricks in discrete probability and shows up constantly in interview questions involving random permutations, matchings, and arrangements.
The result $2n/(2n-1)$ is surprisingly close to 1 for all $n \geq 2$. Intuitively, each couple has about a $1/n$ chance of being adjacent (roughly $2/(2n) = 1/n$), and there are $n$ couples, so the expected count is roughly $n \times 1/n = 1$. The slight excess over 1 comes from the $2n - 1$ denominator being a hair smaller than $2n$. In practice, this kind of "expected count stays bounded even as the number of objects grows" phenomenon appears in many settings -- birthday problems, coupon collector variants, and random graph matching. When you see it in an interview, the interviewer is testing whether you reach for linearity immediately rather than getting bogged down in inclusion-exclusion.