Probability a Group Sits Together at a Circular Table
Fix a positive integer $n$. There are $3n$ people who sit around a circular table with $3n$ seats, uniformly at random. The $3n$ people are divided into 3 groups of $n$ people each.
Find the probability that at least one of the three groups has all $n$ of its members seated adjacent to one another. Rotations of a seating arrangement are considered identical.
Report the answer for $n = 5$, rounded to the nearest ten-thousandth.
Hints
- Let $G_i$ be the event that group $i$ sits together. You want $P(G_1 \cup G_2 \cup G_3)$, and the events overlap -- use inclusion-exclusion.
- To count arrangements where a group sits together in a circle, treat the group as a single block and count circular permutations of the reduced set of objects, then multiply by the internal arrangements within each block.
- There are $(m-1)!$ circular permutations of $m$ distinct objects. When $k$ groups are blocks with $n$ remaining individuals, you have $m = k + n_{\text{remaining}}$ objects to arrange.
Worked Solution
How to Think About It: "At least one" screams inclusion-exclusion. Let $G_i$ be the event that all $n$ people in group $i$ sit in $n$ consecutive seats. We want $P(G_1 \cup G_2 \cup G_3)$. The three events are not independent (knowing one group sits together constrains the remaining seats), but they are symmetric (all groups have the same size), so inclusion-exclusion is clean. The trick for circular permutation problems: when a group sits together, treat them as a single "block" and count circular permutations of the blocks and remaining individuals.
Quick Estimate: For $n = 5$, we have 15 people. The chance that a specific group of 5 sits together in a circle of 15 is roughly $5! \cdot 10! / 14! \approx 0.005$, so by the union bound, $P \leq 3 \times 0.005 = 0.015$. The actual answer should be slightly below this due to the inclusion-exclusion corrections. We will see it comes out to about $0.0147$.
Approach: Apply inclusion-exclusion with three symmetric events, counting circular permutations with blocks.
Formal Solution:
The total number of distinct circular seating arrangements of $3n$ people is $(3n - 1)!$ (fix one person's position to account for rotational equivalence).
By symmetry of the three groups:
$$P(G_1 \cup G_2 \cup G_3) = 3 P(G_1) - 3 P(G_1 \cap G_2) + P(G_1 \cap G_2 \cap G_3)$$
*Computing $P(G_1)$:*
Treat group 1 as a single block. We now arrange $2n + 1$ objects (the block plus $2n$ individuals) in a circle: $(2n)!$ circular permutations. The $n$ people within the block can be arranged in $n!$ ways. So:
$$P(G_1) = \frac{(2n)! \cdot n!}{(3n-1)!}$$
*Computing $P(G_1 \cap G_2)$:*
Treat groups 1 and 2 as two blocks. We arrange $n + 2$ objects ($n$ remaining individuals plus 2 blocks) in a circle: $(n+1)!$ circular permutations. Each block has $n!$ internal arrangements. So:
$$P(G_1 \cap G_2) = \frac{(n+1)! \cdot (n!)^2}{(3n-1)!}$$
*Computing $P(G_1 \cap G_2 \cap G_3)$:*
All three groups are blocks. We arrange 3 blocks in a circle: $2!$ circular permutations. Each block has $n!$ internal arrangements. So:
$$P(G_1 \cap G_2 \cap G_3) = \frac{2 \cdot (n!)^3}{(3n-1)!}$$
*Combining:*
$$P(G_1 \cup G_2 \cup G_3) = \frac{3(2n)! \cdot n! - 3(n+1)!(n!)^2 + 2(n!)^3}{(3n-1)!}$$
*Sanity check ($n = 1$):*
$$P = \frac{3 \cdot 2! \cdot 1! - 3 \cdot 2! \cdot 1 + 2 \cdot 1}{2!} = \frac{6 - 6 + 2}{2} = 1$$
Correct: with $n = 1$, each person is their own group, so every group trivially sits "together."
*Evaluation at $n = 5$:*
- $(3n-1)! = 14!$
- $3(2n)! \cdot n! = 3 \cdot 10! \cdot 5! = 3 \cdot 3628800 \cdot 120 = 1{,}306{,}368{,}000$
- $3(n+1)!(n!)^2 = 3 \cdot 720 \cdot 14400 = 31{,}104{,}000$
- $2(n!)^3 = 2 \cdot 120^3 = 3{,}456{,}000$
Numerator: $1{,}306{,}368{,}000 - 31{,}104{,}000 + 3{,}456{,}000 = 1{,}278{,}720{,}000$
Denominator: $14! = 87{,}178{,}291{,}200$
$$P = \frac{1{,}278{,}720{,}000}{87{,}178{,}291{,}200} \approx 0.01467$$
Rounded to the nearest ten-thousandth: $\approx 0.0147$.
Answer: The probability is
$$P = \frac{3(2n)! \cdot n! - 3(n+1)!(n!)^2 + 2(n!)^3}{(3n-1)!}$$
For $n = 5$, this evaluates to approximately $0.0147$.
Intuition
This is a textbook application of inclusion-exclusion on a circular arrangement. The main technique -- treating a group of adjacent people as a single block -- is the standard trick for circular permutation problems with adjacency constraints. The inclusion-exclusion corrections (subtract pairs, add the triple) are needed because the events are not disjoint: it is possible for two or even all three groups to sit together simultaneously.
The probability is small even for modest $n$ because requiring $n$ specific people to occupy $n$ consecutive seats out of $3n$ is a strong constraint. The denominator grows as $(3n)!$ while the numerator grows much more slowly. For large $n$, the dominant term is $3P(G_1) \approx 3(2n)! \cdot n! / (3n-1)!$, which by Stirling's approximation decays exponentially in $n$. This kind of combinatorial probability calculation -- circular permutations, blocks, inclusion-exclusion -- appears frequently in competition math and in quantitative finance interviews that test combinatorial fluency.